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Minimum in a Sorted and Rotated Array
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Search in a Sorted and Rotated Array

Last Updated : 07 Dec, 2024
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Given a sorted and rotated array arr[] of n distinct elements, the task is to find the index of given key in the array. If the key is not present in the array, return -1.

Examples:  

Input: arr[] = [5, 6, 7, 8, 9, 10, 1, 2, 3], key = 3
Output: 8
Explanation: 3 is present at index 8 in arr[].

Input: arr[] = [3, 5, 1, 2], key = 6
Output: -1
Explanation: 6 is not present in arr[].

Input: arr[] = [33, 42, 72, 99], key = 42
Output: 1
Explanation: 42 is found at index 1.

Table of Content

  • [Naive Approach] Using Linear Search - O(n) Time and O(1) Space
  • [Expected Approach 1] Using Binary Search Twice - O(log n) Time and O(1) Space
  • [Expected Approach 2] Using Single Binary Search - O(log n) Time and O(1) Space

[Naive Approach] Using Linear Search - O(n) Time and O(1) Space

A simple approach is to iterate through the array and check for each element, if it matches the target then return the index, otherwise return -1. To know more about the implementation, please refer Introduction to Linear Search Algorithm.

[Expected Approach 1] Using Binary Search Twice - O(log n) Time and O(1) Space

The idea is to use binary search twice:

1. Find the pivot point (or index of the min element) : For example, the min in [4, 5, 6, 7, 0, 1, 2] is 0 at index 4. Refer Minimum in a Sorted and Rotated Array to find the index of the minimum element.

2. Binary Search in the Sorted Subarray: Once we find the pivot, we can easily divide the given array into two sorted subarrays using the index of the minimum element. For example, the index of the minimum element in [4, 5, 6, 7, 0, 1, 2] is 4, so the two sorted subarrays are [4, 5, 6, 7] and [1, 2]. Following are the cases that arise

  • If the minimum element = key, then return index of the minimum element.
  • If index of minimum element is 0, then the whole array is sorted, we call binary search for the whole array.
  • If index of minimum element > 0, then instead of calling binary search for both sides, we can save one binary search by comparing the given key with the first element on left side. If key is greater than equal to the first element, we do binary search in the first subarray else in the second.
C++ 
// C++ program to search an element in sorted and rotated  // array using binary search twice  #include <iostream> #include <vector> using namespace std;  // An iterative binary search function int binarySearch(vector<int> &arr, int lo, int hi, int x) {     while (lo <= hi) {         int mid = lo + (hi - lo) / 2;         if (arr[mid] == x) return mid;         if (arr[mid] < x) lo = mid + 1;         else hi = mid - 1;     }     return -1; }  // Function to return pivot (index of the smallest element) int findPivot(vector<int> &arr, int lo, int hi) {     while (lo < hi) {                // The current subarray is already sorted,         // the minimum is at the low index         if (arr[lo] <= arr[hi])                     return lo;                  int mid = (lo + hi) / 2;          // The right half is not sorted. So         // the minimum element must be in the         // right half.         if (arr[mid] > arr[hi])             lo = mid + 1;                // The right half is sorted. Note that in         // this case, we do not change high to mid - 1         // but keep it to mid. The mid element         // itself can be the smallest         else             hi = mid;     }      return lo; }  // Searches an element key in a pivoted // sorted array arr of size n int search(vector<int> &arr, int key) {   	int n = arr.size();     int pivot = findPivot(arr, 0, n - 1);      // If we found a pivot, then first compare with pivot     // and then search in two subarrays around pivot     if (arr[pivot] == key)         return pivot;        // If the minimum element is present at index     // 0, then the whole array is sorted     if (pivot == 0)         return binarySearch(arr, 0, n - 1, key);      if (arr[0] <= key)         return binarySearch(arr, 0, pivot - 1, key);     return binarySearch(arr, pivot + 1, n - 1, key); }  int main() {     vector<int> arr = {5, 6, 7, 8, 9, 10, 1, 2, 3};     int key = 3;     cout << search(arr, key);     return 0; }
C 
// C program to search an element in sorted and rotated  // array using binary search twice  #include <stdio.h>  // An iterative binary search function int binarySearch(int arr[], int lo, int hi, int x) {     while (lo <= hi) {         int mid = lo + (hi - lo) / 2;         if (arr[mid] == x) return mid;         if (arr[mid] < x) lo = mid + 1;         else hi = mid - 1;     }     return -1; }  // Function to return pivot (index of the smallest element) int findPivot(int arr[], int lo, int hi) {     while (lo < hi) {                // The current subarray is already sorted,         // the minimum is at the low index         if (arr[lo] <= arr[hi])                     return lo;                  int mid = (lo + hi) / 2;          // The right half is not sorted. So         // the minimum element must be in the         // right half.         if (arr[mid] > arr[hi])             lo = mid + 1;                // The right half is sorted. Note that in         // this case, we do not change high to mid - 1         // but keep it to mid. The mid element         // itself can be the smallest         else             hi = mid;     }      return lo; }  // Searches an element key in a pivoted // sorted array arr of size n int search(int arr[], int n, int key) {     int pivot = findPivot(arr, 0, n - 1);      // If we found a pivot, then first compare with pivot     // and then search in two subarrays around pivot     if (arr[pivot] == key)         return pivot;        // If the minimum element is present at index     // 0, then the whole array is sorted     if (pivot == 0)         return binarySearch(arr, 0, n - 1, key);      if (arr[0] <= key)         return binarySearch(arr, 0, pivot - 1, key);     return binarySearch(arr, pivot + 1, n - 1, key); }  int main() {     int arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 3};     int n = sizeof(arr) / sizeof(arr[0]);     int key = 3;     printf("%d\n", search(arr, n, key));     return 0; }
Java 
// Java program to search an element in sorted and rotated  // array using binary search twice  import java.util.*; class GfG {        // An iterative binary search function     static int binarySearch(int[] arr, int lo, int hi, int x) {         while (lo <= hi) {             int mid = lo + (hi - lo) / 2;             if (arr[mid] == x) return mid;             if (arr[mid] < x) lo = mid + 1;             else hi = mid - 1;         }         return -1;     }      // Function to return pivot (index of the smallest element)     static int findPivot(int[] arr, int lo, int hi) {         while (lo < hi) {                        // The current subarray is already sorted,             // the minimum is at the low index             if (arr[lo] <= arr[hi])                 return lo;              int mid = (lo + hi) / 2;              // The right half is not sorted. So             // the minimum element must be in the             // right half             if (arr[mid] > arr[hi])                 lo = mid + 1;              // The right half is sorted. Note that in             // this case, we do not change high to mid - 1             // but keep it to mid. The mid element             // itself can be the smallest             else                 hi = mid;         }          return lo;     }      // Searches an element key in a pivoted     // sorted array arr of size n     static int search(int[] arr, int key) {         int n = arr.length;         int pivot = findPivot(arr, 0, n - 1);          // If we found a pivot, then first compare with pivot         // and then search in two subarrays around pivot         if (arr[pivot] == key)             return pivot;                // If the minimum element is present at index         // 0, then the whole array is sorted         if (pivot == 0)             return binarySearch(arr, 0, n - 1, key);          if (arr[0] <= key)             return binarySearch(arr, 0, pivot - 1, key);         return binarySearch(arr, pivot + 1, n - 1, key);     }      public static void main(String[] args) {         int[] arr = {5, 6, 7, 8, 9, 10, 1, 2, 3};         int key = 3;         System.out.println(search(arr, key));     } }
Python 
# Python program to search an element in sorted and rotated  # array using binary search twice  # An iterative binary search function def binarySearch(arr, lo, hi, x):     while lo <= hi:         mid = lo + (hi - lo) // 2         if arr[mid] == x:             return mid         if arr[mid] < x:             lo = mid + 1         else:             hi = mid - 1     return -1  # Function to return pivot (index of the smallest element) def findPivot(arr, lo, hi):     while lo < hi:          # The current subarray is already sorted,         # the minimum is at the low index         if arr[lo] <= arr[hi]:             return lo          mid = (lo + hi) // 2          # The right half is not sorted. So         # the minimum element must be in the         # right half.         if arr[mid] > arr[hi]:             lo = mid + 1          # The right half is sorted. Note that in         # this case, we do not change high to mid - 1         # but keep it to mid. The mid element         # itself can be the smallest         else:             hi = mid      return lo  # Searches an element key in a pivoted # sorted array arr of size n def search(arr, key):     n = len(arr)     pivot = findPivot(arr, 0, n - 1)      # If we found a pivot, then first compare with pivot     # and then search in two subarrays around pivot     if arr[pivot] == key:         return pivot          # If the minimum element is present at index     # 0, then the whole array is sorted     if pivot == 0:         return binarySearch(arr, 0, n - 1, key)      if arr[0] <= key:         return binarySearch(arr, 0, pivot - 1, key)     return binarySearch(arr, pivot + 1, n - 1, key)  if __name__ == "__main__":     arr = [5, 6, 7, 8, 9, 10, 1, 2, 3]     key = 3     print(search(arr, key))
C# 
// C# program to search an element in sorted and rotated  // array using  binary search twice  using System; class GfG {      // An iterative binary search function     static int binarySearch(int[] arr, int lo, int hi, int x) {         while (lo <= hi) {             int mid = lo + (hi - lo) / 2;             if (arr[mid] == x) return mid;             if (arr[mid] < x) lo = mid + 1;             else hi = mid - 1;         }         return -1;     }      // Function to return pivot (index of the smallest element)     static int findPivot(int[] arr, int lo, int hi) {         while (lo < hi) {              // The current subarray is already sorted,             // the minimum is at the low index             if (arr[lo] <= arr[hi])                 return lo;              int mid = (lo + hi) / 2;              // The right half is not sorted. So             // the minimum element must be in the             // right half.             if (arr[mid] > arr[hi])                 lo = mid + 1;              // The right half is sorted. Note that in             // this case, we do not change high to mid - 1             // but keep it to mid. The mid element             // itself can be the smallest             else                 hi = mid;         }          return lo;     }      // Searches an element key in a pivoted     // sorted array arr of size n     static int search(int[] arr, int key) {         int n = arr.Length;         int pivot = findPivot(arr, 0, n - 1);          // If we found a pivot, then first compare with pivot         // and then search in two subarrays around pivot         if (arr[pivot] == key)             return pivot;                // If the minimum element is present at index         // 0, then the whole array is sorted         if (pivot == 0)             return binarySearch(arr, 0, n - 1, key);          if (arr[0] <= key)             return binarySearch(arr, 0, pivot - 1, key);         return binarySearch(arr, pivot + 1, n - 1, key);     }      static void Main(string[] args) {         int[] arr = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };         int key = 3;         Console.WriteLine(search(arr, key));     } }
JavaScript 
// JavaScript program to search an element in sorted and rotated  // array using  binary search twice  // An iterative binary search function function binarySearch(arr, lo, hi, x) {     while (lo <= hi) {         let mid = lo + Math.floor((hi - lo) / 2);         if (arr[mid] === x) return mid;         if (arr[mid] < x) lo = mid + 1;         else hi = mid - 1;     }     return -1; }  // Function to return pivot (index of the smallest element) function findPivot(arr, lo, hi) {     while (lo < hi) {          // The current subarray is already sorted,         // the minimum is at the low index         if (arr[lo] <= arr[hi])             return lo;          let mid = Math.floor((lo + hi) / 2);          // The right half is not sorted. So         // the minimum element must be in the         // right half.         if (arr[mid] > arr[hi])             lo = mid + 1;          // The right half is sorted. Note that in         // this case, we do not change high to mid - 1         // but keep it to mid. The mid element         // itself can be the smallest         else             hi = mid;     }      return lo; }  // Searches an element key in a pivoted // sorted array arr of size n function search(arr, key) {     let n = arr.length;     let pivot = findPivot(arr, 0, n - 1);      // If we found a pivot, then first compare with pivot     // and then search in two subarrays around pivot     if (arr[pivot] === key)         return pivot;      // If the minimum element is present at index     // 0, then the whole array is sorted     if (pivot === 0)         return binarySearch(arr, 0, n - 1, key);      if (arr[0] <= key)         return binarySearch(arr, 0, pivot - 1, key);     return binarySearch(arr, pivot + 1, n - 1, key); }  // Driver code let arr = [5, 6, 7, 8, 9, 10, 1, 2, 3]; let key = 3; console.log(search(arr, key));

Output
8

[Expected Approach 2] Using Single Binary Search - O(log n) Time and O(1) Space

The idea is based on the fact that in a sorted and rotated array, if we go to mid, then either the left half would be sorted or the right half (both can also be sorted). For example, arr[] = [5, 6, 0, 1, 2, 3, 4], mid = 3 and we can see that the subarray from mid + 1 to high is sorted. And in [5, 6, 7, 8, 9, 3, 4], we can see that the subarray from low to mid-1 is sorted. We can check which half is sorted by comparing arr[low] and arr[mid] (We could also compare arr[high] and arr[mid]).

  • Find the mid point. If key is same as the mid, return the mid.
  • Find which half is sorted. If the key lies in the sorted half, move to that half. Otherwise move to the other half.

Note that once we find which half is sorted, we can easily check if the key lies here by checking if key lies in the range from smallest to largest in this half.


C++ 
// C++ program to search an element in sorted and rotated  // array using binary search  #include <iostream> #include <vector> using namespace std;  int search(vector<int>& arr, int key) {        // Initialize two pointers, lo and hi, at the start     // and end of the array     int lo = 0, hi = arr.size() - 1;      while (lo <= hi) {         int mid = lo + (hi - lo) / 2;          // If key found, return the index         if (arr[mid] == key)             return mid;          // If Left half is sorted         if (arr[mid] >= arr[lo]) {                        // If the key lies within this sorted half,             // move the hi pointer to mid - 1             if (key >= arr[lo] && key < arr[mid])                 hi = mid - 1;                        // Otherwise, move the lo pointer to mid + 1             else                 lo = mid + 1;         }                // If Right half is sorted         else {                        // If the key lies within this sorted half,             // move the lo pointer to mid + 1             if (key > arr[mid] && key <= arr[hi])                 lo = mid + 1;                        // Otherwise, move the hi pointer to mid - 1             else                 hi = mid - 1;         }     } 	   	// Key not found     return -1;  }  int main() {     vector<int> arr1 = {5, 6, 7, 8, 9, 10, 1, 2, 3};     int key1 = 3;     cout << search(arr1, key1) << endl;      return 0; }
C 
// C program to search an element in sorted and rotated  // array using binary search  #include <stdio.h> int search(int arr[], int n, int key) {        // Initialize two pointers, lo and hi, at the start     // and end of the array     int lo = 0, hi = n - 1;      while (lo <= hi) {         int mid = lo + (hi - lo) / 2;          // If key found, return the index         if (arr[mid] == key)             return mid;          // If Left half is sorted         if (arr[mid] >= arr[lo]) {                        // If the key lies within this sorted half,             // move the hi pointer to mid - 1             if (key >= arr[lo] && key < arr[mid])                 hi = mid - 1;                        // Otherwise, move the lo pointer to mid + 1             else                 lo = mid + 1;         }                // If Right half is sorted         else {                        // If the key lies within this sorted half,             // move the lo pointer to mid + 1             if (key > arr[mid] && key <= arr[hi])                 lo = mid + 1;                        // Otherwise, move the hi pointer to mid - 1             else                 hi = mid - 1;         }     } 	   	// Key not found     return -1;  }  int main() {     int arr1[] = {5, 6, 7, 8, 9, 10, 1, 2, 3};     int n1 = sizeof(arr1) / sizeof(arr1[0]);     int key1 = 3;     printf("%d\n", search(arr1, n1, key1));      return 0; }
Java 
// Java program to search an element in sorted and rotated  // array using binary search  import java.util.*; class GfG {        static int search(int[] arr, int key) {                // Initialize two pointers, lo and hi, at the start         // and end of the array         int lo = 0, hi = arr.length - 1;          while (lo <= hi) {             int mid = lo + (hi - lo) / 2;              // If key found, return the index             if (arr[mid] == key)                 return mid;              // If Left half is sorted             if (arr[mid] >= arr[lo]) {                                // If the key lies within this sorted half,                 // move the hi pointer to mid - 1                 if (key >= arr[lo] && key < arr[mid])                     hi = mid - 1;                                // Otherwise, move the lo pointer to mid + 1                 else                     lo = mid + 1;             }                        // If Right half is sorted             else {                                // If the key lies within this sorted half,                 // move the lo pointer to mid + 1                 if (key > arr[mid] && key <= arr[hi])                     lo = mid + 1;                                // Otherwise, move the hi pointer to mid - 1                 else                     hi = mid - 1;             }         }                // Key not found         return -1;      }      public static void main(String[] args) {         int[] arr1 = {5, 6, 7, 8, 9, 10, 1, 2, 3};         int key1 = 3;         System.out.println(search(arr1, key1));     } }
Python 
# Python program to search an element in sorted and rotated  # array using binary search  def search(arr, key):        # Initialize two pointers, lo and hi, at the start     # and end of the array     lo = 0     hi = len(arr) - 1      while lo <= hi:         mid = lo + (hi - lo) // 2          # If key found, return the index         if arr[mid] == key:             return mid          # If Left half is sorted         if arr[mid] >= arr[lo]:                        # If the key lies within this sorted half,             # move the hi pointer to mid - 1             if key >= arr[lo] and key < arr[mid]:                 hi = mid - 1                            # Otherwise, move the lo pointer to mid + 1             else:                 lo = mid + 1                    # If Right half is sorted         else:                        # If the key lies within this sorted half,             # move the lo pointer to mid + 1             if key > arr[mid] and key <= arr[hi]:                 lo = mid + 1                            # Otherwise, move the hi pointer to mid - 1             else:                 hi = mid - 1      # Key not found     return -1  if __name__ == "__main__":     arr1 = [5, 6, 7, 8, 9, 10, 1, 2, 3]     key1 = 3     print(search(arr1, key1))
C# 
// C# program to search an element in sorted and rotated  // array using binary search  using System; class GfG {        static int search(int[] arr, int key) {                // Initialize two pointers, lo and hi, at the start         // and end of the array         int lo = 0, hi = arr.Length - 1;          while (lo <= hi) {             int mid = lo + (hi - lo) / 2;              // If key found, return the index             if (arr[mid] == key)                 return mid;              // If Left half is sorted             if (arr[mid] >= arr[lo]) {                                // If the key lies within this sorted half,                 // move the hi pointer to mid - 1                 if (key >= arr[lo] && key < arr[mid])                     hi = mid - 1;                                // Otherwise, move the lo pointer to mid + 1                 else                     lo = mid + 1;             }                        // If Right half is sorted             else {                                // If the key lies within this sorted half,                 // move the lo pointer to mid + 1                 if (key > arr[mid] && key <= arr[hi])                     lo = mid + 1;                                // Otherwise, move the hi pointer to mid - 1                 else                     hi = mid - 1;             }         }          // Key not found         return -1;      }      static void Main(string[] args) {         int[] arr1 = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };         int key1 = 3;         Console.WriteLine(search(arr1, key1));     } }
JavaScript 
// JavaScript program to search an element in sorted and rotated  // array using binary search  function search(arr, key) {        // Initialize two pointers, lo and hi, at the start     // and end of the array     let lo = 0, hi = arr.length - 1;      while (lo <= hi) {         let mid = lo + Math.floor((hi - lo) / 2);          // If key found, return the index         if (arr[mid] === key)             return mid;          // If Left half is sorted         if (arr[mid] >= arr[lo]) {                        // If the key lies within this sorted half,             // move the hi pointer to mid - 1             if (key >= arr[lo] && key < arr[mid])                 hi = mid - 1;                        // Otherwise, move the lo pointer to mid + 1             else                 lo = mid + 1;         }                // If Right half is sorted         else {                        // If the key lies within this sorted half,             // move the lo pointer to mid + 1             if (key > arr[mid] && key <= arr[hi])                 lo = mid + 1;                        // Otherwise, move the hi pointer to mid - 1             else                 hi = mid - 1;         }     }      // Key not found     return -1;  }  // Driver code let arr1 = [5, 6, 7, 8, 9, 10, 1, 2, 3]; let key1 = 3; console.log(search(arr1, key1));

Output
8



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    Why is Binary Search preferred over Ternary Search?
    The following is a simple recursive Binary Search function in C++ taken from here.  C++ // CPP program for the above approach #include <bits/stdc++.h> using namespace std; // A recursive binary search function. It returns location of x in // given array arr[l..r] is present, otherwise -1 int b
    11 min read
    Is Sentinel Linear Search better than normal Linear Search?
    Sentinel Linear search is a type of linear search where the element to be searched is placed in the last position and then all the indices are checked for the presence of the element without checking for the index out of bound case.The number of comparisons is reduced in this search as compared to a
    8 min read

    Library implementations of Searching algorithms

    Binary Search functions in C++ STL (binary_search, lower_bound and upper_bound)
    In C++, STL provide various functions like std::binary_search(), std::lower_bound(), and std::upper_bound() which uses the the binary search algorithm for different purposes. These function will only work on the sorted data.There are the 3 binary search function in C++ STL:Table of Contentbinary_sea
    3 min read
    Arrays.binarySearch() in Java with Examples | Set 1
    In Java, the Arrays.binarySearch() method searches the specified array of the given data type for the specified value using the binary search algorithm. The array must be sorted by the Arrays.sort() method before making this call. If it is not sorted, the results are undefined. Example:Below is a si
    3 min read
    Arrays.binarySearch() in Java with examples | Set 2 (Search in subarray)
    Arrays.binarySearch()| Set 1 Covers how to find an element in a sorted array in Java. This set will cover "How to Search a key in an array within a given range including only start index". Syntax : public static int binarySearch(data_type[] arr, int fromIndex, int toIndex, data_type key) Parameters
    5 min read
    Collections.binarySearch() in Java with Examples
    java.util.Collections.binarySearch() method is a java.util.Collections class method that returns the position of an object in a sorted list.// Returns index of key in a sorted list sorted in// ascending orderpublic static int binarySearch(List slist, T key)// Returns index of key in a sorted list so
    4 min read

    Easy problems on Searching algorithms

    Find the Missing Number
    Given an array arr[] of size n-1 with distinct integers in the range of [1, n]. This array represents a permutation of the integers from 1 to n with one element missing. Find the missing element in the array.Examples: Input: arr[] = [8, 2, 4, 5, 3, 7, 1]Output: 6Explanation: All the numbers from 1 t
    12 min read
    Find the first repeating element in an array of integers
    Given an array of integers arr[], The task is to find the index of first repeating element in it i.e. the element that occurs more than once and whose index of the first occurrence is the smallest. Examples: Input: arr[] = {10, 5, 3, 4, 3, 5, 6}Output: 5 Explanation: 5 is the first element that repe
    8 min read
    Missing and Repeating in an Array
    Given an unsorted array of size n. Array elements are in the range of 1 to n. One number from set {1, 2, ...n} is missing and one number occurs twice in the array. The task is to find these two numbers.Examples: Input: arr[] = {3, 1, 3}Output: 3, 2Explanation: In the array, 2 is missing and 3 occurs
    15+ min read
    Count 1's in a sorted binary array
    Given a binary array arr[] of size n, which is sorted in non-increasing order, count the number of 1's in it. Examples: Input: arr[] = [1, 1, 0, 0, 0, 0, 0]Output: 2Explanation: Count of the 1's in the given array is 2.Input: arr[] = [1, 1, 1, 1, 1, 1, 1]Output: 7Input: arr[] = [0, 0, 0, 0, 0, 0, 0]
    7 min read
    Two Sum - Pair Closest to 0
    Given an integer array arr[], the task is to find the maximum sum of two elements such that sum is closest to zero. Note: In case if we have two of more ways to form sum of two elements closest to zero return the maximum sum.Examples:Input: arr[] = [-8, 5, 2, -6]Output: -1Explanation: The min absolu
    15+ min read
    Pair with the given difference
    Given an unsorted array and an integer x, the task is to find if there exists a pair of elements in the array whose absolute difference is x. Examples: Input: arr[] = [5, 20, 3, 2, 50, 80], x = 78Output: YesExplanation: The pair is {2, 80}.Input: arr[] = [90, 70, 20, 80, 50], x = 45Output: NoExplana
    14 min read
    Kth smallest element in a row-wise and column-wise sorted 2D array
    Given an n x n matrix, every row and column is sorted in non-decreasing order. Given a number K where K lies in the range [1, n*n], find the Kth smallest element in the given 2D matrix.Example:Input: mat =[[10, 20, 30, 40], [15, 25, 35, 45], [24, 29, 37, 48], [32, 33, 39, 50]]K = 3Output: 20Explanat
    15+ min read
    Find common elements in three sorted arrays
    Given three sorted arrays in non-decreasing order, print all common elements in non-decreasing order across these arrays. If there are no such elements return an empty array. In this case, the output will be -1.Note: In case of duplicate common elements, print only once.Examples: Input: arr1[] = [1,
    12 min read
    Ceiling in a sorted array
    Given a sorted array and a value x, find index of the ceiling of x. The ceiling of x is the smallest element in an array greater than or equal to x. Note: In case of multiple occurrences of ceiling of x, return the index of the first occurrence.Examples : Input: arr[] = [1, 2, 8, 10, 10, 12, 19], x
    13 min read
    Floor in a Sorted Array
    Given a sorted array and a value x, find the element of the floor of x. The floor of x is the largest element in the array smaller than or equal to x.Examples:Input: arr[] = [1, 2, 8, 10, 10, 12, 19], x = 5Output: 1Explanation: Largest number less than or equal to 5 is 2, whose index is 1Input: arr[
    9 min read
    Bitonic Point - Maximum in Increasing Decreasing Array
    Given an array arr[] of integers which is initially strictly increasing and then strictly decreasing, the task is to find the bitonic point, that is the maximum value in the array. Note: Bitonic Point is a point in bitonic sequence before which elements are strictly increasing and after which elemen
    10 min read
    Given Array of size n and a number k, find all elements that appear more than n/k times
    Given an array of size n and an integer k, find all elements in the array that appear more than n/k times. Examples:Input: arr[ ] = [3, 4, 2, 2, 1, 2, 3, 3], k = 4Output: [2, 3]Explanation: Here n/k is 8/4 = 2, therefore 2 appears 3 times in the array that is greater than 2 and 3 appears 3 times in
    15+ min read

    Medium problems on Searching algorithms

    3 Sum - Find All Triplets with Zero Sum
    Given an array arr[], the task is to find all possible indices {i, j, k} of triplet {arr[i], arr[j], arr[k]} such that their sum is equal to zero and all indices in a triplet should be distinct (i != j, j != k, k != i). We need to return indices of a triplet in sorted order, i.e., i < j < k.Ex
    11 min read
    Find the element before which all the elements are smaller than it, and after which all are greater
    Given an array, find an element before which all elements are equal or smaller than it, and after which all the elements are equal or greater.Note: Print -1, if no such element exists.Examples:Input: arr[] = [5, 1, 4, 3, 6, 8, 10, 7, 9]Output: 6 Explanation: 6 is present at index 4. All elements on
    14 min read
    Largest pair sum in an array
    Given an unsorted of distinct integers, find the largest pair sum in it. For example, the largest pair sum is 74. If there are less than 2 elements, then we need to return -1.Input : arr[] = {12, 34, 10, 6, 40}, Output : 74Input : arr[] = {10, 10, 10}, Output : 20Input arr[] = {10}, Output : -1[Naiv
    10 min read
    K’th Smallest Element in Unsorted Array
    Given an array arr[] of N distinct elements and a number K, where K is smaller than the size of the array. Find the K'th smallest element in the given array. Examples:Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 3 Output: 7Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 4 Output: 10 Table of Content[Naive Ap
    15 min read
    Search in a Sorted and Rotated Array
    Given a sorted and rotated array arr[] of n distinct elements, the task is to find the index of given key in the array. If the key is not present in the array, return -1. Examples: Input: arr[] = [5, 6, 7, 8, 9, 10, 1, 2, 3], key = 3Output: 8Explanation: 3 is present at index 8 in arr[].Input: arr[]
    15+ min read
    Minimum in a Sorted and Rotated Array
    Given a sorted array of distinct elements arr[] of size n that is rotated at some unknown point, the task is to find the minimum element in it. Examples: Input: arr[] = [5, 6, 1, 2, 3, 4]Output: 1Explanation: 1 is the minimum element present in the array.Input: arr[] = [3, 1, 2]Output: 1Explanation:
    9 min read
    Find a Fixed Point (Value equal to index) in a given array
    Given an array of n distinct integers sorted in ascending order, the task is to find the First Fixed Point in the array. Fixed Point in an array is an index i such that arr[i] equals i. Note that integers in the array can be negative. Note: If no Fixed Point is present in the array, print -1.Example
    7 min read
    K Mmost Frequent Words in a File
    Given a book of words and an integer K. Assume you have enough main memory to accommodate all words. Design a dynamic data structure to find the top K most frequent words in a book. The structure should allow new words to be added in main memory.Examples:Input: fileData = "Welcome to the world of Ge
    15+ min read
    Closest K Elements in a Sorted Array
    You are given a sorted array arr[] containing unique integers, a number k, and a target value x. Your goal is to return exactly k elements from the array that are closest to x, excluding x itself if it is present in the array.An element a is closer to x than b if:|a - x| < |b - x|, or|a - x| == |
    15+ min read
    2 Sum - Pair Sum Closest to Target using Binary Search
    Given an array arr[] of n integers and an integer target, the task is to find a pair in arr[] such that it’s sum is closest to target.Note: Return the pair in sorted order and if there are multiple such pairs return the pair with maximum absolute difference. If no such pair exists return an empty ar
    10 min read
    Find the closest pair from two sorted arrays
    Given two arrays arr1[0...m-1] and arr2[0..n-1], and a number x, the task is to find the pair arr1[i] + arr2[j] such that absolute value of (arr1[i] + arr2[j] - x) is minimum. Example: Input: arr1[] = {1, 4, 5, 7}; arr2[] = {10, 20, 30, 40}; x = 32Output: 1 and 30Input: arr1[] = {1, 4, 5, 7}; arr2[]
    15+ min read
    Find three closest elements from given three sorted arrays
    Given three sorted arrays A[], B[] and C[], find 3 elements i, j and k from A, B and C respectively such that max(abs(A[i] - B[j]), abs(B[j] - C[k]), abs(C[k] - A[i])) is minimized. Here abs() indicates absolute value. Example : Input : A[] = {1, 4, 10} B[] = {2, 15, 20} C[] = {10, 12} Output: 10 15
    15+ min read
    Search in an Array of Rational Numbers without floating point arithmetic
    Given a sorted array of rational numbers, where each rational number is represented in the form p/q (where p is the numerator and q is the denominator), the task is to find the index of a given rational number x in the array. If the number does not exist in the array, return -1.Examples: Input: arr[
    9 min read

    Hard problems on Searching algorithms

    Median of two sorted arrays of same size
    Given 2 sorted arrays a[] and b[], each of size n, the task is to find the median of the array obtained after merging a[] and b[]. Note: Since the size of the merged array will always be even, the median will be the average of the middle two numbers.Input: a[] = [1, 12, 15, 26, 38], b[] = [2, 13, 17
    15+ min read
    Search in an almost sorted array
    Given a sorted integer array arr[] consisting of distinct elements, where some elements of the array are moved to either of the adjacent positions, i.e. arr[i] may be present at arr[i-1] or arr[i+1].Given an integer target. You have to return the index ( 0-based ) of the target in the array. If targ
    7 min read
    Find position of an element in a sorted array of infinite numbers
    Given a sorted array arr[] of infinite numbers. The task is to search for an element k in the array.Examples:Input: arr[] = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170], k = 10Output: 4Explanation: 10 is at index 4 in array.Input: arr[] = [2, 5, 7, 9], k = 3Output: -1Explanation: 3 is not present i
    15+ min read
    Pair Sum in a Sorted and Rotated Array
    Given an array arr[] of size n, which is sorted and then rotated around an unknown pivot, the task is to check whether there exists a pair of elements in the array whose sum is equal to a given target value.Examples : Input: arr[] = [11, 15, 6, 8, 9, 10], target = 16Output: trueExplanation: There is
    10 min read
    K’th Smallest/Largest Element in Unsorted Array | Worst case Linear Time
    Given an array of distinct integers arr[] and an integer k. The task is to find the k-th smallest element in the array. For better understanding, k refers to the element that would appear in the k-th position if the array were sorted in ascending order. Note: k will always be less than the size of t
    15 min read
    K'th largest element in a stream
    Given an input stream of n integers, represented as an array arr[], and an integer k. After each insertion of an element into the stream, you need to determine the kth largest element so far (considering all elements including duplicates). If k elements have not yet been inserted, return -1 for that
    15+ min read
    Best First Search (Informed Search)
    Best First Search is a heuristic search algorithm that selects the most promising node for expansion based on an evaluation function. It prioritizes nodes in the search space using a heuristic to estimate their potential. By iteratively choosing the most promising node, it aims to efficiently naviga
    13 min read
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