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Find a Fixed Point (Value equal to index) in a given array
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Minimum in a Sorted and Rotated Array

Last Updated : 13 Dec, 2024
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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: 1
Explanation: 1 is the minimum element present in the array.

Input: arr[] = [3, 1, 2]
Output: 1
Explanation: 1 is the minimum element present in the array.

Input: arr[] = [4, 2, 3]
Output: 2
Explanation: 2 is the only minimum element in the array.

Table of Content

  • [Naive Approach] Linear Search – O(n) Time and O(1) Space
  • [Expected Approach] Binary Search – O(log n) Time and O(1) Space

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

A simple solution is to use linear search to traverse the complete array and find a minimum. 

C++ 
// C++ program to find minimum element in a  // sorted rotated array using linear search  #include <iostream> #include <vector> using namespace std;  int findMin(vector<int>& arr) {        int res = arr[0];      // Traverse over arr[] to find minimum element     for (int i = 1; i < arr.size(); i++)          res = min(res, arr[i]);      return res; }  int main() {     vector<int> arr = {5, 6, 1, 2, 3, 4};     int n = arr.size();      cout << findMin(arr) << endl;     return 0; }
C 
// C program to find minimum element in a  // sorted rotated array using linear search  #include <stdio.h>  int findMin(int arr[], int n) {     int res = arr[0];      // Traverse over arr[] to find minimum element     for (int i = 1; i < n; i++) {         if (arr[i] < res) {             res = arr[i];         }     }      return res; }  int main() {     int arr[] = {5, 6, 1, 2, 3, 4};     int n = sizeof(arr) / sizeof(arr[0]);      printf("%d\n", findMin(arr, n));     return 0; }
Java 
// Java program to find minimum element in a  // sorted rotated array using linear search  import java.util.*;  class GfG {     static int findMin(int[] arr) {         int res = arr[0];          // Traverse over arr[] to find minimum element         for (int i = 1; i < arr.length; i++) {             res = Math.min(res, arr[i]);         }          return res;     }      public static void main(String[] args) {         int[] arr = {5, 6, 1, 2, 3, 4};         System.out.println(findMin(arr));     } }
Python 
# Python program to find minimum element in a  # sorted rotated array using linear search  def findMin(arr):     res = arr[0]      # Traverse over arr[] to find minimum element     for i in range(1, len(arr)):         res = min(res, arr[i])      return res  if __name__ == "__main__":     arr = [5, 6, 1, 2, 3, 4]     print(findMin(arr))
C# 
// C# program to find minimum element in a  // sorted rotated array using linear search  using System; using System.Collections.Generic;  class GfG {     static int FindMin(int[] arr) {         int res = arr[0];          // Traverse over arr[] to find minimum element         for (int i = 1; i < arr.Length; i++) {             res = Math.Min(res, arr[i]);         }          return res;     }      static void Main() {         int[] arr = { 5, 6, 1, 2, 3, 4 };         Console.WriteLine(FindMin(arr));     } }
JavaScript 
// JavaScript program to find minimum element in a  // sorted rotated array using linear search  function findMin(arr) {     let res = arr[0];  	// Traverse over arr[] to find minimum element     for (let i = 1; i < arr.length; i++) {         res = Math.min(res, arr[i]);     }      return res; }  // Driver code const arr = [5, 6, 1, 2, 3, 4]; console.log(findMin(arr));

Output
1

[Expected Approach] Binary Search – O(log n) Time and O(1) Space

We can optimize the minimum element searching by using Binary Search where we find the mid element and then decide whether to stop or to go to left half or right half:

  • If arr[mid] > arr[high], it means arr[low … mid] is sorted and we need to search in the right half. So we change low = mid + 1.
  • If arr[mid] <= arr[high], it means arr[mid … high] is sorted and we need to search in the left half. So we change high = mid. (Note: Current mid might be the minimum element).

How do we terminate the search? One way could be to check if the mid is smaller than both of its adjacent, then we return mid. This would require a lot of condition checks like if adjacent indexes are valid or not and then comparing mid with both. We use an interesting fact here: If arr[low] < arr[high], then the current subarray is sorted, So we return arr[low].


C++ 
// C++ program to find minimum element in a  // sorted and rotated array using binary search  #include <iostream> #include <vector> using namespace std;  int findMin(vector<int> &arr) {     int lo = 0, hi = arr.size() - 1;      while (lo < hi) {                // The current subarray is already sorted,          // the minimum is at the low index         if (arr[lo] < arr[hi])                     return arr[lo];                     // We reach here when we have at least         // two elements and the current subarray         // is rotated                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. As the mid element         // itself can be the smallest         else             hi = mid;     }      return arr[lo];  }  int main() {     vector<int> arr = {5, 6, 1, 2, 3, 4};     cout << findMin(arr) << endl;     return 0; }
C 
// C program to find minimum element in a  // sorted and rotated array using binary search  #include <stdio.h>  int findMin(int arr[], int n) {     int lo = 0, hi = n - 1;      while (lo < hi) {                // The current subarray is already sorted,          // the minimum is at the low index         if (arr[lo] < arr[hi])                     return arr[lo];                     // We reach here when we have at least         // two elements and the current subarray         // is rotated                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. As the mid element         // itself can be the smallest         else             hi = mid;     }      return arr[lo];  }  int main() {     int arr[] = {5, 6, 1, 2, 3, 4};     int n = sizeof(arr) / sizeof(arr[0]);        printf("%d\n", findMin(arr, n));     return 0; }
Java 
// Java program to find minimum element in a  // sorted and rotated array using binary search  import java.util.*; class GfG {     static int findMin(int[] arr) {         int lo = 0, hi = arr.length - 1;          while (lo < hi) {                        // The current subarray is already sorted,              // the minimum is at the low index             if (arr[lo] < arr[hi])                         return arr[lo];                             // We reach here when we have at least             // two elements and the current subarray             // is rotated                        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. As the mid element             // itself can be the smallest             else                 hi = mid;         }          return arr[lo];      }      public static void main(String[] args) {         int[] arr = {5, 6, 1, 2, 3, 4};         System.out.println(findMin(arr));     } }
Python 
# Python program to find minimum element in a  # sorted and rotated array using binary search  def findMin(arr):     lo, hi = 0, len(arr) - 1      while lo < hi:                # The current subarray is already sorted,          # the minimum is at the low index         if arr[lo] < arr[hi]:             return arr[lo]                     # We reach here when we have at least         # two elements and the current subarray         # is rotated                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. As the mid element         # itself can be the smallest         else:             hi = mid      return arr[lo]  if __name__ == "__main__":     arr = [5, 6, 1, 2, 3, 4]     print(findMin(arr))
C# 
// C# program to find minimum element in a  // sorted and rotated array using binary search  using System; class GfG {     static int findMin(int[] arr) {         int lo = 0, hi = arr.Length - 1;          while (lo < hi) {                        // The current subarray is already sorted,              // the minimum is at the low index             if (arr[lo] < arr[hi])                         return arr[lo];                             // We reach here when we have at least             // two elements and the current subarray             // is rotated                        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. As the mid element             // itself can be the smallest             else                 hi = mid;         }          return arr[lo];      }      static void Main() {         int[] arr = {5, 6, 1, 2, 3, 4};         Console.WriteLine(findMin(arr));     } }
JavaScript 
// JavaScript program to find minimum element in a  // sorted and rotated array using binary search  function findMin(arr) {     let lo = 0, hi = arr.length - 1;      while (lo < hi) {                // The current subarray is already sorted,          // the minimum is at the low index         if (arr[lo] < arr[hi])                     return arr[lo];                     // We reach here when we have at least         // two elements and the current subarray         // is rotated                const 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. As the mid element         // itself can be the smallest         else             hi = mid;     }      return arr[lo];  }  // Driver Code const arr = [5, 6, 1, 2, 3, 4]; console.log(findMin(arr));

Output
1




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    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, 1
      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 presen
      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 i
      10 min read
    • K’th Smallest/Largest Element in Unsorted Array | Worst case Linear Time
      Given an array and a number k where k is smaller than the size of the array, we need to find the k’th smallest element in the given array. It is given that all array elements are distinct.Examples: Input: arr[] = {7, 10, 4, 3, 20, 15} k = 3Output: 7Input: arr[] = {7, 10, 4, 3, 20, 15} k = 4Output: 1
      15+ min read
    • K'th largest element in a stream
      Given an infinite stream of integers, find the Kth largest element at any point of time. Note: Here we have a stream instead of a whole array and we are allowed to store only K elements. Examples: Input: stream[] = {10, 20, 11, 70, 50, 40, 100, 5, . . .}, K = 3Output: {_, _, 10, 11, 20, 40, 50, 50,
      14 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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