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K'th largest element in a stream
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K’th Smallest/Largest Element in Unsorted Array | Worst case Linear Time

Last Updated : 10 May, 2025
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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 the array.

Examples:

Input: arr[] = [7, 10, 4, 3, 20, 15], k = 3
Output: 7
Explanation: The sorted array is [3, 4, 7, 10, 15, 20]. The 3rd smallest element is 7.

Input: arr[] = [12, 3, 5, 7, 19], k = 2
Output: 5
Explanation: The sorted array is [3, 5, 7, 12, 19]. The 2nd smallest element is 5.

Input: arr[] = [1, 5, 2, 8, 3], k = 4
Output: 5

In the previous post, we explored an algorithm with expected linear time complexity. In this post, a worst-case linear time we method is discussed.

Approach:

The intuition of this code starts with the same base idea as QuickSelect(), to find the k-th smallest element by partitioning the array around a pivot. But unlike QuickSelect, which may choose a bad pivot and degrade to O(n²) in the worst case, this algorithm ensures worst-case linear time by carefully choosing a pivot using the Median of Medians technique.
We want a pivot that guarantees a reasonably balanced partition, not perfectly balanced, but not extremely skewed either. That means the pivot should ensure that a significant portion of the array lies on both sides of it. This is where the Median of Medians strategy comes in

Steps to implement the above idea:

  • To find a good pivot, we divide the array into groups of 5 elements. This size (5) is a key observation, it is small enough to allow fast sorting and large enough to ensure a mathematically provable balance during partitioning.
  • Each group is sorted independently and its median is collected into a new list called medians.
  • Once we gather all medians, we recursively find the median of this median list. This value becomes our pivot. The purpose is to avoid bad pivots by using a value that is likely close to the true median of the entire array.
  • After determining the pivot, we partition the array using the standard logic (elements <= pivot on the left, > pivot on the right). The function partitionAroundPivot() moves the pivot to its correct position and returns that position.
  • Now, we compare the position of this pivot with the desired k-th index. If it matches, we return that value directly as our answer.
  • If not, we decide whether to recurse on the left or right side of the pivot:
    • If the pivot lies after the k-th position, the answer lies in the left subarray.
    • If the pivot lies before the k-th position, we adjust k accordingly and recurse on the right subarray.
C++ 
// C++ implementation of the Worst Case Linear Time algorithm // to find the k-th smallest element using Median of Medians #include <bits/stdc++.h> using namespace std;  // Returns median of a small group (size <= 5) int getMedian(vector<int> &group) {     sort(group.begin(), group.end());     return group[group.size() / 2]; }  // Function to Partition array from index  // l to r around  the pivot value x int partitionAroundPivot(vector<int> &arr,                           int l, int r, int x) {          // Move pivot x to end     int i;     for (i = l; i < r; i++) {         if (arr[i] == x) break;     }     swap(arr[i], arr[r]);      // Standard partition logic     i = l;     for (int j = l; j < r; j++) {         if (arr[j] <= x) {             swap(arr[i], arr[j]);             i++;         }     }      swap(arr[i], arr[r]);          // Final position of pivot     return i;  }  // Recursively finds the k-th smallest element in arr[l..r] int selectKthSmallest(vector<int> &arr, int l, int r, int k) {     if (k > 0 && k <= r - l + 1) {         int n = r - l + 1;         vector<int> medians;         int i;          // Divide array into groups of 5 and store their medians         for (i = 0; i < n / 5; i++) {             vector<int> group(arr.begin() + l + i * 5,                               arr.begin() + l + i * 5 + 5);             medians.push_back(getMedian(group));         }          // Handle the last group with less than 5 elements         if (i * 5 < n) {             vector<int> lastGroup(arr.begin() + l + i * 5,                                   arr.begin() + l + i * 5 + (n % 5));             medians.push_back(getMedian(lastGroup));         }          // Find median of medians         int pivot;         if (medians.size() == 1) {             pivot = medians[0];         } else {             pivot = selectKthSmallest(medians, 0, medians.size() - 1,                                       medians.size() / 2);         }          // Partition array and get position of pivot         int pos = partitionAroundPivot(arr, l, r, pivot);          // If position matches k, return result         if (pos - l == k - 1) return arr[pos];          // Recur on left or right part accordingly         if (pos - l > k - 1)             return selectKthSmallest(arr, l, pos - 1, k);          return selectKthSmallest(arr, pos + 1, r, k - pos + l - 1);     }      return INT_MAX;  }  // Function to find kth Smallest in Array int kthSmallest(vector<int> &arr, int k) {     return selectKthSmallest(arr, 0, arr.size() - 1, k); }  // Driver code int main() {     vector<int> arr = {7, 10, 4, 3, 20, 15};     int k = 3;      cout << kthSmallest(arr, k);      return 0; }
Java 
// Java implementation of the Worst Case Linear Time algorithm // to find the k-th smallest element using Median of Medians import java.util.*;  class GfG {      // Returns median of a small group (size <= 5)     static int getMedian(int[] group) {         Arrays.sort(group);         return group[group.length / 2];     }      // Function to Partition array from index      // l to r around  the pivot value x     static int partitionAroundPivot(int[] arr, int l, int r, int x) {                  // Move pivot x to end         int i;         for (i = l; i < r; i++) {             if (arr[i] == x) break;         }         int temp = arr[i];         arr[i] = arr[r];         arr[r] = temp;          // Standard partition logic         i = l;         for (int j = l; j < r; j++) {             if (arr[j] <= x) {                 int t = arr[i];                 arr[i] = arr[j];                 arr[j] = t;                 i++;             }         }          int t = arr[i];         arr[i] = arr[r];         arr[r] = t;          // Final position of pivot         return i;     }      // Recursively finds the k-th smallest element in arr[l..r]     static int selectKthSmallest(int[] arr, int l, int r, int k) {         if (k > 0 && k <= r - l + 1) {             int n = r - l + 1;             ArrayList<Integer> medians = new ArrayList<>();             int i;              // Divide array into groups of 5 and store their medians             for (i = 0; i < n / 5; i++) {                 int[] group = Arrays.copyOfRange(arr, l + i * 5, l + i * 5 + 5);                 medians.add(getMedian(group));             }              // Handle the last group with less than 5 elements             if (i * 5 < n) {                 int[] lastGroup = Arrays.copyOfRange(arr, l + i * 5, l + i * 5 + (n % 5));                 medians.add(getMedian(lastGroup));             }              // Find median of medians             int pivot;             if (medians.size() == 1) {                 pivot = medians.get(0);             } else {                 int[] medArr = new int[medians.size()];                 for (int j = 0; j < medians.size(); j++) medArr[j] = medians.get(j);                 pivot = selectKthSmallest(medArr, 0, medArr.length - 1, medArr.length / 2);             }              // Partition array and get position of pivot             int pos = partitionAroundPivot(arr, l, r, pivot);              // If position matches k, return result             if (pos - l == k - 1) return arr[pos];              // Recur on left or right part accordingly             if (pos - l > k - 1)                 return selectKthSmallest(arr, l, pos - 1, k);              return selectKthSmallest(arr, pos + 1, r, k - pos + l - 1);         }          return Integer.MAX_VALUE;     }      // Function to find kth Smallest in Array     static int kthSmallest(int[] arr, int k) {         return selectKthSmallest(arr, 0, arr.length - 1, k);     }      public static void main(String[] args) {         int[] arr = {7, 10, 4, 3, 20, 15};         int k = 3;          System.out.println(kthSmallest(arr, k));     } }
Python 
# Python implementation of the Worst Case Linear Time algorithm # to find the k-th smallest element using Median of Medians  # Returns median of a small group (size <= 5) def getMedian(group):     group.sort()     return group[len(group) // 2]  # Function to Partition array from index  # l to r around  the pivot value x def partitionAroundPivot(arr, l, r, x):      # Move pivot x to end     for i in range(l, r):         if arr[i] == x:             break     arr[i], arr[r] = arr[r], arr[i]      # Standard partition logic     i = l     for j in range(l, r):         if arr[j] <= x:             arr[i], arr[j] = arr[j], arr[i]             i += 1      arr[i], arr[r] = arr[r], arr[i]      # Final position of pivot     return i  # Recursively finds the k-th smallest element in arr[l..r] def selectKthSmallest(arr, l, r, k):     if k > 0 and k <= r - l + 1:         n = r - l + 1         medians = []          # Divide array into groups of 5 and store their medians         i = 0         while i < n // 5:             group = arr[l + i * 5: l + i * 5 + 5]             medians.append(getMedian(group))             i += 1          # Handle the last group with less than 5 elements         if i * 5 < n:             lastGroup = arr[l + i * 5: l + i * 5 + (n % 5)]             medians.append(getMedian(lastGroup))          # Find median of medians         if len(medians) == 1:             pivot = medians[0]         else:             pivot = selectKthSmallest(medians, 0, len(medians) - 1,                                       len(medians) // 2 + 1)          # Partition array and get position of pivot         pos = partitionAroundPivot(arr, l, r, pivot)          # If position matches k, return result         if pos - l == k - 1:             return arr[pos]          # Recur on left or right part accordingly         if pos - l > k - 1:             return selectKthSmallest(arr, l, pos - 1, k)          return selectKthSmallest(arr, pos + 1, r,                                  k - pos + l - 1)      return float('inf')  # Function to find kth Smallest in Array def kthSmallest(arr, k):     return selectKthSmallest(arr, 0, len(arr) - 1, k)  # Driver code if __name__ == "__main__":     arr = [7, 10, 4, 3, 20, 15]     k = 3      print(kthSmallest(arr, k))
C# 
// C# implementation of the Worst Case Linear Time algorithm // to find the k-th smallest element using Median of Medians using System; using System.Collections.Generic;  class GfG {      // Returns median of a small group (size <= 5)     static int getMedian(int[] group) {         Array.Sort(group);         return group[group.Length / 2];     }      // Function to Partition array from index      // l to r around  the pivot value x     static int partitionAroundPivot(int[] arr, int l, int r, int x) {                  // Move pivot x to end         int i;         for (i = l; i < r; i++) {             if (arr[i] == x) break;         }         int temp = arr[i];         arr[i] = arr[r];         arr[r] = temp;          // Standard partition logic         i = l;         for (int j = l; j < r; j++) {             if (arr[j] <= x) {                 int t = arr[i];                 arr[i] = arr[j];                 arr[j] = t;                 i++;             }         }          int t2 = arr[i];         arr[i] = arr[r];         arr[r] = t2;          // Final position of pivot         return i;     }      // Recursively finds the k-th smallest element in arr[l..r]     static int selectKthSmallest(int[] arr, int l, int r, int k) {         if (k > 0 && k <= r - l + 1) {             int n = r - l + 1;             List<int> medians = new List<int>();             int i;              // Divide array into groups of 5 and store their medians             for (i = 0; i < n / 5; i++) {                 int[] group = new int[5];                 Array.Copy(arr, l + i * 5, group, 0, 5);                 medians.Add(getMedian(group));             }              // Handle the last group with less than 5 elements             if (i * 5 < n) {                 int len = n % 5;                 int[] lastGroup = new int[len];                 Array.Copy(arr, l + i * 5, lastGroup, 0, len);                 medians.Add(getMedian(lastGroup));             }              // Find median of medians             int pivot;             if (medians.Count == 1) {                 pivot = medians[0];             } else {                 int[] medArr = medians.ToArray();                 pivot = selectKthSmallest(medArr, 0, medArr.Length - 1, medArr.Length / 2);             }              // Partition array and get position of pivot             int pos = partitionAroundPivot(arr, l, r, pivot);              // If position matches k, return result             if (pos - l == k - 1) return arr[pos];              // Recur on left or right part accordingly             if (pos - l > k - 1)                 return selectKthSmallest(arr, l, pos - 1, k);              return selectKthSmallest(arr, pos + 1, r, k - pos + l - 1);         }          return int.MaxValue;     }      // Function to find kth Smallest in Array     static int kthSmallest(int[] arr, int k) {         return selectKthSmallest(arr, 0, arr.Length - 1, k);     }      static void Main() {         int[] arr = {7, 10, 4, 3, 20, 15};         int k = 3;          Console.WriteLine(kthSmallest(arr, k));     } }
JavaScript 
// JavaScript implementation of the Worst Case Linear Time algorithm // to find the k-th smallest element using Median of Medians  // Returns median of a small group (size <= 5) function getMedian(group) {     group.sort((a, b) => a - b);     return group[Math.floor(group.length / 2)]; }  // Function to Partition array from index  // l to r around  the pivot value x function partitionAroundPivot(arr, l, r, x) {          // Move pivot x to end     let i;     for (i = l; i < r; i++) {         if (arr[i] === x) break;     }     [arr[i], arr[r]] = [arr[r], arr[i]];      // Standard partition logic     i = l;     for (let j = l; j < r; j++) {         if (arr[j] <= x) {             [arr[i], arr[j]] = [arr[j], arr[i]];             i++;         }     }      [arr[i], arr[r]] = [arr[r], arr[i]];      // Final position of pivot     return i; }  // Recursively finds the k-th smallest element in arr[l..r] function selectKthSmallest(arr, l, r, k) {     if (k > 0 && k <= r - l + 1) {         let n = r - l + 1;         let medians = [];          // Divide array into groups of 5 and store their medians         let i;         for (i = 0; i < Math.floor(n / 5); i++) {             let group = arr.slice(l + i * 5, l + i * 5 + 5);             medians.push(getMedian(group));         }          // Handle the last group with less than 5 elements         if (i * 5 < n) {             let lastGroup = arr.slice(l + i * 5, l + i * 5 + (n % 5));             medians.push(getMedian(lastGroup));         }          // Find median of medians         let pivot;         if (medians.length === 1) {             pivot = medians[0];         } else {             pivot = selectKthSmallest(medians, 0, medians.length - 1,                                       Math.floor(medians.length / 2));         }          // Partition array and get position of pivot         let pos = partitionAroundPivot(arr, l, r, pivot);          // If position matches k, return result         if (pos - l === k - 1) return arr[pos];          // Recur on left or right part accordingly         if (pos - l > k - 1)             return selectKthSmallest(arr, l, pos - 1, k);          return selectKthSmallest(arr, pos + 1, r, k - pos + l - 1);     }      return Infinity; }  // Function to find kth Smallest in Array function kthSmallest(arr, k) {     return selectKthSmallest(arr, 0, arr.length - 1, k); }  // Driver code let arr = [7, 10, 4, 3, 20, 15]; let k = 3;  console.log(kthSmallest(arr, k));

Output
7

Time Complexity: O(n), Worst-case linear time for selection
Space Complexity: O(n), Extra space for storing medians recursively in each level of selection calls.

Detailed Time Complexity Analysis

We analyze the worst-case time complexity of the Median of Medians algorithm step by step:

  • Dividing the array into groups of 5. There are n/5 such groups. Finding the median of each group takes O(1) time since the size is constant. So, the total time for this step is O(n).
  • Finding the median of medians. Recursively finding the median of the n/5 medians takes T(n/5) time.
  • Partitioning the array around the pivot (median of medians) using a standard partition operation takes O(n) time.
  • After partitioning, the recursive call proceeds into one side (either left or right) depending on where the k-th smallest element lies.

To understand the size of the recursive call, we analyze how many elements are guaranteed to be greater or smaller than the pivot. At least half of the n/5 medians are greater than or equal to the median of medians. Each of those groups contributes at least 3 elements greater than the median of medians. Therefore, the number of elements greater than or equal to the pivot is at least (3 * ceil(1/2 * ceil(n/5)) - 6), which is at least (3n/10 - 6). Similarly, the number of elements smaller than the pivot is at least 3n/10 - 6.

So, in the worst case, the recursive call goes to at most n - (3n/10 - 6) = 7n/10 + 6 elements.

The recurrence relation becomes:

T(n) <= Θ(1), if n <= 80
T(n) <= T(ceil(n/5)) + T(7n/10 + 6) + O(n), if n > 80

We prove T(n) = O(n) by substitution. Assume T(n) <= cn for some constant c and for all n > 80.

Substituting into the recurrence:

T(n) <= c(n/5) + c(7n/10 + 6) + O(n)
= cn/5 + 7cn/10 + 6c + O(n)
= 9cn/10 + 6c + O(n)

We can choose c large enough such that cn/10 >= 6c + O(n), so T(n) <= cn.

Hence, the worst-case running time is linear, O(n).

Conclusion

Although this algorithm is linear in the worst case, the constants involved are large due to recursive overhead and multiple passes. In practice, a randomized QuickSelect is much faster and is generally preferred despite its average-case nature.


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    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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