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2 Sum - Pair Sum Closest to Target using Binary Search
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Closest K Elements in a Sorted Array

Last Updated : 11 Jun, 2025
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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| == |b - x| and a > b (i.e., prefer the larger element if tied)

Examples: 

Input: arr[] = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56], k = 4, x = 35
Output: 39 30 42 45
Explanation: First closest element to 35 is 39.
Second closest element to 35 is 30.
Third closest element to 35 is 42.
And fourth closest element to 35 is 45.

Input: arr[] = [1, 3, 4, 10, 12], k = 2, x = 4
Output: 3 1
Explanation: 4 is excluded, Closest elements to 4 are: 3 (1), 1 (3). So, the 2 closest elements are: 3 1

Table of Content

  • [Naive Aprroach] Using Absolute difference and Custom Sorting O(n*log(n)) Time and O(k) Space
  • [Better Approach] Using Linear Search- O(n) time and O(k) space
  • [Expected Approach] Using Binary Search - O(k + log n) time and O(k) space

[Naive Aprroach] Using Absolute difference and Custom Sorting O(n*log(n)) Time and O(k) Space

The idea is to use the absolute difference between each element and the target value to measure how close they are. Then, we apply custom sorting: elements with smaller differences come first, and in case of a tie, the larger element is preferred. we take the k element from get from custrom sorted array then return it.

C++ 
#include <iostream> #include <vector> #include <algorithm> #include <cmath> using namespace std;  // Function to return k closest elements to x vector<int> printKClosest(vector<int> &arr, int k, int x) {          // Custom comparator using absolute difference and tie-breaker     sort(arr.begin(), arr.end(), [x](int a, int b) {         int diffA = abs(a - x);         int diffB = abs(b - x);         // If differences are equal, prefer the larger element         if (diffA == diffB)             return a > b;         return diffA < diffB;     });      vector<int> result;     int count = 0;      // Pick first k elements which are not equal to x     for (int num : arr) {         // skip if element is equal to x         if (num == x) continue;          result.push_back(num);         count++;         if (count == k)             break;     }      return result; }  // Main function with example int main() {     vector<int> arr = {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};     int k = 4;     int x = 35;      vector<int> closest = printKClosest(arr, k, x);          for (int num : closest) {         cout << num << " ";     }      return 0; }
Java 
import java.util.*;  public class Main {      // Function to return k closest elements to x     public static int[] printKClosest(int[] arr, int k, int x) {         // Convert array to list for easier sorting with custom comparator         List<Integer> list = new ArrayList<>();         for (int num : arr) {             list.add(num);         }          // Custom sort using absolute difference and tie-breaking         list.sort((a, b) -> {             int diffA = Math.abs(a - x);             int diffB = Math.abs(b - x);             if (diffA == diffB) {                 // prefer larger element                 return b - a;              }             return diffA - diffB;         });          List<Integer> result = new ArrayList<>();         int count = 0;          // Pick first k elements not equal to x         for (int num : list) {              // skip if element is equal to x             if (num == x) continue;             result.add(num);             count++;             if (count == k) break;         }          // Convert list to array         int[] closest = new int[k];         for (int i = 0; i < k; i++) {             closest[i] = result.get(i);         }          return closest;     }      // Main function with example     public static void main(String[] args) {         int[] arr = {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};         int k = 4;         int x = 35;          int[] closest = printKClosest(arr, k, x);         for (int num : closest) {             System.out.print(num + " ");         }     } }
Python 
def printKClosest(arr, k, x):     # Custom sort using absolute difference and tie-breaking     # -a to prefer larger on tie     arr.sort(key=lambda a: (abs(a - x), -a))        result = []     count = 0      # Pick first k elements not equal to x     for num in arr:         if num == x:             continue  # skip if element is equal to x         result.append(num)         count += 1         if count == k:             break      return result  if __name__ == "__main__":     arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]     k = 4     x = 35          closest = printKClosest(arr, k, x)     print( *closest)
C# 
using System; using System.Collections.Generic; using System.Linq;  class Program {     // Function to return k closest elements to x     public static int[] printKClosest(int[] arr, int k, int x){                  // Convert to list for sorting with custom comparator         List<int> list = arr.ToList();          // Custom sort using absolute difference and tie-breaking         list.Sort((a, b) => {                          int diffA = Math.Abs(a - x);             int diffB = Math.Abs(b - x);             if (diffA == diffB)                 // prefer larger element                 return b - a;              return diffA - diffB;         });          List<int> result = new List<int>();         int count = 0;          // Pick first k elements not equal to x         foreach (int num in list){                          // skip if element is equal to x             if (num == x) continue;              result.Add(num);             count++;             if (count == k) break;         }          return result.ToArray();     }      // Main function with example     static void Main(){                  int[] arr = { 12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56 };         int k = 4;         int x = 35;          int[] closest = printKClosest(arr, k, x);         foreach (int num in closest){                          Console.Write(num + " ");         }     } }
JavaScript 
function printKClosest(arr, k, x) {     // Custom sort using absolute difference and tie-breaking     arr.sort((a, b) => {         let diffA = Math.abs(a - x);         let diffB = Math.abs(b - x);         if (diffA === diffB) return b - a; // prefer larger element         return diffA - diffB;     });      let result = [];     let count = 0;      // Pick first k elements not equal to x     for (let num of arr) {         if (num === x) continue;         result.push(num);         count++;         if (count === k) break;     }      return result; }  // Cotume code let arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]; let k = 4; let x = 35;  let closest = printKClosest(arr, k, x); console.log(...closest);

Output
39 30 42 45

[Better Approach] Using Linear Search- O(n) time and O(k) space

The idea is to first go through the array to find the last element that is less than or equal to the target value, skipping the target if it's present. Then, we use two pointers to choose the k closest elements by comparing their differences, while following the tie-breaking rules.

C++ 
// C++ program to find k closest elements to a given value #include <bits/stdc++.h> using namespace std;  // Function to find k closest elements to a given value vector<int> printKClosest(vector<int> &arr, int k, int x) {     int n = arr.size();     int i = 0;          // Find index of element just less than x     while (i < n && arr[i] < x) i++;     int left = i - 1, right = i;          // If value at right index is x, increment      if (arr[right] == x) right++;          vector<int> res;          while (left >=0 && right < n && res.size() < k) {         int leftDiff = abs(arr[left] - x);         int rightDiff = abs(arr[right] - x);                  if (leftDiff < rightDiff) {             res.push_back(arr[left]);             left--;         }         else {             res.push_back(arr[right]);             right++;         }     }          // If k elements are not filled      while (left >=0 && res.size() < k) {         res.push_back(arr[left]);         left--;     }          while (right < n && res.size() < k) {         res.push_back(arr[right]);         right++;     }          return res; } int main() {     vector<int> arr =      {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};     int k = 4, x = 35;     vector<int> res = printKClosest(arr, k, x);     for (int val: res) cout << val << " ";     cout << endl;      return 0; }
Java 
// Java program to find k closest elements to a given value  class GfG {          // Function to find k closest elements to a given value     static int[] printKClosest(int[] arr, int k, int x) {         int n = arr.length;         int i = 0;                  // Find index of element just less than x         while (i < n && arr[i] < x) i++;         int left = i - 1, right = i;                  // If value at right index is x, increment          if (right < n && arr[right] == x) right++;                  int[] res = new int[k];         int count = 0;                  while (left >= 0 && right < n && count < k) {             int leftDiff = Math.abs(arr[left] - x);             int rightDiff = Math.abs(arr[right] - x);                          if (leftDiff < rightDiff) {                 res[count++] = arr[left];                 left--;             }             else {                 res[count++] = arr[right];                 right++;             }         }                  // If k elements are not filled          while (left >= 0 && count < k) {             res[count++] = arr[left];             left--;         }                  while (right < n && count < k) {             res[count++] = arr[right];             right++;         }                  return res;     }      public static void main(String[] args) {         int[] arr =          {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};         int k = 4, x = 35;         int[] res = printKClosest(arr, k, x);         for (int val : res) System.out.print(val + " ");         System.out.println();     } }
Python 
# Python program to find k closest elements to a given value  # Function to find k closest elements to a given value def printKClosest(arr, k, x):     n = len(arr)     i = 0          # Find index of element just less than x     while i < n and arr[i] < x:         i += 1     left = i - 1     right = i          # If value at right index is x, increment      if right < n and arr[right] == x:         right += 1          res = []          while left >= 0 and right < n and len(res) < k:         leftDiff = abs(arr[left] - x)         rightDiff = abs(arr[right] - x)                  if leftDiff < rightDiff:             res.append(arr[left])             left -= 1         else:             res.append(arr[right])             right += 1          # If k elements are not filled      while left >= 0 and len(res) < k:         res.append(arr[left])         left -= 1          while right < n and len(res) < k:         res.append(arr[right])         right += 1          return res  if __name__ == "__main__":     arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]     k = 4     x = 35     res = printKClosest(arr, k, x)     print(' '.join(map(str, res)))
C# 
// C# program to find k closest elements to a given value  using System;  class GfG {          // Function to find k closest elements to a given value     static int[] printKClosest(int[] arr, int k, int x) {         int n = arr.Length;         int i = 0;                  // Find index of element just less than x         while (i < n && arr[i] < x) i++;         int left = i - 1, right = i;                  // If value at right index is x, increment          if (right < n && arr[right] == x) right++;                  int[] res = new int[k];         int count = 0;                  while (left >= 0 && right < n && count < k) {             int leftDiff = Math.Abs(arr[left] - x);             int rightDiff = Math.Abs(arr[right] - x);                          if (leftDiff < rightDiff) {                 res[count++] = arr[left];                 left--;             }             else {                 res[count++] = arr[right];                 right++;             }         }                  // If k elements are not filled          while (left >= 0 && count < k) {             res[count++] = arr[left];             left--;         }                  while (right < n && count < k) {             res[count++] = arr[right];             right++;         }                  return res;     }      static void Main() {         int[] arr =          {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};         int k = 4, x = 35;         int[] res = printKClosest(arr, k, x);         foreach (int val in res) Console.Write(val + " ");         Console.WriteLine();     } }
JavaScript 
// JavaScript program to find k closest elements to a given value  // Function to find k closest elements to a given value function printKClosest(arr, k, x) {     let n = arr.length;     let i = 0;          // Find index of element just less than x     while (i < n && arr[i] < x) i++;     let left = i - 1, right = i;          // If value at right index is x, increment      if (right < n && arr[right] === x) right++;          let res = [];          while (left >= 0 && right < n && res.length < k) {         let leftDiff = Math.abs(arr[left] - x);         let rightDiff = Math.abs(arr[right] - x);                  if (leftDiff < rightDiff) {             res.push(arr[left]);             left--;         }         else {             res.push(arr[right]);             right++;         }     }          // If k elements are not filled      while (left >= 0 && res.length < k) {         res.push(arr[left]);         left--;     }          while (right < n && res.length < k) {         res.push(arr[right]);         right++;     }          return res; }  let arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]; let k = 4, x = 35; let res = printKClosest(arr, k, x); console.log(res.join(' '));

Output
39 30 42 45

[Expected Approach] Using Binary Search - O(k + log n) time and O(k) space

The idea is to first use binary search to quickly find the last element in the array that is less than or equal to the target (skipping the target itself if it exists). Then, we use a two-pointer approach to select the k closest elements, following the tie-breaking rules.

C++ 
// C++ program to find k closest elements to a given value #include <bits/stdc++.h> using namespace std;  // Function to find k closest elements to a given value vector<int> printKClosest(vector<int> &arr, int k, int x) {     int n = arr.size();     int i = 0;          int low = 0, high = n - 1, pos = -1;      // Binary search to find last element less than x     while (low <= high) {         int mid = (low + high) / 2;         if (arr[mid] < x) {             pos = mid;             low = mid + 1;         } else {             high = mid - 1;         }     }      int left = pos, right = pos + 1;          // If value at right index is x, increment      if (arr[right] == x) right++;          vector<int> res;           // Use two-pointer technique to pick k closest elements     while (left >=0 && right < n && res.size() < k) {         int leftDiff = abs(arr[left] - x);         int rightDiff = abs(arr[right] - x);                  if (leftDiff < rightDiff) {             res.push_back(arr[left]);             left--;         }         else {             res.push_back(arr[right]);             right++;         }     }          // If k elements are not filled      while (left >=0 && res.size() < k) {         res.push_back(arr[left]);         left--;     }          while (right < n && res.size() < k) {         res.push_back(arr[right]);         right++;     }          return res; }  int main() {     vector<int> arr =      {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};     int k = 4, x = 35;     vector<int> res = printKClosest(arr, k, x);     for (int val: res) cout << val << " ";     cout << endl;      return 0; }
Java 
// Java program to find k closest elements to a given value  class GfG {          // Function to find k closest elements to a given value     static int[] printKClosest(int[] arr, int k, int x) {         int n = arr.length;         int low = 0, high = n - 1, pos = -1;          // Binary search to find last element less than x         while (low <= high) {             int mid = (low + high) / 2;             if (arr[mid] < x) {                 pos = mid;                 low = mid + 1;             } else {                 high = mid - 1;             }         }          int left = pos, right = pos + 1;                  // If value at right index is x, increment          if (right < n && arr[right] == x) right++;                  int[] res = new int[k];         int count = 0;                  // Use two-pointer technique to pick k closest elements         while (left >= 0 && right < n && count < k) {             int leftDiff = Math.abs(arr[left] - x);             int rightDiff = Math.abs(arr[right] - x);                          if (leftDiff < rightDiff) {                 res[count++] = arr[left];                 left--;             }             else {                 res[count++] = arr[right];                 right++;             }         }                  // If k elements are not filled          while (left >= 0 && count < k) {             res[count++] = arr[left];             left--;         }                  while (right < n && count < k) {             res[count++] = arr[right];             right++;         }                  return res;     }      public static void main(String[] args) {         int[] arr =          {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};         int k = 4, x = 35;         int[] res = printKClosest(arr, k, x);         for (int val : res) System.out.print(val + " ");         System.out.println();     } }
Python 
# Python program to find k closest elements to a given value  # Function to find k closest elements to a given value def printKClosest(arr, k, x):     n = len(arr)     low, high, pos = 0, n - 1, -1      # Binary search to find last element less than x     while low <= high:         mid = (low + high) // 2         if arr[mid] < x:             pos = mid             low = mid + 1         else:             high = mid - 1      left, right = pos, pos + 1          # If value at right index is x, increment      if right < n and arr[right] == x:         right += 1          res = []          # Use two-pointer technique to pick k closest elements     while left >= 0 and right < n and len(res) < k:         leftDiff = abs(arr[left] - x)         rightDiff = abs(arr[right] - x)                  if leftDiff < rightDiff:             res.append(arr[left])             left -= 1         else:             res.append(arr[right])             right += 1          # If k elements are not filled      while left >= 0 and len(res) < k:         res.append(arr[left])         left -= 1          while right < n and len(res) < k:         res.append(arr[right])         right += 1          return res  if __name__ == "__main__":     arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]     k = 4     x = 35     res = printKClosest(arr, k, x)     print(' '.join(map(str, res)))
C# 
// C# program to find k closest elements to a given value  using System;  class GfG {          // Function to find k closest elements to a given value     static int[] printKClosest(int[] arr, int k, int x) {         int n = arr.Length;         int low = 0, high = n - 1, pos = -1;          // Binary search to find last element less than x         while (low <= high) {             int mid = (low + high) / 2;             if (arr[mid] < x) {                 pos = mid;                 low = mid + 1;             } else {                 high = mid - 1;             }         }          int left = pos, right = pos + 1;                  // If value at right index is x, increment          if (right < n && arr[right] == x) right++;                  int[] res = new int[k];         int count = 0;                  // Use two-pointer technique to pick k closest elements         while (left >= 0 && right < n && count < k) {             int leftDiff = Math.Abs(arr[left] - x);             int rightDiff = Math.Abs(arr[right] - x);                          if (leftDiff < rightDiff) {                 res[count++] = arr[left];                 left--;             }             else {                 res[count++] = arr[right];                 right++;             }         }                  // If k elements are not filled          while (left >= 0 && count < k) {             res[count++] = arr[left];             left--;         }                  while (right < n && count < k) {             res[count++] = arr[right];             right++;         }                  return res;     }      static void Main() {         int[] arr =          {12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56};         int k = 4, x = 35;         int[] res = printKClosest(arr, k, x);         foreach (int val in res) Console.Write(val + " ");         Console.WriteLine();     } }
JavaScript 
// JavaScript program to find k closest elements to a given value  // Function to find k closest elements to a given value function printKClosest(arr, k, x) {     let n = arr.length;     let low = 0, high = n - 1, pos = -1;      // Binary search to find last element less than x     while (low <= high) {         let mid = Math.floor((low + high) / 2);         if (arr[mid] < x) {             pos = mid;             low = mid + 1;         } else {             high = mid - 1;         }     }      let left = pos, right = pos + 1;          // If value at right index is x, increment      if (right < n && arr[right] === x) right++;          let res = [];          // Use two-pointer technique to pick k closest elements     while (left >= 0 && right < n && res.length < k) {         let leftDiff = Math.abs(arr[left] - x);         let rightDiff = Math.abs(arr[right] - x);                  if (leftDiff < rightDiff) {             res.push(arr[left]);             left--;         }         else {             res.push(arr[right]);             right++;         }     }          // If k elements are not filled      while (left >= 0 && res.length < k) {         res.push(arr[left]);         left--;     }          while (right < n && res.length < k) {         res.push(arr[right]);         right++;     }          return res; }  let arr = [12, 16, 22, 30, 35, 39, 42, 45, 48, 50, 53, 55, 56]; let k = 4, x = 35; let res = printKClosest(arr, k, x); console.log(res.join(' '));

Output
39 30 42 45

Next Article
2 Sum - Pair Sum Closest to Target using Binary Search

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    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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