Maximize the minimum difference between any element pair by selecting K elements from given Array
Given an array of N integers the task is to select K elements out of these N elements in such a way that the minimum difference between each of the K numbers is the Largest. Return the largest minimum difference after choosing any K elements.
Examples:
Input: N = 4, K = 3, arr = [2, 6, 2, 5]Output: 1
Explanation: 3 elements out of 4 elements are to be selected with a minimum difference as large as possible. Selecting 2, 2, 5 will result in minimum difference as 0. Selecting 2, 5, 6 will result in minimum difference as 6-5=1Input: N = 7, K = 4, arr = [1, 4, 9, 0, 2, 13, 3]Output: 4
Explanation: Selecting 0, 4, 9, 13 will result in minimum difference of 4, which is the largest minimum difference possible
Naive Approach: Generate all the subsets of size K and find the minimum difference in all of them. Then return the maximum among the differences.
Efficient Approach: Given problem can be solved efficiently using binary search on answer technique. Below steps can be followed to solve the problem:
- Sort the array in ascending order
- Initialize minimum answer ans to 1
- Binary search is used on the range 1 to maximum element in array arr
- Variable dif is used to store the largest minimum difference at every iteration
- Helper function is used to check if selection of K elements is possible with minimum difference greater than dif calculated in previous iteration. If possible then true is returned or else false is returned.
- If above function returns:
- True then update ans to dif and left to dif + 1
- False then update right to dif – 1
Below is the implementation of the above binary search approach
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// To check if selection of K elements// is possible such that difference// between them is greater than difbool isPossibleToSelect(int arr[], int N, int dif, int K){ // Selecting first element in the // sorted array int count = 1; // prev is the previously selected // element initially at index 0 as // first element is already selected int prev = arr[0]; // Check if selection of K-1 elements // from array with a minimum // difference of dif is possible for (int i = 1; i < N; i++) { // If the current element is // atleast dif difference apart // from the previously selected // element then select the current // element and increase the count if (arr[i] >= (prev + dif)) { count++; // If selection of K elements // with a min difference of dif // is possible then return true if (count == K) return true; // Prev will become current // element for the next iteration prev = arr[i]; } } // If selection of K elements with minimum // difference of dif is not possible // then return false return false;}int binarySearch(int arr[], int left, int right, int K, int N){ // Minimum largest difference // possible is 1 int ans = 1; while (left <= right) { int dif = left + (right - left) / 2; // Check if selection of K elements // is possible with a minimum // difference of dif if (isPossibleToSelect(arr, N, dif, K)) { // If dif is greater than // previous ans we update ans ans = max(ans, dif); // Continue to search for better // answer. Try finding greater dif left = dif + 1; } // K elements cannot be selected else right = dif - 1; } return ans;}// Driver codeint main(){ int N, K; N = 7, K = 4; int arr[] = { 1, 4, 9, 0, 2, 13, 3 }; // arr should be in a sorted order sort(arr, arr + N); cout << binarySearch(arr, 0, arr[N - 1], K, N) << "\n"; return 0;} |
Java
// Java implementation for the above approachimport java.io.*;import java.util.Arrays;class GFG{ // To check if selection of K elements // is possible such that difference // between them is greater than dif static boolean isPossibleToSelect(int []arr, int N, int dif, int K) { // Selecting first element in the // sorted array int count = 1; // prev is the previously selected // element initially at index 0 as // first element is already selected int prev = arr[0]; // Check if selection of K-1 elements // from array with a minimum // difference of dif is possible for (int i = 1; i < N; i++) { // If the current element is // atleast dif difference apart // from the previously selected // element then select the current // element and increase the count if (arr[i] >= (prev + dif)) { count++; // If selection of K elements // with a min difference of dif // is possible then return true if (count == K) return true; // Prev will become current // element for the next iteration prev = arr[i]; } } // If selection of K elements with minimum // difference of dif is not possible // then return false return false; } static int binarySearch(int []arr, int left, int right, int K, int N) { // Minimum largest difference // possible is 1 int ans = 1; while (left <= right) { int dif = left + (right - left) / 2; // Check if selection of K elements // is possible with a minimum // difference of dif if (isPossibleToSelect(arr, N, dif, K)) { // If dif is greater than // previous ans we update ans ans = Math.max(ans, dif); // Continue to search for better // answer. Try finding greater dif left = dif + 1; } // K elements cannot be selected else right = dif - 1; } return ans; } // Driver code public static void main(String[] args) { int N, K; N = 7; K = 4; int []arr = { 1, 4, 9, 0, 2, 13, 3 }; // arr should be in a sorted order Arrays.sort(arr); System.out.println(binarySearch(arr, 0, arr[N - 1], K, N)); }}// This code is contributed by shivanisinghss2110 |
Python3
# Python 3 implementation for the above approach# To check if selection of K elements# is possible such that difference# between them is greater than difdef isPossibleToSelect(arr, N, dif, K): # Selecting first element in the # sorted array count = 1 # prev is the previously selected # element initially at index 0 as # first element is already selected prev = arr[0] # Check if selection of K-1 elements # from array with a minimum # difference of dif is possible for i in range(1, N): # If the current element is # atleast dif difference apart # from the previously selected # element then select the current # element and increase the count if (arr[i] >= (prev + dif)): count += 1 # If selection of K elements # with a min difference of dif # is possible then return true if (count == K): return True # Prev will become current # element for the next iteration prev = arr[i] # If selection of K elements with minimum # difference of dif is not possible # then return false return Falsedef binarySearch(arr, left, right, K, N): # Minimum largest difference # possible is 1 ans = 1 while (left <= right): dif = left + (right - left) // 2 # Check if selection of K elements # is possible with a minimum # difference of dif if (isPossibleToSelect(arr, N, dif, K)): # If dif is greater than # previous ans we update ans ans = max(ans, dif) # Continue to search for better # answer. Try finding greater dif left = dif + 1 # K elements cannot be selected else: right = dif - 1 return ans# Driver codeif __name__ == "__main__": N = 7 K = 4 arr = [1, 4, 9, 0, 2, 13, 3] # arr should be in a sorted order arr.sort() print(binarySearch(arr, 0, arr[N - 1], K, N) ) # This code is contributed by ukasp. |
C#
// C# implementation for the above approachusing System;using System.Collections.Generic;class GFG{// To check if selection of K elements// is possible such that difference// between them is greater than difstatic bool isPossibleToSelect(int []arr, int N, int dif, int K){ // Selecting first element in the // sorted array int count = 1; // prev is the previously selected // element initially at index 0 as // first element is already selected int prev = arr[0]; // Check if selection of K-1 elements // from array with a minimum // difference of dif is possible for (int i = 1; i < N; i++) { // If the current element is // atleast dif difference apart // from the previously selected // element then select the current // element and increase the count if (arr[i] >= (prev + dif)) { count++; // If selection of K elements // with a min difference of dif // is possible then return true if (count == K) return true; // Prev will become current // element for the next iteration prev = arr[i]; } } // If selection of K elements with minimum // difference of dif is not possible // then return false return false;}static int binarySearch(int []arr, int left, int right, int K, int N){ // Minimum largest difference // possible is 1 int ans = 1; while (left <= right) { int dif = left + (right - left) / 2; // Check if selection of K elements // is possible with a minimum // difference of dif if (isPossibleToSelect(arr, N, dif, K)) { // If dif is greater than // previous ans we update ans ans = Math.Max(ans, dif); // Continue to search for better // answer. Try finding greater dif left = dif + 1; } // K elements cannot be selected else right = dif - 1; } return ans;}// Driver codepublic static void Main(){ int N, K; N = 7; K = 4; int []arr = { 1, 4, 9, 0, 2, 13, 3 }; // arr should be in a sorted order Array.Sort(arr); Console.Write(binarySearch(arr, 0, arr[N - 1], K, N));}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript Program to implement // the above approach // To check if selection of K elements // is possible such that difference // between them is greater than dif function isPossibleToSelect(arr, N, dif, K) { // Selecting first element in the // sorted array let count = 1; // prev is the previously selected // element initially at index 0 as // first element is already selected let prev = arr[0]; // Check if selection of K-1 elements // from array with a minimum // difference of dif is possible for (let i = 1; i < N; i++) { // If the current element is // atleast dif difference apart // from the previously selected // element then select the current // element and increase the count if (arr[i] >= (prev + dif)) { count++; // If selection of K elements // with a min difference of dif // is possible then return true if (count == K) return true; // Prev will become current // element for the next iteration prev = arr[i]; } } // If selection of K elements with minimum // difference of dif is not possible // then return false return false; } function binarySearch(arr, left, right, K, N) { // Minimum largest difference // possible is 1 let ans = 1; while (left <= right) { let dif = left + Math.floor((right - left) / 2); // Check if selection of K elements // is possible with a minimum // difference of dif if (isPossibleToSelect(arr, N, dif, K)) { // If dif is greater than // previous ans we update ans ans = Math.max(ans, dif); // Continue to search for better // answer. Try finding greater dif left = dif + 1; } // K elements cannot be selected else right = dif - 1; } return ans; } // Driver code let N, K; N = 7, K = 4; let arr = [1, 4, 9, 0, 2, 13, 3]; // arr should be in a sorted order arr.sort(function (a, b) { return a - b }) document.write(binarySearch(arr, 0, arr[N - 1], K, N) + '<br>'); // This code is contributed by Potta Lokesh </script> |
Output
4
Time complexity: O(N * log N)
Space complexity: O(1)
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