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Maximize the minimum difference between any element pair by selecting K elements from given Array

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  • Difficulty Level : Medium
  • Last Updated : 23 Sep, 2021
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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=1 

Input: 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 dif
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;
}
 
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 code
int 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 approach
import 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 dif
def 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 False
 
 
def 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 code
if __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 approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// To check if selection of K elements
// is possible such that difference
// between them is greater than dif
static 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 code
public 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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