Maximise difference between largest and smallest in Array by shifting X value in pairs after at most K operations
Last Updated :
11 Feb, 2022
Given an array arr[] of size N and a positive integer K, the task is to maximise difference between the largest element and the smallest element in the array by performing at most K operations, wherein each operation:
- select two integers arr[i] and arr[j] from the array and
- update arr[i] to arr[i] – X and arr[j] to arr[j] + X, where X is any integer in range [0, min(arr[i], arr[j])].
Examples:
Input: arr[] = {3, 3, 2, 3, 3}, K = 1
Output: 6
Explanation: In the 1st operation select the integers (arr[1], arr[0]) and X as 3. Hence arr[1] = arr[1] – X = 3 – 3 = 0. Similarly, arr[0] = arr[0] + X = 3 + 3 = 6. Hence, arr[] = {6, 0, 2, 3, 3} and the diffecence between smallest and largest element is 6 which is the maximum possible.
Input: arr[] = {7, 4, 8, 11, 2, 23, 67, 22, 5, 29, 6, 4, 56}, N = 13, X = 5
Output: 208
Approach: The given problem can be solved by using a greedy approach. The idea is to maximize the value of the largest element of the array. It can also be observed that the smallest element that can be achieved in the array is 0. Below are the steps to follow:
- Sort the given array arr[] in descending order.
- Iterate through the array in the range [1, N) and perform the given operation on (arr[i], arr[0]) for X = arr[i].
- After K operations have been performed, arr[0] will be the required answer, except for the case where K =0.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maxDifference( int arr[], int N, int K)
{
sort(arr, arr + N, greater< int >());
if (K == 0) {
return arr[0] - arr[N - 1];
}
for ( int i = 1; i < N && K != 0; ++i) {
arr[0] += arr[i];
arr[i] = 0;
K--;
}
return arr[0];
}
int main()
{
int arr[] = { 3, 3, 2, 3, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
int K = 1;
cout << maxDifference(arr, N, K);
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
static int maxDifference(Integer[] arr, int N, int K)
{
Arrays.sort(arr, Collections.reverseOrder());
if (K == 0 ) {
return arr[ 0 ] - arr[N - 1 ];
}
for ( int i = 1 ; i < N && K != 0 ; ++i) {
arr[ 0 ] += arr[i];
arr[i] = 0 ;
K--;
}
return arr[ 0 ];
}
public static void main (String[] args) {
Integer[] arr = { 3 , 3 , 2 , 3 , 3 };
int N = arr.length;
int K = 1 ;
System.out.print(maxDifference(arr, N, K));
}
}
|
Python3
def maxDifference(arr, N, K):
arr.sort(reverse = True )
if (K = = 0 ):
return arr[ 0 ] - arr[N - 1 ]
for i in range ( 1 , N):
if (K ! = 0 ):
arr[ 0 ] = arr[ 0 ] + arr[i]
arr[i] = 0
K = K - 1
return arr[ 0 ]
arr = [ 3 , 3 , 2 , 3 , 3 ]
N = len (arr)
K = 1
print (maxDifference(arr, N, K))
|
C#
using System;
class GFG
{
static int maxDifference( int [] arr, int N, int K)
{
Array.Sort(arr);
Array.Reverse(arr);
if (K == 0) {
return arr[0] - arr[N - 1];
}
for ( int i = 1; i < N && K != 0; ++i) {
arr[0] += arr[i];
arr[i] = 0;
K--;
}
return arr[0];
}
public static void Main()
{
int [] arr = { 3, 3, 2, 3, 3 };
int n = arr.Length;
int k = 1;
Console.Write(maxDifference(arr, n, k));
}
}
|
Javascript
<script>
function maxDifference(arr, N, K)
{
arr.sort( function (a, b) { return b - a })
if (K == 0) {
return arr[0] - arr[N - 1];
}
for (let i = 1; i < N && K != 0; ++i) {
arr[0] += arr[i];
arr[i] = 0;
K--;
}
return arr[0];
}
let arr = [3, 3, 2, 3, 3];
let N = arr.length;
let K = 1;
document.write(maxDifference(arr, N, K));
</script>
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Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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