Minimum increment operations to make K elements equal
Last Updated :
28 Feb, 2023
Given an array arr[] of N elements and an integer K, the task is to make any K elements of the array equal by performing only increment operations i.e. in one operation, any element can be incremented by 1. Find the minimum number of operations required to make any K elements equal.
Examples:
Input: arr[] = {3, 1, 9, 100}, K = 3
Output: 14
Increment 3 six times and 1 eight times for a total of
14 operations to make 3 elements equal to 9.
Input: arr[] = {5, 3, 10, 5}, K = 2
Output: 0
No operations are required as first and last
elements are already equal.
Naive approach:
- Sort the array in increasing order.
- Now select K elements and make them equal.
- Choose the ith value as the largest value and make all elements just smaller than it equal to the ith element.
- Calculate the number of operations needed to make K elements equal to the ith element for all i.
- The answer will be the minimum of all the possibilities.
C++14
#include <bits/stdc++.h>
using namespace std;
int minOperations(vector< int > ar, int & n, int & k)
{
sort(ar.begin(), ar.end());
int opsneeded, ans = INT_MAX;
for ( int i = k; i < n; i++) {
opsneeded = 0;
for ( int j = i - k; j < i; j++)
opsneeded += ar[i - 1] - ar[j];
ans = min(ans, opsneeded);
}
return ans;
}
int main()
{
vector< int > arr = { 3, 1, 9, 100 };
int n = arr.size();
int k = 3;
cout << minOperations(arr, n, k);
return 0;
}
|
Java
import java.util.*;
class GFG {
public static int minOperations(ArrayList<Integer> ar,
int n, int k)
{
Collections.sort(ar);
int opsneeded, ans = Integer.MAX_VALUE;
for ( int i = k; i < n; i++) {
opsneeded = 0 ;
for ( int j = i - k; j < i; j++)
opsneeded += ar.get(i - 1 ) - ar.get(j);
ans = Math.min(ans, opsneeded);
}
return ans;
}
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<Integer>(
Arrays.asList( 3 , 1 , 9 , 100 ));
int n = arr.size();
int k = 3 ;
System.out.print(minOperations(arr, n, k));
}
}
|
Python3
import math
def minOperations(ar, n, k):
ar.sort()
opsneeded, ans = math.inf, math.inf
for i in range (k, n):
opsneeded = 0
for j in range (i - k, i):
opsneeded + = ar[i - 1 ] - ar[j]
ans = min (ans, opsneeded)
return ans
arr = [ 3 , 1 , 9 , 100 ]
n = len (arr)
k = 3
print (minOperations(arr, n, k))
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
public static int minOperations(List< int > ar, int n, int k) {
ar.Sort();
int opsneeded, ans = int .MaxValue;
for ( int i = k; i < n; i++) {
opsneeded = 0;
for ( int j = i - k; j < i; j++)
opsneeded += ar[i - 1] - ar[j];
ans = Math.Min(ans, opsneeded);
}
return ans;
}
public static void Main( string [] args) {
List< int > arr = new List< int >( new int [] {3, 1, 9, 100});
int n = arr.Count;
int k = 3;
Console.Write(minOperations(arr, n, k));
}
}
|
Javascript
function minOperations(ar,n,k){
ar.sort((a,b) => a-b);
let opsneeded;
let ans=Infinity;
for (let i = k; i < n; i++) {
opsneeded = 0
for (let j = i - k; j < i; j++) {
opsneeded += ar[i - 1] - ar[j]
}
ans = Math.min(ans, opsneeded)
}
return ans;
}
let arr = [3, 1, 9, 100]
let k = 3
console.log(minOperations(arr,arr.length,k));
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Time Complexity: Depends on sorting, it will be either O(n^2+n*K) or O(n log n+n*K)
Auxiliary Space: Depends on sorting, it will be either O(n) or O(1)
Efficient approach: the naive approach can be modified to calculate the minimum operations needed to make K elements equal to the ith element faster than O(K) using the sliding window technique in constant time given that the operations required for making the 1st K elements equal to the Kth element are known.
Let C be the operations needed or cost for making the elements in the range [l, l + K – 1] equal to the (l + K – 1)th element. Now to find the cost for the range [l + 1, l + K], the solution for the range [l, l + K – 1] can be used.
Let C’ be the cost for the range [l + 1, l + K].
- Since we increment lth element to (l + K – 1)th element, C includes the cost element(l + k – 1) – element(l) but C’ does not need to include this cost.
So, C’ = C – (element(l + k – 1) – element(l))
- Now C’ represents the cost of making all the elements in the range [l + 1, l + K – 1] equal to (l + K – 1)th element.
Since, we need to make all elements equal to the (l + K)th element instead of the (l + K – 1)th element, we can increment these k – 1 elements to the (l + K)th element which makes C’ = C’ + (k – 1) * (element(l + k) – element(l + k -1))
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int minOperations(vector< int > ar, int k)
{
sort(ar.begin(), ar.end());
int opsNeeded = 0;
for ( int i = 0; i < k; i++) {
opsNeeded += ar[k - 1] - ar[i];
}
int ans = opsNeeded;
for ( int i = k; i < ar.size(); i++) {
opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k]);
opsNeeded += (k - 1) * (ar[i] - ar[i - 1]);
ans = min(ans, opsNeeded);
}
return ans;
}
int main()
{
vector< int > arr = { 3, 1, 9, 100 };
int n = arr.size();
int k = 3;
cout << minOperations(arr, k);
return 0;
}
|
Java
import java.util.Arrays;
class geeksforgeeks {
static int minOperations( int ar[], int k)
{
Arrays.sort(ar);
int opsNeeded = 0 ;
for ( int i = 0 ; i < k; i++) {
opsNeeded += ar[k - 1 ] - ar[i];
}
int ans = opsNeeded;
for ( int i = k; i < ar.length; i++) {
opsNeeded = opsNeeded - (ar[i - 1 ] - ar[i - k]);
opsNeeded += (k - 1 ) * (ar[i] - ar[i - 1 ]);
ans = Math.min(ans, opsNeeded);
}
return ans;
}
public static void main(String[] args)
{
int [] arr = { 3 , 1 , 9 , 100 };
int n = arr.length;
int k = 3 ;
System.out.printf( "%d" ,minOperations(arr, k));
}
}
|
Python3
def minOperations(ar, k):
ar = sorted (ar)
opsNeeded = 0
for i in range (k):
opsNeeded + = ar[k - 1 ] - ar[i]
ans = opsNeeded
for i in range (k, len (ar)):
opsNeeded = opsNeeded - (ar[i - 1 ] - ar[i - k])
opsNeeded + = (k - 1 ) * (ar[i] - ar[i - 1 ])
ans = min (ans, opsNeeded)
return ans
arr = [ 3 , 1 , 9 , 100 ]
n = len (arr)
k = 3
print (minOperations(arr, k))
|
C#
using System;
class geeksforgeeks {
static int minOperations( int [] ar, int k)
{
Array.Sort(ar);
int opsNeeded = 0;
for ( int i = 0; i < k; i++) {
opsNeeded += ar[k - 1] - ar[i];
}
int ans = opsNeeded;
for ( int i = k; i < ar.Length; i++) {
opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k]);
opsNeeded += (k - 1) * (ar[i] - ar[i - 1]);
ans = Math.Min(ans, opsNeeded);
}
return ans;
}
public static void Main()
{
int [] arr = { 3, 1, 9, 100 };
int n = arr.Length;
int k = 3;
Console.Write(minOperations(arr, k));
}
}
|
Javascript
<script>
function minOperations(ar,k)
{
ar.sort( function (a,b){ return a-b});
let opsNeeded = 0;
for (let i = 0; i < k; i++) {
opsNeeded += ar[k - 1] - ar[i];
}
let ans = opsNeeded;
for (let i = k; i < ar.length; i++) {
opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k]);
opsNeeded += (k - 1) * (ar[i] - ar[i - 1]);
ans = Math.min(ans, opsNeeded);
}
return ans;
}
let arr=[3, 1, 9, 100 ];
let n = arr.length;
let k = 3;
document.write(minOperations(arr, k));
</script>
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Time Complexity: Depends on sorting, it will be either O(n^2) or O(n log n)
Auxiliary Space: Depends on sorting, it will be either O(n) or O(1)
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