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# Minimum increment operations to make K elements equal

• Difficulty Level : Medium
• Last Updated : 15 Jul, 2022

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:
No operations are required as first and last

Naive approach:

1. Sort the array in increasing order.
2. Now select K elements and make them equal.
3. Choose the ith value as the largest value and make all elements just smaller than it equal to the ith element.
4. Calculate the number of operations needed to make K elements equal to the ith element for all i.
5. The answer will be the minimum of all the possibilities.

## C++14

 `#include `` ` `using` `namespace` `std;`` ` `int` `minOperations(vector<``int``> ar, ``int``& n, ``int``& k)``{``    ``// Sort the array in increasing order``    ``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;``}``// this code is contributed by prophet1999`

## Java

 `// JAVA code for the above approach``import` `java.util.*;``class` `GFG {``  ``public` `static` `int` `minOperations(ArrayList ar,``                                  ``int` `n, ``int` `k)``  ``{`` ` `    ``// Sort the array in increasing order``    ``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 arr = ``new` `ArrayList(``      ``Arrays.asList(``3``, ``1``, ``9``, ``100``));``    ``int` `n = arr.size();``    ``int` `k = ``3``;`` ` `    ``System.out.print(minOperations(arr, n, k));``  ``}``}`` ` `// This code is contributed by Taranpreet`

Output

`14`

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]

1. 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))

2. 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++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the minimum number of``// increment operations required to make``// any k elements of the array equal``int` `minOperations(vector<``int``> ar, ``int` `k)``{``    ``// Sort the array in increasing order``    ``sort(ar.begin(), ar.end());`` ` `    ``// Calculate the number of operations``    ``// needed to make 1st k elements equal to``    ``// the kth element i.e. the 1st window``    ``int` `opsNeeded = 0;``    ``for` `(``int` `i = 0; i < k; i++) {``        ``opsNeeded += ar[k - 1] - ar[i];``    ``}`` ` `    ``// Answer will be the minimum of all``    ``// possible k sized windows``    ``int` `ans = opsNeeded;`` ` `    ``// Find the operations needed to make``    ``// k elements equal to ith element``    ``for` `(``int` `i = k; i < ar.size(); i++) {`` ` `        ``// Slide the window to the right and``        ``// subtract increments spent on leftmost``        ``// element of the previous window``        ``opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k]);`` ` `        ``// Add increments needed to make the 1st k-1``        ``// elements of this window equal to the``        ``// kth element of the current window``        ``opsNeeded += (k - 1) * (ar[i] - ar[i - 1]);``        ``ans = min(ans, opsNeeded);``    ``}``    ``return` `ans;``}`` ` `// Driver code``int` `main()``{``    ``vector<``int``> arr = { 3, 1, 9, 100 };``    ``int` `n = arr.size();``    ``int` `k = 3;`` ` `    ``cout << minOperations(arr, k);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach ``import` `java.util.Arrays; `` ` `class` `geeksforgeeks {`` ` `// Function to return the minimum number of ``// increment operations required to make ``// any k elements of the array equal ``static` `int` `minOperations(``int` `ar[], ``int` `k) ``{ ``    ``// Sort the array in increasing order ``    ``Arrays.sort(ar); `` ` `    ``// Calculate the number of operations ``    ``// needed to make 1st k elements equal to ``    ``// the kth element i.e. the 1st window ``    ``int` `opsNeeded = ``0``; ``    ``for` `(``int` `i = ``0``; i < k; i++) { ``        ``opsNeeded += ar[k - ``1``] - ar[i]; ``    ``} `` ` `    ``// Answer will be the minimum of all ``    ``// possible k sized windows ``    ``int` `ans = opsNeeded; `` ` `    ``// Find the operations needed to make ``    ``// k elements equal to ith element ``    ``for` `(``int` `i = k; i < ar.length; i++) { `` ` `        ``// Slide the window to the right and ``        ``// subtract increments spent on leftmost ``        ``// element of the previous window ``        ``opsNeeded = opsNeeded - (ar[i - ``1``] - ar[i - k]); `` ` `        ``// Add increments needed to make the 1st k-1 ``        ``// elements of this window equal to the ``        ``// kth element of the current window ``        ``opsNeeded += (k - ``1``) * (ar[i] - ar[i - ``1``]); ``        ``ans = Math.min(ans, opsNeeded); ``    ``} ``    ``return` `ans; ``} `` ` `// Driver code ``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)); ``} ``}`` ` `// This code is contributed by Atul_kumar_Shrivastava`

## Python3

 `# Python3 implementation of the approach`` ` `# Function to return the minimum number of``# increment operations required to make``# any k elements of the array equal``def` `minOperations(ar, k):`` ` `    ``# Sort the array in increasing order``    ``ar ``=` `sorted``(ar)`` ` `    ``# Calculate the number of operations``    ``# needed to make 1st k elements equal to``    ``# the kth element i.e. the 1st window``    ``opsNeeded ``=` `0``    ``for` `i ``in` `range``(k):``        ``opsNeeded ``+``=` `ar[k ``-` `1``] ``-` `ar[i]`` ` `    ``# Answer will be the minimum of all``    ``# possible k sized windows``    ``ans ``=` `opsNeeded`` ` `    ``# Find the operations needed to make``    ``# k elements equal to ith element``    ``for` `i ``in` `range``(k, ``len``(ar)):`` ` `        ``# Slide the window to the right and``        ``# subtract increments spent on leftmost``        ``# element of the previous window``        ``opsNeeded ``=` `opsNeeded ``-` `(ar[i ``-` `1``] ``-` `ar[i ``-` `k])`` ` `        ``# Add increments needed to make the 1st k-1``        ``# elements of this window equal to the``        ``# kth element of the current window``        ``opsNeeded ``+``=` `(k ``-` `1``) ``*` `(ar[i] ``-` `ar[i ``-` `1``])``        ``ans ``=` `min``(ans, opsNeeded)`` ` `    ``return` `ans`` ` `# Driver code``arr ``=` `[``3``, ``1``, ``9``, ``100``]``n ``=` `len``(arr)``k ``=` `3`` ` `print``(minOperations(arr, k))`` ` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach ``using` `System; `` ` `class` `geeksforgeeks {`` ` `// Function to return the minimum number of ``// increment operations required to make ``// any k elements of the array equal ``static` `int` `minOperations(``int``[] ar, ``int` `k) ``{ ``    ``// Sort the array in increasing order ``    ``Array.Sort(ar); `` ` `    ``// Calculate the number of operations ``    ``// needed to make 1st k elements equal to ``    ``// the kth element i.e. the 1st window ``    ``int` `opsNeeded = 0; ``    ``for` `(``int` `i = 0; i < k; i++) { ``        ``opsNeeded += ar[k - 1] - ar[i]; ``    ``} `` ` `    ``// Answer will be the minimum of all ``    ``// possible k sized windows ``    ``int` `ans = opsNeeded; `` ` `    ``// Find the operations needed to make ``    ``// k elements equal to ith element ``    ``for` `(``int` `i = k; i < ar.Length; i++) { `` ` `        ``// Slide the window to the right and ``        ``// subtract increments spent on leftmost ``        ``// element of the previous window ``        ``opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k]); `` ` `        ``// Add increments needed to make the 1st k-1 ``        ``// elements of this window equal to the ``        ``// kth element of the current window ``        ``opsNeeded += (k - 1) * (ar[i] - ar[i - 1]); ``        ``ans = Math.Min(ans, opsNeeded); ``    ``} ``    ``return` `ans; ``} `` ` `// Driver code ``public` `static` `void` `Main() ``{ ``    ``int``[] arr = { 3, 1, 9, 100 }; ``    ``int` `n = arr.Length; ``    ``int` `k = 3; `` ` `    ``Console.Write(minOperations(arr, k)); ``} ``}`` ` `// This code is contributed by AbhiThakur`

## Javascript

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Output

`14`

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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