# Remove Minimum coins such that absolute difference between any two piles is less than K

Given an array, arr[] of size N and an integer K which means there are N piles of coins and the ith contains arr[i] coins. The task is to adjust the number of coins in each pile such that for any two piles if a be the number of coins in the first pile and b be the number of coins in the second pile then |a – b| ? K
One can remove coins from different piles to decrease the number of coins in those piles but cannot increase the number of coins in a pile by adding more coins. Find the minimum number of coins to be removed in order to satisfy the given condition.

Examples:

Input: arr[] = {2, 2, 2, 2}, K = 0
Output:
For any two piles the difference in the number of coins is ? 0.
So, no need to remove any coins.
Input: arr[] = {1, 5, 1, 2, 5, 1}, K = 3
Output:
If we remove one coin each from both the piles containing
5 coins, then for any two piles the absolute difference
in the number of coins is ? 3.

Approach: Since we cannot increase the number of coins in a pile. So, the minimum number of coins in any pile will remain the same as they can’t be removed, and increasing them will add to the operations which we need to minimize. Now, find the minimum coins in a pile, and for every other pile if the difference between the coins in the current pile and the minimum coin pile is greater than K then remove the extra coins from the current pile.

Algorithm:

Step 1: Define a static function min_val that takes an integer array an as input and returns the minimum element of the array.
Step 2: Define another static function “minimumCoins” that takes an integer array a, integer n, and integer k as inputs.
Step 3: Set the value of the integer variable “cnt” to 0 to serve as a counter for the number of coins to be taken out.
Step 4: Find the minimum value in the array a using the min_val function and store it in “minVal”.
Step 5: Use a for loop to iterate through the array, and then carry out the following actions for each entry in turn:
a. Compute and save in the variable diff the difference between the current pile and the minimum coin pile.
b. If the difference between the current pile and the minimum coin pile is larger than k, then add the difference between the current pile                       and k to the “cnt” variable to calculate the number of extra coins that need to be removed.
Step 6: Return the “cnt” variable.

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 coins that need to be removed``int` `minimumCoins(``int` `a[], ``int` `n, ``int` `k)``{``    ``// To store the coins needed to be removed``    ``int` `cnt = 0;` `    ``// Minimum value from the array``    ``int` `minVal = *min_element(a, a + n);` `    ``// Iterate over the array``    ``// and remove extra coins``    ``for` `(``int` `i = 0; i < n; i++) ``    ``{``        ``int` `diff = a[i] - minVal;` `        ``// If the difference between``        ``//  the current pile and the ``        ``// minimum coin pile is greater than k``        ``if` `(diff > k) ``        ``{` `            ``// Count the extra coins to be removed``            ``cnt += (diff - k);``        ``}``    ``}` `    ``// Return the required count``    ``return` `cnt;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 1, 5, 1, 2, 5, 1 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);``    ``int` `k = 3;` `    ``cout << minimumCoins(a, n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG {` `    ``// Function to return the minimum number``    ``// of coins that need to be removed``    ``static` `int` `min_val(``int``[] a)``    ``{``        ``int` `min = ``2147483647``;``        ``for` `(``int` `el : a) {``            ``if` `(el < min) {``                ``min = el;``            ``}``        ``}``        ``return` `min;``    ``}``    ``static` `int` `minimumCoins(``int` `a[], ``int` `n, ``int` `k)``    ``{``        ``// To store the coins needed to be removed``        ``int` `cnt = ``0``;` `        ``// Minimum value from the array``        ``int` `minVal = min_val(a);` `        ``// Iterate over the array and remove extra coins``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``int` `diff = a[i] - minVal;` `            ``// If the difference between the current pile``            ``// and the minimum coin pile is greater than k``            ``if` `(diff > k) {``                ``// Count the extra coins to be removed``                ``cnt += (diff - k);``            ``}``        ``}` `        ``// Return the required count``        ``return` `cnt;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``1``, ``5``, ``1``, ``2``, ``5``, ``1` `};``        ``int` `n = a.length;``        ``int` `k = ``3``;``        ``System.out.println(minimumCoins(a, n, k));``    ``}``}` `// This code is contributed by jit_t`

## Python3

 `# Python implementation of the approach` `# Function to return the minimum number``# of coins that need to be removed``def` `minimumCoins(a, n, k):``    ``# To store the coins needed to be removed``    ``cnt ``=` `0``;` `    ``# Minimum value from the array``    ``minVal ``=` `min``(a);` `    ``# Iterate over the array and remove extra coins``    ``for` `i ``in` `range``(n):``        ``diff ``=` `a[i] ``-` `minVal;` `        ``# If the difference between the current pile and``        ``# the minimum coin pile is greater than k``        ``if` `(diff > k):``            ``# Count the extra coins to be removed``            ``cnt ``+``=` `(diff ``-` `k);``    ``# Return the required count``    ``return` `cnt;` `# Driver code``a ``=` `[``1``, ``5``, ``1``, ``2``, ``5``, ``1``];``n ``=` `len``(a);``k ``=` `3``;``print``(minimumCoins(a, n, k));` `    ` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG {``  ` `  ` `    ``// Function to return the minimum number``    ``// of coins that need to be removed``    ``static` `int` `min_val(``int``[] a, ``int` `n)``    ``{``        ``int` `min = 2147483647;``        ``for` `(``int` `i = 0;i k) ``            ``{``                ``// Count the extra coins to be removed``                ``cnt += (diff - k);``            ``}``        ``}` `        ``// Return the required count``        ``return` `cnt;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] a = { 1, 5, 1, 2, 5, 1 };``        ``int` `n = a.Length;``        ``int` `k = 3;``        ``Console.WriteLine(minimumCoins(a, n, k));``    ``}``}` `/* This code is contributed by PrinciRaj1992 */`

## Javascript

 ``

Output
`2`

Time Complexity: O(n), where n is the size of the array
Auxiliary Space: O(1), as no extra space is used

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