# 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 **i ^{th}** 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:0

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:2

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.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `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); ` ` ` ` ` `// Itereate 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; ` `} ` |

*chevron_right*

*filter_none*

## 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` `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 = ` `1` `; ` ` ` ` ` `// Itereate 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 ` |

*chevron_right*

*filter_none*

## 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 ` `=` `1` `; ` ` ` ` ` `# Itereate 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 ` |

*chevron_right*

*filter_none*

## 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` `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 = 1; ` ` ` ` ` `// Itereate 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; ` ` ` `Console.WriteLine(minimumCoins(a, n, k)); ` `} ` `} ` ` ` `/* This code is contributed by PrinciRaj1992 */` |

*chevron_right*

*filter_none*

**Output:**

2

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

## Recommended Posts:

- Sum of minimum absolute difference of each array element
- Minimum absolute difference of XOR values of two subarrays
- Minimum sum of absolute difference of pairs of two arrays
- Minimum value of maximum absolute difference of all adjacent pairs in an Array
- Minimum absolute difference of adjacent elements in a circular array
- Missing occurrences of a number in an array such that maximum absolute difference of adjacent elements is minimum
- Minimum number of coins that can generate all the values in the given range
- Find out the minimum number of coins required to pay total amount
- Maximum sum of absolute difference of any permutation
- Triplets in array with absolute difference less than k
- Sort an array according to absolute difference with given value
- Maximum absolute difference between sum of two contiguous sub-arrays
- Sort an array according to absolute difference with given value using Functors
- Count pairs in an array such that the absolute difference between them is ≥ K
- k-th smallest absolute difference of two elements in an array

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.