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# Minimum decrement operations to make Array elements equal by only decreasing K each time

• Last Updated : 22 Apr, 2021

Given an array arr[] of size N consisting of positive integers and an integer K, the task is to find the minimum number of steps required to make all the elements of the array equal such that at each step, one value from the array can be selected and decremented by K. Print -1 if the array can’t be made equal.
Examples:

Input: arr[] = {12, 9, 15}, K = 3
Output:
Explanation:
Initially: {12, 9, 15}
After decreasing K from 15 at position 3: [12, 9, 12]
After decreasing K from 12 at position 1: [9, 9, 12]
After decreasing K from 12 at position 3: [9, 9, 9]
Input: arr[] = {10, 9}, K = 2
Output: -1
Explanation:
It is impossible to equalize all elements

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Approach: The idea is to keep the minimum valued elements unaffected and count the number of decrement operations taken by the other elements to reach this minimum value. The following steps can be followed to compute the result:

1. Find the minimum element minx in the array.
2. Once the minimum value is found, a variable decrements is maintained and initialized to 0.
3. Then a loop is run over all elements, adding (arr[i]-minx)/K to the decrements variable.
4. If any arr[i] is encountered such that arr[i]-minx is not divisible by K, then return -1 as it can’t be decreased to the minimum value.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach` `#include ``using` `namespace` `std;` `#define lli long long int` `lli solve(lli arr[], lli n, lli k)``{``    ``lli i, minx = INT_MAX;` `    ``// Finding the minimum element``    ``for` `(i = 0; i < n; i++) {``        ``minx = min(minx, arr[i]);``    ``}` `    ``lli decrements = 0;` `    ``// Loop over all the elements``    ``// and find the difference``    ``for` `(i = 0; i < n; i++) {``        ``if` `((arr[i] - minx) % k != 0) {``            ``return` `-1;``        ``}``        ``else` `{``            ``decrements += ((arr[i] - minx) / k);``        ``}``    ``}``    ``// Solution found and returned``    ``return` `decrements;``}` `// Driver code``int` `main()``{``    ``lli n, k;``    ``n = 3;``    ``k = 3;``    ``lli arr[n] = { 12, 9, 15 };` `    ``cout << solve(arr, n, k);``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{` `    ``static` `int` `INT_MAX = Integer.MAX_VALUE ;``    ` `    ``static` `int` `solve(``int` `arr[], ``int` `n, ``int` `k)``    ``{``        ``int` `minx = INT_MAX;``        ``int` `i;``        ` `        ``// Finding the minimum element``        ``for` `(i = ``0``; i < n; i++)``        ``{``            ``minx = Math.min(minx, arr[i]);``        ``}``    ` `        ``int` `decrements = ``0``;``    ` `        ``// Loop over all the elements``        ``// and find the difference``        ``for` `(i = ``0``; i < n; i++)``        ``{``            ``if` `((arr[i] - minx) % k != ``0``)``            ``{``                ``return` `-``1``;``            ``}``            ``else``            ``{``                ``decrements += ((arr[i] - minx) / k);``            ``}``        ``}``        ` `        ``// Solution found and returned``        ``return` `decrements;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n, k;``        ``n = ``3``;``        ``k = ``3``;``        ``int` `arr[] = { ``12``, ``9``, ``15` `};``    ` `        ``System.out.println(solve(arr, n, k));``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the above approach``import` `sys` `def` `solve(arr, n, k) :` `    ``minx ``=` `sys.maxsize;` `    ``# Finding the minimum element``    ``for` `i ``in` `range``(n) :``        ``minx ``=` `min``(minx, arr[i]);` `    ``decrements ``=` `0``;` `    ``# Loop over all the elements``    ``# and find the difference``    ``for` `i ``in` `range``(n) :``        ``if` `((arr[i] ``-` `minx) ``%` `k !``=` `0``) :``            ``return` `-``1``;``        ` `        ``else` `:``            ``decrements ``+``=` `((arr[i] ``-` `minx) ``/``/` `k);``    ` `    ``# Solution found and returned``    ``return` `decrements;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `3``;``    ``k ``=` `3``;``    ``arr ``=` `[ ``12``, ``9``, ``15` `];` `    ``print``(solve(arr, n, k));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``    ``static` `int` `INT_MAX = ``int``.MaxValue ;``    ` `    ``static` `int` `solve(``int` `[]arr, ``int` `n, ``int` `k)``    ``{``        ``int` `minx = INT_MAX;``        ``int` `i;``        ` `        ``// Finding the minimum element``        ``for` `(i = 0; i < n; i++)``        ``{``            ``minx = Math.Min(minx, arr[i]);``        ``}``    ` `        ``int` `decrements = 0;``    ` `        ``// Loop over all the elements``        ``// and find the difference``        ``for` `(i = 0; i < n; i++)``        ``{``            ``if` `((arr[i] - minx) % k != 0)``            ``{``                ``return` `-1;``            ``}``            ``else``            ``{``                ``decrements += ((arr[i] - minx) / k);``            ``}``        ``}``        ` `        ``// Solution found and returned``        ``return` `decrements;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n, k;``        ``n = 3;``        ``k = 3;``        ``int` `[]arr = { 12, 9, 15 };``    ` `        ``Console.WriteLine(solve(arr, n, k));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`3`

Time complexity: O(N)

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