# Minimum number of subtract operation to make an array decreasing

• Difficulty Level : Medium
• Last Updated : 14 Apr, 2021

You are given a sequence of numbers arr, arr, …, arr[N – 1] and a positive integer K. In each operation, you may subtract K from any element of the array. You are required to find the minimum number of operations to make the given array decreasing.
An array is called decreasing if for each i: Input : N = 4, K = 5, arr[] = {1, 1, 2, 3}
Output : 3
Explanation :
Since arr == arr so no subtraction is required for arr. For arr, since arr > arr (2 > 1) so we have to subtract arr by k and after the one subtraction value of arr is -3 which is less than the value of arr, so number of subtraction required only 1 and now value of arr has been updated by -3.
Similarly for arr, since arr > arr (3 > -3) so for this we have to subtract arr by k two times to make the value of arr lesser than arr, the number of subtraction required 2 and the updated value of arr is -7. Now count total number of subtraction /operation required by adding number of operation on each step and that is = 0+1+2 = 3.
Input : N = 5, K = 2, arr[] = {5, 4, 3, 2, 1}
Output : 0

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Approach :

```1. Traverse each element of array from 1 to n-1.
2. Check if (arr[i] > arr[i-1]) then
Find noOfSubtraction;
noOfSubtraction = If ( (arr[i] - arr[i-1]) % k == 0 )
then noOfSubtraction++
Modify arr[i];
arr[i] = ```

Below is implementation of above approach :

## CPP

 `// CPP program to make an array decreasing``#include ``using` `namespace` `std;` `// Function to count minimum no of operation``int` `min_noOf_operation(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `noOfSubtraction;``    ``int` `res = 0;``    ``for` `(``int` `i = 1; i < n; i++) {``        ``noOfSubtraction = 0;` `        ``if` `(arr[i] > arr[i - 1]) {` `            ``// Count how many times we have to subtract.``            ``noOfSubtraction = (arr[i] - arr[i - 1]) / k;` `            ``// Check an additional subtraction is``            ``// required or not.``            ``if` `((arr[i] - arr[i - 1]) % k != 0)``                ``noOfSubtraction++;` `            ``// Modify the value of arr[i].``            ``arr[i] = arr[i] - k * noOfSubtraction;``        ``}` `        ``// Count total no of operation/subtraction .``        ``res = res + noOfSubtraction;``    ``}` `    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 1, 2, 3 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `k = 5;``    ``cout << min_noOf_operation(arr, N, k) << endl;``    ``return` `0;``}`

## Java

 `// Java program to make an``// array decreasing``import` `java.util.*;``import` `java.lang.*;` `public` `class` `GfG{``    ` `    ``// Function to count minimum no of operation``    ``public` `static` `int` `min_noOf_operation(``int` `arr[],``                                      ``int` `n, ``int` `k)``    ``{``        ``int` `noOfSubtraction;``        ``int` `res = ``0``;``        ` `        ``for` `(``int` `i = ``1``; i < n; i++) {``            ``noOfSubtraction = ``0``;` `            ``if` `(arr[i] > arr[i - ``1``]) {``    ` `                ``// Count how many times``                ``// we have to subtract.``                ``noOfSubtraction = (arr[i] - arr[i - ``1``]) / k;` `                ``// Check an additional subtraction``                ``// is required or not.``                ``if` `((arr[i] - arr[i - ``1``]) % k != ``0``)``                    ``noOfSubtraction++;` `                ``// Modify the value of arr[i]``                ``arr[i] = arr[i] - k * noOfSubtraction;``            ``}` `            ``// Count total no of subtraction``            ``res = res + noOfSubtraction;``        ``}` `        ``return` `res;``    ``}``    ` `    ``// driver function``    ``public` `static` `void` `main(String argc[]){``        ``int` `arr = { ``1``, ``1``, ``2``, ``3` `};``        ``int` `N = ``4``;``        ``int` `k = ``5``;``        ``System.out.println(min_noOf_operation(arr,``                                           ``N, k));``    ``}``    ` `}` `/* This code is contributed by Sagar Shukla */`

## Python3

 `# Python program to make an array decreasing` `# Function to count minimum no of operation``def` `min_noOf_operation(arr, n, k):` `    ``res ``=` `0``    ``for` `i ``in` `range``(``1``,n):``        ``noOfSubtraction ``=` `0` `        ``if` `(arr[i] > arr[i ``-` `1``]):` `            ``# Count how many times we have to subtract.``            ``noOfSubtraction ``=` `(arr[i] ``-` `arr[i ``-` `1``]) ``/` `k;` `            ``# Check an additional subtraction is``            ``# required or not.``            ``if` `((arr[i] ``-` `arr[i ``-` `1``]) ``%` `k !``=` `0``):``                ``noOfSubtraction``+``=``1` `            ``# Modify the value of arr[i].``            ``arr[i] ``=` `arr[i] ``-` `k ``*` `noOfSubtraction``        `  `        ``# Count total no of operation/subtraction .``        ``res ``=` `res ``+` `noOfSubtraction``    `  `    ``return` `int``(res)`  `# Driver Code``arr ``=` `[ ``1``, ``1``, ``2``, ``3` `]``N ``=` `len``(arr)``k ``=` `5``print``(min_noOf_operation(arr, N, k))` `# This code is contributed by``# Smitha Dinesh Semwal`

## C#

 `// C# program to make an``// array decreasing``using` `System;` `public` `class` `GfG{``    ` `    ``// Function to count minimum no of operation``    ``public` `static` `int` `min_noOf_operation(``int` `[]arr,``                                       ``int` `n, ``int` `k)``    ``{``        ``int` `noOfSubtraction;``        ``int` `res = 0;``        ` `        ``for` `(``int` `i = 1; i < n; i++) {``            ``noOfSubtraction = 0;` `            ``if` `(arr[i] > arr[i - 1]) {``    ` `                ``// Count how many times``                ``// we have to subtract.``                ``noOfSubtraction = (arr[i] - arr[i - 1]) / k;` `                ``// Check an additional subtraction``                ``// is required or not.``                ``if` `((arr[i] - arr[i - 1]) % k != 0)``                    ``noOfSubtraction++;` `                ``// Modify the value of arr[i]``                ``arr[i] = arr[i] - k * noOfSubtraction;``            ``}` `            ``// Count total no of subtraction``            ``res = res + noOfSubtraction;``        ``}` `        ``return` `res;``    ``}``    ` `    ``// driver function``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 1, 1, 2, 3 };``        ``int` `N = 4;``        ``int` `k = 5;``        ``Console.WriteLine(min_noOf_operation(arr,``                                        ``N, k));``    ``}``    ` `}` `// This code is contributed by vt_m`

## PHP

 ` ``\$arr``[``\$i` `- 1])``        ``{` `            ``// Count how many times we``            ``// have to subtract.``            ``\$noOfSubtraction` `= (``\$arr``[``\$i``] -``                      ``\$arr``[``\$i` `- 1]) / ``\$k``;` `            ``// Check an additional subtraction``            ``// is required or not.``            ``if` `((``\$arr``[``\$i``] - ``\$arr``[``\$i` `- 1])``                               ``% ``\$k` `!= 0)``                ``\$noOfSubtraction``++;` `            ``// Modify the value of arr[i].``            ``\$arr``[``\$i``] = ``\$arr``[``\$i``] - ``\$k` `*``                      ``\$noOfSubtraction``;``        ``}` `        ``// Count total no of``        ``// operation/subtraction .``        ``\$res` `= ``\$res` `+ ``\$noOfSubtraction``;``    ``}` `    ``return` `floor``(``\$res``);``}` `    ``// Driver Code``    ``\$arr` `= ``array``(1, 1, 2, 3);``    ``\$N` `= ``count``(``\$arr``);``    ``\$k` `= 5;``    ``echo` `min_noOf_operation(``\$arr``, ``\$N``, ``\$k``) ;` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output :

` 3`

Time Complexity : O(N).

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