# Minimum increments of Non-Decreasing Subarrays required to make Array Non-Decreasing

Given an array arr[] consisting of N integers, the task is to find the minimum number of operations required to make the array non-decreasing, where, each operation involves incrementing all elements of a non-decreasing subarray from the given array by 1.

Examples:

Input: arr[] = {1, 3, 1, 2, 4}
Output:
Explanation:
Operation 1: Incrementing arr[2] modifies array to {1, 3, 2, 2, 4}
Operation 2: Incrementing subarray {arr[2], arr[3]} modifies array to {1, 3, 3, 3, 4}
Therefore, the final array is non-decreasing.
Input: arr[] = {1, 3, 5, 10}
Output:
Explanation: The array is already non-decreasing.

Approach: Follow the steps below to solve the problem:

• If the array is already a non-decreasing array, then no changes required.
• Otherwise, for any index i where 0 ? i < N, if arr[i] > arr[i+1], add the difference to ans.
• Finally, print ans as answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to return to the minimum` `// number of operations required to` `// make the array non-decreasing` `int` `getMinOps(``int` `arr[], ``int` `n)` `{` `    `  `    ``// Stores the count of operations` `    ``int` `ans = 0;` `    ``for``(``int` `i = 0; i < n - 1; i++)` `    ``{`   `        ``// If arr[i] > arr[i + 1], add` `        ``// arr[i] - arr[i + 1] to the answer` `        ``// Otherwise, add 0` `        ``ans += max(arr[i] - arr[i + 1], 0);` `    ``}` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 3, 1, 2, 4 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    `  `    ``cout << (getMinOps(arr, n));` `}`   `// This code is contributed by PrinciRaj1992`

## Java

 `// Java Program to implement the` `// above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to return to the minimum` `    ``// number of operations required to` `    ``// make the array non-decreasing` `    ``public` `static` `int` `getMinOps(``int``[] arr)` `    ``{` `        ``// Stores the count of operations` `        ``int` `ans = ``0``;` `        ``for` `(``int` `i = ``0``; i < arr.length - ``1``; i++) {`   `            ``// If arr[i] > arr[i + 1], add` `            ``// arr[i] - arr[i + 1] to the answer` `            ``// Otherwise, add 0` `            ``ans += Math.max(arr[i] - arr[i + ``1``], ``0``);` `        ``}`   `        ``return` `ans;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] arr = { ``1``, ``3``, ``1``, ``2``, ``4` `};`   `        ``System.out.println(getMinOps(arr));` `    ``}` `}`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to return to the minimum ` `# number of operations required to` `# make the array non-decreasing` `def` `getMinOps(arr):`   `    ``# Stores the count of operations` `    ``ans ``=` `0` `    `  `    ``for` `i ``in` `range``(``len``(arr) ``-` `1``):`   `        ``# If arr[i] > arr[i + 1], add` `        ``# arr[i] - arr[i + 1] to the answer` `        ``# Otherwise, add 0` `        ``ans ``+``=` `max``(arr[i] ``-` `arr[i ``+` `1``], ``0``)`   `    ``return` `ans`   `# Driver Code`   `# Given array arr[]` `arr ``=` `[ ``1``, ``3``, ``1``, ``2``, ``4` `]`   `# Function call` `print``(getMinOps(arr))`   `# This code is contributed by Shivam Singh`

## C#

 `// C# Program to implement the` `// above approach` `using` `System;` `class` `GFG` `{`   `  ``// Function to return to the minimum` `  ``// number of operations required to` `  ``// make the array non-decreasing` `  ``public` `static` `int` `getMinOps(``int``[] arr)` `  ``{` `    ``// Stores the count of operations` `    ``int` `ans = 0;` `    ``for` `(``int` `i = 0; i < arr.Length - 1; i++) ` `    ``{`   `      ``// If arr[i] > arr[i + 1], add` `      ``// arr[i] - arr[i + 1] to the answer` `      ``// Otherwise, add 0` `      ``ans += Math.Max(arr[i] - arr[i + 1], 0);` `    ``}` `    ``return` `ans;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(String[] args)` `  ``{` `    ``int``[] arr = { 1, 3, 1, 2, 4 };`   `    ``Console.WriteLine(getMinOps(arr));` `  ``}` `}`   `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(1)

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