# Count of suffix increment/decrement operations to construct a given array

Given an array of non-negative integers. We need to construct given array from an array of all zeros. We are allowed to do following operation.

• Choose any index of say i and add 1 to all the elements or subtract 1 from all the elements from index i to last index. We basically increase/decrease a suffix by 1.

Examples :

Input : brr[] = {1, 2, 3, 4, 5}
Output : 5
Here, we can successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.

Input : brr[] = {1, 2, 2, 1}
Output : 3
Here, we choose indices 1 and 2 and adds 1 to corresponding suffixes, then we choose index 4 and subtract 1.

Let brr[] be given array and arr[] be current array (which is initially 0).
The approach is simple:

• To make first element equal we have to make |brr| operations. Once this is done, arr, arr, arr, … arr[n] = brr.
• To make Second element equal we have to make |brr – brr| operations. Once this is done, arr, arr, arr, … arr[n] = brr.

In general, to make arr[i] = brr[i] we need to make |brr[i] – b[i – 1]| operations. So in total we have to make |b| + |b – b| + |b – b| + … + |b[n] – b[n – 1]| operations.

Below is CPP and Java implementation of the above approach:

## C++

 `// CPP program to find minimum number of steps` `// to make the array equal to the given array.` `#include ` `using` `namespace` `std;`   `// function to calculate min_Steps` `int` `minSteps(``int` `arr[], ``int` `n)` `{` `    ``int` `min_Steps = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(i > 0) ` `            ``min_Steps += ``abs``(arr[i] - arr[i - 1]);` `        `  `        ``// first element of arr.` `        ``else` `            ``min_Steps += ``abs``(arr[i]);` `    ``}` `    ``return` `min_Steps;` `}`   `// driver function` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 2, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << minSteps(arr, n) << endl;` `}`

## Java

 `// Java program to find minimum number of steps` `// to make the array equal to the given array.` `import` `java.util.*;` `import` `java.lang.*;`   `public` `class` `GfG {` `    ``// function to calculate min_Steps` `    ``public` `static` `int` `minSteps(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `min_Steps = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(i > ``0``) ` `                ``min_Steps += ` `                    ``Math.abs(arr[i] - arr[i - ``1``]);` `            `  `            ``// first element of arr.` `            ``else` `                ``min_Steps += Math.abs(arr[i]);` `        ``}` `        ``return` `min_Steps;` `    ``}`   `    ``// driver function` `    ``public` `static` `void` `main(String argc[])` `    ``{` `        ``int``[] arr = ``new` `int``[] { ``1``, ``2``, ``2``, ``1` `};` `        ``int` `n = ``4``;` `        ``System.out.println(minSteps(arr, n));` `    ``}` `}`

## Python3

 `# Python 3 program to find minimum number` `# of steps to make the array equal to the` `# given array.`   `# function to calculate min_Steps` `def` `minSteps(arr, n):` `    ``min_Steps ``=` `0` `    ``for` `i ``in` `range``(n):` `        ``if` `(i > ``0``): ` `            ``min_Steps ``+``=` `abs``(arr[i] ``-` `                             ``arr[i ``-` `1``])` `        `  `        ``# first element of arr.` `        ``else``:` `            ``min_Steps ``+``=` `abs``(arr[i])` `    ``return` `min_Steps`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[ ``1``, ``2``, ``2``, ``1` `]` `    ``n ``=` `len``(arr)` `    ``print``(minSteps(arr, n))`   `# This code is contributed ` `# by PrinciRaj19992`

## C#

 `// C# program to find minimum number of steps` `// to make the array equal to the given array.` `using` `System;`   `public` `class` `GfG {` `    `  `    ``// function to calculate min_Steps` `    ``public` `static` `int` `minSteps(``int``[] arr, ``int` `n)` `    ``{` `        ``int` `min_Steps = 0;` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(i > 0)` `                ``min_Steps += Math.Abs(arr[i] - arr[i - 1]);`   `            ``// first element of arr.` `            ``else` `                ``min_Steps += Math.Abs(arr[i]);` `        ``}` `        ``return` `min_Steps;` `    ``}`   `    ``// driver function` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = ``new` `int``[] { 1, 2, 2, 1 };` `        ``int` `n = 4;` `        ``Console.WriteLine(minSteps(arr, n));` `    ``}` `}`   `// This code is contributed by vt_m`

## PHP

 ` 0) ` `            ``\$min_Steps` `+= ``abs``(``\$arr``[``\$i``] -` `                              ``\$arr``[``\$i` `- 1]);` `        `  `        ``// first element of arr.` `        ``else` `            ``\$min_Steps` `+= ``abs``(``\$arr``[``\$i``]);` `    ``}` `    ``return` `\$min_Steps``;` `}`   `// Driver Code` `\$arr` `= ``array``( 1, 2, 2, 1 );` `\$n` `= sizeof(``\$arr``) ;`   `echo` `minSteps(``\$arr``, ``\$n``),``"\n"``;`   `// This code is contributed by ajit` `?>`

## Javascript

 ``

Output

`3`

Time complexity: O(n), where N is the number of elements in the given array.
Auxiliary space: O(1) because it is using constant space

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