# 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `?> `

Output :

```3
```

Time complexity = O(n).

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