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# Minimum number of steps required to obtain the given Array by the given operations

• Last Updated : 23 Apr, 2021

Given an array arr[] of N positive integers, the task is to find the minimum number of operations required of the following types to obtain the array arr[] from an array of zeroes only.

• Select any index i and increment all the elements at the indices [i, N – 1] by 1.
• Select any index i and decrease all the elements at the indices [i, N – 1] by 1.

Examples:

Input: arr[]={1,1,2,2,1}
Output:
Explanation:
Initially arr[] = {0,0,0,0,0}
Step 1: arr[]={1,1,1,1,1}(Operation 1 on index 0)
Step 2: arr[]={1,1,2,2,2}(Operation 1 on index 2)
Step 3: arr[]={1,1,2,2,1}(Operation 2 on index 4)
Input: arr[]={1,2,3,4}
Output:

Naive Approach: The simplest approach is to convert each element of the array result array, brr[] to arr[] by performing one of the above operations over the indices [i, N – 1] and increase the count of each operation performed.
Time Complexity: O(N2
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Greedy Approach. Follow the steps below to solve the problem:

• For the 0th index, convert the number 0 to arr. Therefore, the number of steps required will always be a. Hence, add arr to the answer.
• For all the other indices, the common greedy observation is to use the increase or the decrease operation by abs(a[i]-a[i-1]) times.
• The intuition behind this approach is, if the number is less than a[i-1], then increase everything from ((i-1).. n-1) by a[i], and then decrease (a[i-1] – a[i]) to get a[i].
• If a[i] > a[i-1], the approach is to use the increase operation from ((i-1)..n-1) by a[i-1], and for the remaining value, we increase it by (a[i]-a[i-1]) operations from (i..(n-1)).
• Therefore, traverse the array and for every element after the first, add the absolute difference of consecutive pairs to the answer.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to calculate the minimum``// steps to obtain the desired array``int` `min_operation(``int` `a[], ``int` `n)``{``    ``// Initialize variable``    ``int` `ans = 0;` `    ``// Iterate over the array arr[]``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Check if i > 0``        ``if` `(i > 0)` `            ``// Update the answer``            ``ans += ``abs``(a[i] - a[i - 1]);` `        ``else``            ``ans += ``abs``(a[i]);``    ``}` `    ``// Return the result``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << min_operation(arr, n);` `    ``return` `0;``}`

## Java

 `// Java Program to implement``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to calculate the minimum``// steps to obtain the desired array``static` `int` `min_operation(``int` `a[], ``int` `n)``{``    ``// Initialize variable``    ``int` `ans = ``0``;` `    ``// Iterate over the array arr[]``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Check if i > 0``        ``if` `(i > ``0``)` `            ``// Update the answer``            ``ans += Math.abs(a[i] - a[i - ``1``]);` `        ``else``            ``ans += Math.abs(a[i]);``    ``}` `    ``// Return the result``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4` `};``    ``int` `n = arr.length;` `    ``System.out.print(min_operation(arr, n));``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to calculate the minimum``# steps to obtain the desired array``def` `min_operation(a, n):` `    ``# Initialize variable``    ``ans ``=` `0` `    ``# Iterate over the array arr[]``    ``for` `i ``in` `range``(n):` `        ``# Check if i > 0``        ``if` `(i > ``0``):` `            ``# Update the answer``            ``ans ``+``=` `abs``(a[i] ``-` `a[i ``-` `1``])``        ``else``:``            ``ans ``+``=` `abs``(a[i])` `    ``# Return the result``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4` `]``    ``n ``=` `len``(arr)` `    ``print``(min_operation(arr, n))` `# This code is contributed by chitranayal`

## C#

 `// C# Program to implement``// the above approach``using` `System;``class` `GFG{` `// Function to calculate the minimum``// steps to obtain the desired array``static` `int` `min_operation(``int` `[]a, ``int` `n)``{``    ``// Initialize variable``    ``int` `ans = 0;` `    ``// Iterate over the array []arr``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// Check if i > 0``        ``if` `(i > 0)` `            ``// Update the answer``            ``ans += Math.Abs(a[i] - a[i - 1]);` `        ``else``            ``ans += Math.Abs(a[i]);``    ``}` `    ``// Return the result``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 2, 3, 4 };``    ``int` `n = arr.Length;` `    ``Console.Write(min_operation(arr, n));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`4`

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

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