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[1]| operations. Once this is done, arr[2], arr[3], arr[4], … arr[n] = brr[1].
- To make Second element equal we have to make |brr[2] – brr[1]| operations. Once this is done, arr[3], arr[4], arr[5], … arr[n] = brr[2].
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[1]| + |b[2] – b[1]| + |b[3] – b[2]| + … + |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 <bits/stdc++.h>using namespace std;// function to calculate min_Stepsint 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 functionint main(){ int arr[] = { 1, 2, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); 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_Stepsdef 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 Codeif __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
<?php// PHP program to find minimum // number of steps to make the // array equal to the given array.// function to calculate min_Stepsfunction minSteps($arr, $n){ $min_Steps = 0; for ($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 $min_Steps;}// Driver Code$arr = array( 1, 2, 2, 1 );$n = sizeof($arr) ;echo minSteps($arr, $n),"\n";// This code is contributed by ajit?> |
Javascript
<script> // Javascript program to find minimum number of steps // to make the array equal to the given array. // function to calculate min_Steps function minSteps(arr, n) { let min_Steps = 0; for (let 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; } let arr = [ 1, 2, 2, 1 ]; let n = arr.length; document.write(minSteps(arr, n)); // This code is contributed by divyeshrabadiya07.</script> |
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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