Given an array of size N and X. Find minimum moves required to make the array in increasing order. In each move one can add X to any element in the array.
Examples:
Input : a = { 1, 3, 3, 2 }, X = 2
Output : 3
Explanation : Modified array is { 1, 3, 5, 6 }
Input : a = { 3, 5, 6 }, X = 5
Output : 0
Observation:
Let’s take two numbers a and b. a >= b and convert this into a < b by adding some number X.
so, a < b + k*X
( a – b ) / x < k
so, the minimum possible value of k is ( a – b ) / x + 1.
Approach :
Iterate over the given array and take two numbers when a[i] >= a[i-1] and apply above observation.
Below is the implementation of the above approach:
C++
// C++ program to find minimum moves required// to make the array in increasing order#include <bits/stdc++.h>using namespace std;// function to find minimum moves required// to make the array in increasing orderint MinimumMoves(int a[], int n, int x){ // to store answer int ans = 0; // iterate over an array for (int i = 1; i < n; i++) { // non- increasing order if (a[i] <= a[i - 1]) { int p = (a[i - 1] - a[i]) / x + 1; // add moves to answer ans += p; // increase the element a[i] += p * x; } } // return required answer return ans;}// Driver codeint main(){ int arr[] = { 1, 3, 3, 2 }; int x = 2; int n = sizeof(arr) / sizeof(arr[0]); cout << MinimumMoves(arr, n, x); return 0;} |
C
// C program to find minimum moves required// to make the array in increasing order#include <stdio.h>// function to find minimum moves required// to make the array in increasing orderint MinimumMoves(int a[], int n, int x){ // to store answer int ans = 0; // iterate over an array for (int i = 1; i < n; i++) { // non- increasing order if (a[i] <= a[i - 1]) { int p = (a[i - 1] - a[i]) / x + 1; // add moves to answer ans += p; // increase the element a[i] += p * x; } } // return required answer return ans;}// Driver codeint main(){ int arr[] = { 1, 3, 3, 2 }; int x = 2; int n = sizeof(arr) / sizeof(arr[0]); printf("%d",MinimumMoves(arr, n, x)); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java program to find minimum moves required// to make the array in increasing orderimport java.util.*;import java.lang.*;import java.io.*;class GFG{// function to find minimum moves required// to make the array in increasing orderstatic int MinimumMoves(int a[], int n, int x){ // to store answer int ans = 0; // iterate over an array for (int i = 1; i < n; i++) { // non- increasing order if (a[i] <= a[i - 1]) { int p = (a[i - 1] - a[i]) / x + 1; // add moves to answer ans += p; // increase the element a[i] += p * x; } } // return required answer return ans;} // Driver codepublic static void main(String args[]){ int arr[] = { 1, 3, 3, 2 }; int x = 2; int n = arr.length; System.out.println(MinimumMoves(arr, n, x)); }} |
Python3
# Python3 program to find minimum # moves required to make the array # in increasing order # function to find minimum moves required # to make the array in increasing order def MinimumMoves(a, n, x) : # to store answer ans = 0 # iterate over an array for i in range(1, n) : # non- increasing order if a[i] <= a[i - 1] : p = (a[i - 1] - a[i]) // x + 1 # add moves to answer ans += p # increase the element a[i] += p * x # return required answer return ans # Driver code if __name__ == "__main__" : arr = [1, 3, 3, 2] x = 2 n = len(arr) print(MinimumMoves(arr, n, x))# This code is contributed by ANKITRAI1 |
C#
// C# program to find minimum moves required // to make the array in increasing order using System;class GFG { // function to find minimum moves required // to make the array in increasing order static int MinimumMoves(int[] a, int n, int x) { // to store answer int ans = 0; // iterate over an array for (int i = 1; i < n; i++) { // non- increasing order if (a[i] <= a[i - 1]) { int p = (a[i - 1] - a[i]) / x + 1; // add moves to answer ans += p; // increase the element a[i] += p * x; } } // return required answer return ans; } // Driver code public static void Main() { int[] arr = {1, 3, 3, 2}; int x = 2; int n = arr.Length; Console.Write(MinimumMoves(arr, n, x)); } } // This code is contributed by ChitraNayal |
PHP
<?php// PHP program to find minimum // moves required to make the // array in increasing order// function to find minimum // moves required to make the // array in increasing orderfunction MinimumMoves(&$a, $n, $x){ // to store answer $ans = 0; // iterate over an array for ($i = 1; $i < $n; $i++) { // non- increasing order if ($a[$i] <= $a[$i - 1]) { $p = ($a[$i - 1] - $a[$i]) / $x + 1; // add moves to answer $ans += $p; // increase the element $a[$i] += $p * $x; } } // return required answer return $ans;}// Driver code$arr = array(1, 3, 3, 2 );$x = 2;$n = sizeof($arr);echo ((int)MinimumMoves($arr, $n, $x));// This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script>// Javascript program to find minimum // moves required to make the array in// increasing order // Function to find minimum moves required// to make the array in increasing orderfunction MinimumMoves(a, n, x){ // To store answer var ans = 0; // Tterate over an array for(i = 1; i < n; i++) { // Non- increasing order if (a[i] <= a[i - 1]) { var p = parseInt((a[i - 1] - a[i]) / x + 1); // Add moves to answer ans += p; // Increase the element a[i] += p * x; } } // Return required answer return ans;}// Driver codevar arr = [ 1, 3, 3, 2 ];var x = 2;var n = arr.length;document.write(MinimumMoves(arr, n, x));// This code is contributed by aashish1995 </script> |
Output
3
Complexity Analysis:
- Time Complexity: O(n), to iterate over the array where n is the size of the array
- Auxiliary Space: O(1), as no extra space is required
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