CSES Solutions Increasing Array (original) (raw)

Last Updated : 18 Jul, 2024

Given an array **arr[] of size N, you want to modify the array so that it is increasing, i.e., every element is at least as large as the previous element. On each move, you may increase the value of any element by one. Find the minimum number of moves required.

**Examples:

**Input: N = 5, arr[] = {3, 2, 5, 1, 7}
**Output: 5
**Explanation:

Increase arr[1] by 1, so arr[] becomes {3, 3, 5, 1, 7}.

Increase arr[3] by 4, so arr[] becomes {3, 3, 5, 5, 7}.

**Input: N = 3, arr[] = {1, 2, 3}
**Output: 0
**Explanation: No moves are needed as the array is already in increasing order.

**Algorithm: To solve the problem, follow the below idea:

The problem can be solved by traversing the array and comparing each element with its previous element. If the current element is smaller than the previous element so we need to increase the current element till it becomes equal to the previous element. This can be done by applying increment operations ****(previous element - current element)** number of times. Sum of all these operations will be the **minimum number of moves required.

**Step-by-step algorithm:

Initialize a variable ans = 0 to store the minimum number of moves.
Start a loop from the second element of the array.
For every element, if the previous element is greater than the current element, then add (previous element - current element) to ans and update the current element = previous element.
After iterating over all the elements, return ans.

Below is the implementation of the algorithm:

C++ `

#include <bits/stdc++.h> #define ll long long using namespace std;

// Function to find the minimum number of moves ll solve(ll* arr, ll N) { // variable to store the final answer ll ans = 0; // Iterate over all the elements and compare them with // the previous element for (ll i = 1; i < N; i++) { if (arr[i - 1] > arr[i]) { ans += (arr[i - 1] - arr[i]); arr[i] = arr[i - 1]; } } return ans; }

int main() { // Sample Input ll N = 5; ll arr[N] = { 3, 2, 5, 1, 7 };

cout << solve(arr, N) << "\n";
return 0;

}

Java

import java.util.Arrays;

public class Main {

// Function to find the minimum number of moves
static long solve(long[] arr, int N)
{
    // variable to store the final answer
    long ans = 0;
    // Iterate over all the elements and compare them
    // with the previous element
    for (int i = 1; i < N; i++) {
        if (arr[i - 1] > arr[i]) {
            ans += (arr[i - 1] - arr[i]);
            arr[i] = arr[i - 1];
        }
    }
    return ans;
}

public static void main(String[] args)
{
    // Sample Input
    int N = 5;
    long[] arr = { 3, 2, 5, 1, 7 };

    System.out.println(solve(arr, N));
}

}

// This code is contributed by rambabuguphka

Python

Function to find the minimum number of moves

def solve(arr, N): # variable to store the final answer ans = 0 # Iterate over all the elements and compare them with # the previous element for i in range(1, N): if arr[i - 1] > arr[i]: ans += (arr[i - 1] - arr[i]) arr[i] = arr[i - 1] return ans

def main(): # Sample Input N = 5 arr = [3, 2, 5, 1, 7]

print(solve(arr, N))

if name == "main": main()

C#

using System;

class GFG { // Function to find the minimum number of moves static long Solve(long[] arr, long N) { // Variable to store the final answer long ans = 0; // Iterate over all the elements and compare them with // previous element for (long i = 1; i < N; i++) { if (arr[i - 1] > arr[i]) { ans += (arr[i - 1] - arr[i]); arr[i] = arr[i - 1]; } } return ans; } public static void Main(string[] args) { // Sample Input long N = 5; long[] arr = { 3, 2, 5, 1, 7 }; // Call the Solve function and // print the result Console.WriteLine(Solve(arr, N)); } }

JavaScript

// Function to find the minimum number of moves function solve(arr, N) { // variable to store the final answer let ans = 0;

// Iterate over all the elements and compare them
// with the previous element
for (let i = 1; i < N; i++) {
    if (arr[i - 1] > arr[i]) {
        ans += (arr[i - 1] - arr[i]);
        arr[i] = arr[i - 1];
    }
}
return ans;

}

// Main function function main() { // Sample Input const N = 5; const arr = [3, 2, 5, 1, 7];

// Call the solve function and print the result
console.log(solve(arr, N));

}

// Call the main function to execute the program main();

**Time Complexity: O(N), where **N is the size of input array **arr[].
**Auxiliary Space: O(1)