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Minimum decrements to make an Array at most 0 such that all array elements are cyclically decremented after a number is reduced to 0

  • Difficulty Level : Easy
  • Last Updated : 18 Nov, 2021

Given a circular array arr[] of N integers and an array cost[], the task is to calculate the minimum number of operations required to make all elements of the array equal to 0, where at each operation decrement the value of an index i by 1. If the value of an index becomes 0, decrement the value of arr[i+1] by cost[i], decrement the value of arr[i + 2] by cost[i + 1] and so on.

Note: If an element becomes less than 0, it is considered to be 0.

Example:

Input: arr[] = {7, 2, 5}, cost[] = {8, 9, 3}
Output: 6
Explanation: Decrements can be made in the following way:

  • Decrement the value of arr[1] twice. Hence, the value of arr[2] will be decremented by cost[1] and the value of arr[0] will be decremented by cost[2]. Therefore, the final array will be arr[] = {4, 0, 0}.
  • Now, decrement the value of arr[0], 4 times to make it 0. Therefore, the array becomes arr[] = {0, 0, 0}.

Hence, the number of required operations to make all the elements of the array equal to zero is 6 which is the minimum possible.

Input: arr[] = {6, 7, 7, 10, 8, 2}, cost[] = {5, 10, 1, 4, 7, 7}
Output: 16

 

Approach: The given problem can be solved using a Greedy Approach by the following observations:

  • The last remaining non-zero element of the array arr[], suppose x will require x decrement operations.
  • Suppose after the decrement operations, the value of arr[i] becomes 0, then for arr[i], the required number of decrements are arr[i], for arr[i+1], the required number of decrements will be arr[i+1] – max(0, arr[i+1] – cost[i]) and so on.

Below are the steps to follow:

  • Traverse the given array arr[] in the range [0, N) using a variable i.
  • The number of operations required the make the ith index of the array equal to zero are arr[i] – min(arr[i], cost[i-1]). Therefore, maintain the sum of this value over all the indices in a variable ans.
  • Increment the value of ans by the minimum value of arr[i] or cost[i] over all the indices in the range [0, N).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum decrements
// required to make all array elements 0
int minDecrements(int arr[], int powr[], int N)
{
    // Variable to store the answer
    int ans = 0;
    int mn = INT_MAX;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int idx = (i + 1) % N;
        int val = min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 7, 7, 10, 8, 2 };
    int powr[] = { 5, 10, 1, 4, 7, 7 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << minDecrements(arr, powr, N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
public class GFG
{
// Function to find minimum decrements
// required to make all array elements 0
static int minDecrements(int []arr, int []powr, int N)
{
    // Variable to store the answer
    int ans = 0;
    int mn = Integer.MAX_VALUE;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int idx = (i + 1) % N;
        int val = Math.min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = Math.min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int []arr = { 6, 7, 7, 10, 8, 2 };
    int []powr = { 5, 10, 1, 4, 7, 7 };
    int N = arr.length;
 
    System.out.println(minDecrements(arr, powr, N));
}
}
// This code is contributed by Samim Hossain Mondal.

Python3




# Python program for the above approach
 
# Function to find minimum decrements
# required to make all array elements 0
def minDecrements(arr, powr, N):
 
    # Variable to store the answer
    ans = 0
    mn = 99999999
 
    # Traverse the array
    for i in range(N):
        idx = (i + 1) % N
        val = min(arr[idx], powr[i])
 
        ans += arr[idx] - val
 
       # Store the minimum one
        mn = min(mn, val)
    ans += mn
 
    # Return the ans
    return ans
 
# Driver Code
if __name__ == "__main__":
    arr = [6, 7, 7, 10, 8, 2]
    powr = [5, 10, 1, 4, 7, 7]
    N = len(arr)
    print(minDecrements(arr, powr, N))
 
# This code is contributed by Potta Lokesh

C#




// C# program for the above approach
using System;
using System.Collections;
 
class GFG
{
// Function to find minimum decrements
// required to make all array elements 0
static int minDecrements(int []arr, int []powr, int N)
{
    // Variable to store the answer
    int ans = 0;
    int mn = Int32.MaxValue;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        int idx = (i + 1) % N;
        int val = Math.Min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = Math.Min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 6, 7, 7, 10, 8, 2 };
    int []powr = { 5, 10, 1, 4, 7, 7 };
    int N = arr.Length;
 
    Console.Write(minDecrements(arr, powr, N));
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
// Javascript program for the above approach
 
// Function to find minimum decrements
// required to make all array elements 0
function minDecrements(arr, powr, N)
{
    // Variable to store the answer
    let ans = 0;
    let mn = Number.MAX_SAFE_INTEGER
 
    // Traverse the array
    for (let i = 0; i < N; i++) {
        let idx = (i + 1) % N;
        let val = Math.min(arr[idx], powr[i]);
 
        ans += arr[idx] - val;
 
        // Store the minimum one
        mn = Math.min(mn, val);
    }
    ans += mn;
 
    // Return the ans
    return ans;
}
 
// Driver Code
 
let arr = [ 6, 7, 7, 10, 8, 2 ];
let powr = [ 5, 10, 1, 4, 7, 7 ];
let N = arr.length;
 
document.write(minDecrements(arr, powr, N));
 
// This code is contributed by Samim Hossain Mondal.
</script>
Output
16

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


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