Maximum product of an Array after subtracting 1 from any element N times

Last Updated : 06 Dec, 2021

Given an array arr[] of positive integers of size M and an integer N, the task is to maximize the product of array after subtracting 1 from any element N number of times

Examples:

Input: M = 5, arr[] = {5, 1, 7, 8, 3}, N = 2
Output: 630
Explanation: After subtracting 1 from arr[3] 2 times, array becomes {5, 1, 7, 6, 3} with product = 630
Input: M = 2, arr[] = {2, 2}, N = 4
Output: 0
Explanation: After subtracting 2 from arr[0] and arr[1] 2 times each, array becomes {0, 0} with product = 0

Approach: The task can be solved with the help of max-heap Follow the below steps to solve the problem:

Insert all the elements inside a max-heap
Pop the top element from the max- heap, and decrement 1 from it, also decrement N
Insert the popped element back to max-heap
Continue this process till N > 0
The maximum product will be the product of all the elements inside the max-heap

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 the maximum
// product of the array
int getMax(int m, int arr[], int n)
{

    // Max-heap
    priority_queue<int> q;

    // Store all the elements inside max-heap
    for (int i = 0; i < m; i++)
        q.push(arr[i]);

    // n operations
    while (n--) {
        int x = q.top();
        q.pop();

        // Decrement x
        --x;

        // Push back x inside the heap
        q.push(x);
    }

    // Store the max product possible
    int ans = 1;
    while (!q.empty()) {
        ans *= q.top();
        q.pop();
    }

    return ans;
}

// Driver Code
int main()
{
    int M = 5;
    int arr[5] = { 5, 1, 7, 8, 3 };
    int N = 2;

    cout << getMax(M, arr, N);
}

Java

// Java program for the above approach
import java.util.*;

class GFG
{
  
    // Function to find the maximum
    // product of the array
    static Integer getMax(int m, Integer arr[], int n)
    {

        // Max-heap
        PriorityQueue<Integer> q
            = new PriorityQueue<Integer>(
                Collections.reverseOrder());

        // Store all the elements inside max-heap
        for (int i = 0; i < m; i++)
            q.add(arr[i]);

        // n operations
        while (n-- != 0) {
            Integer x = q.poll();

            // Decrement x
            --x;

            // Push back x inside the heap
            q.add(x);
        }

        // Store the max product possible
        Integer ans = 1;
        while (q.size() != 0) {
            ans *= q.poll();
        }

        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int M = 5;
        Integer arr[] = { 5, 1, 7, 8, 3 };
        int N = 2;

        System.out.println(getMax(M, arr, N));
    }
}

// This code is contributed by Potta Lokesh

Python3

# python program for the above approach
from queue import PriorityQueue

# Function to find the maximum
# product of the array
def getMax(m, arr, n):

        # Max-heap
    q = PriorityQueue()

    # Store all the elements inside max-heap
    for i in range(0, m):
        q.put(-arr[i])

        # n operations
    while (n):
        n -= 1
        x = -q.get()

        # Decrement x
        x -= 1

        # Push back x inside the heap
        q.put(-x)

        # Store the max product possible
    ans = 1
    while (not q.empty()):
        ans *= -q.get()

    return ans

# Driver Code
if __name__ == "__main__":

    M = 5
    arr = [5, 1, 7, 8, 3]
    N = 2

    print(getMax(M, arr, N))

    # This code is contributed by rakeshsahni

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG
{
  
    // Function to find the maximum
    // product of the array
    static int getMax(int m, int []arr, int n)
    {

        // Max-heap
        List<int> q
            = new List<int>();

        // Store all the elements inside max-heap
        for (int i = 0; i < m; i++){
            q.Add(arr[i]);
        }
        q.Sort();
        q.Reverse();
        // n operations
        while (n-- != 0) {
            int x = q[0];
            q.RemoveAt(0);
            // Decrement x
            --x;

            // Push back x inside the heap
            q.Add(x);
            q.Sort();
            q.Reverse();
        }

        // Store the max product possible
        int ans = 1;
        while (q.Count != 0) {
            ans *= q[0];
            q.RemoveAt(0);
        }

        return ans;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int M = 5;
        int []arr = { 5, 1, 7, 8, 3 };
        int N = 2;

        Console.WriteLine(getMax(M, arr, N));
    }
}

// This code is contributed by shikhasingrajput

JavaScript

<script>
// Javascript program for the above approach

// Function to find the maximum
// product of the array
function getMax(m, arr, n)
{

    // Max-heap
    let q = [];

    // Store all the elements inside max-heap
    for (let i = 0; i < m; i++) {
        q.push(arr[i]);
    }
    q.sort();
    q.reverse();
    
    // n operations
    while (n-- != 0) {
        let x = q[0];
        q.splice(0, 1);
        
        // Decrement x
        --x;

        // Push back x inside the heap
        q.push(x);
        q.sort();
        q.reverse();
    }

    // Store the max product possible
    let ans = 1;
    while (q.length != 0) {
        ans *= q[0];
        q.splice(0, 1);
    }
    return ans;
}

// Driver Code
let M = 5;
let arr = [5, 1, 7, 8, 3];
let N = 2;

document.write(getMax(M, arr, N));

// This code is contributed by gfgking
</script>

Output

Time Complexity: O(nlogm)
Auxiliary Space: O(m)

Maximum product of an Array after subtracting 1 from any element N times

saurabh15899

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