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Maximum volume of cube for every person when edge of N cubes are given

  • Last Updated : 09 Aug, 2021
Geek Week

Given an array of N integers which denotes the edges of N cubical structures respectively. Also given are M integers which denote the number of peoples. The task is to find the maximum amount of volume of a cube that can be given to every person. 
Note: Cubes can be cut off any shape from any of the N cubes. 
Examples:  

Input: a[] = {1, 1, 1, 2, 2}, m = 3 
Output:
All three people get a slice of volume 4 each 
Person 1 gets a slice of volume 4 from the last cube. 
Person 2 gets a slice of volume 4 from the last cube. 
Person 3 gets a slice of volume 4 from the second last cube. 
Input: a[] = {2, 2, 2, 2, 2}, m = 4 
Output:
 

Naive Approach: A naive approach is to first calculate the volume of all the cubes and then linearly check for every volume that it can be distributed among all M people or not and find the maximum volume among all such volumes.
Time Complexity: O(N2)

Efficient Approach: An efficient approach is to use binary search to find the answer. Since the edge lengths are given in the array, convert them to the volume of the respective cubes. Find the maximum volume among the volumes of all the cubes. Say, the maximum volume is maxVolume. Now, perform binary search on the range [0, maxVolume].  

  • Calculate the middle value of the range, say mid.
  • Now, calculate the total number of cubes that can be cut of all the cubes of volume mid.
  • If the total cubes that can be cut exceed the number of persons, then that amount of volume of cubes can be cut for every person, hence we check for a larger value in the range [mid+1, maxVolume].
  • If the total cubes do not exceed the number of persons, then we check for an answer in the range [low, mid-1].

Below is the implementation of the above approach: 
 



C++




// C++ program to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to get the maximum volume that
// every person can get
int getMaximumVloume(int a[], int n, int m)
{
    int maxVolume = 0;
 
    // Convert the length to respective volumes
    // and find the maximum volumes
    for (int i = 0; i < n; i++) {
        a[i] = a[i] * a[i] * a[i];
 
        maxVolume = max(a[i], maxVolume);
    }
 
    // Apply binary search with initial
    // low as 0 and high as the maximum most
    int low = 0, high = maxVolume;
 
    // Initial answer is 0 slices
    int maxVol = 0;
 
    // Apply binary search
    while (low <= high) {
 
        // Get the mid element
        int mid = (low + high) >> 1;
 
        // Count the slices of volume mid
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            cnt += a[i] / mid;
        }
 
        // If the slices of volume
        // exceeds the number of persons
        // then every person can get volume mid
        if (cnt >= m) {
 
            // Then check for larger in the right half
            low = mid + 1;
 
            // Replace tbe answer with
            // current maximum i.e., mid
            maxVol = max(maxVol, mid);
        }
 
        // else traverse in the left half
        else
            high = mid - 1;
    }
 
    return maxVol;
}
 
// Driver code
int main()
{
    int a[] = { 1, 1, 1, 2, 2 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = 3;
 
    cout << getMaximumVloume(a, n, m);
 
    return 0;
}

Java




// Java program to implement the above approach
class GFG
{
 
    // Function to get the maximum volume that
    // every person can get
    static int getMaximumVloume(int a[], int n, int m)
    {
        int maxVolume = 0;
 
        // Convert the length to respective volumes
        // and find the maximum volumes
        for (int i = 0; i < n; i++)
        {
            a[i] = a[i] * a[i] * a[i];
 
            maxVolume = Math.max(a[i], maxVolume);
        }
 
        // Apply binary search with initial
        // low as 0 and high as the maximum most
        int low = 0, high = maxVolume;
 
        // Initial answer is 0 slices
        int maxVol = 0;
 
        // Apply binary search
        while (low <= high)
        {
 
            // Get the mid element
            int mid = (low + high) >> 1;
 
            // Count the slices of volume mid
            int cnt = 0;
            for (int i = 0; i < n; i++)
            {
                cnt += a[i] / mid;
            }
 
            // If the slices of volume
            // exceeds the number of persons
            // then every person can get volume mid
            if (cnt >= m)
            {
 
                // Then check for larger in the right half
                low = mid + 1;
 
                // Replace tbe answer with
                // current maximum i.e., mid
                maxVol = Math.max(maxVol, mid);
            }
             
            // else traverse in the left half
            else
            {
                high = mid - 1;
            }
        }
 
        return maxVol;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = {1, 1, 1, 2, 2};
        int n = a.length;
        int m = 3;
 
        System.out.println(getMaximumVloume(a, n, m));
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 program to implement
# the above approach
 
# Function to get the maximum volume
# that every person can get
def getMaximumVloume(a, n, m):
    maxVolume = 0
 
    # Convert the length to respective
    # volumes and find the maximum volumes
    for i in range(n):
        a[i] = a[i] * a[i] * a[i]
 
        maxVolume = max(a[i], maxVolume)
 
    # Apply binary search with initial
    # low as 0 and high as the maximum most
    low = 0
    high = maxVolume
 
    # Initial answer is 0 slices
    maxVol = 0
 
    # Apply binary search
    while (low <= high):
         
        # Get the mid element
        mid = (low + high) >> 1
 
        # Count the slices of volume mid
        cnt = 0
        for i in range(n):
            cnt += int(a[i] / mid)
 
        # If the slices of volume
        # exceeds the number of persons
        # then every person can get volume mid
        if (cnt >= m):
             
            # Then check for larger in the right half
            low = mid + 1
 
            # Replace tbe answer with
            # current maximum i.e., mid
            maxVol = max(maxVol, mid)
 
        # else traverse in the left half
        else:
            high = mid - 1
 
    return maxVol
 
# Driver code
if __name__ == '__main__':
    a = [1, 1, 1, 2, 2]
    n = len(a)
    m = 3
 
    print(getMaximumVloume(a, n, m))
 
# This code is contributed
# by Surendra_Gangwar

C#




// C# program to implement the above approach
using System;
 
class GFG
{
 
    // Function to get the maximum volume that
    // every person can get
    static int getMaximumVloume(int []a, int n, int m)
    {
        int maxVolume = 0;
 
        // Convert the length to respective volumes
        // and find the maximum volumes
        for (int i = 0; i < n; i++)
        {
            a[i] = a[i] * a[i] * a[i];
 
            maxVolume = Math.Max(a[i], maxVolume);
        }
 
        // Apply binary search with initial
        // low as 0 and high as the maximum most
        int low = 0, high = maxVolume;
 
        // Initial answer is 0 slices
        int maxVol = 0;
 
        // Apply binary search
        while (low <= high)
        {
 
            // Get the mid element
            int mid = (low + high) >> 1;
 
            // Count the slices of volume mid
            int cnt = 0;
            for (int i = 0; i < n; i++)
            {
                cnt += a[i] / mid;
            }
 
            // If the slices of volume
            // exceeds the number of persons
            // then every person can get volume mid
            if (cnt >= m)
            {
 
                // Then check for larger in the right half
                low = mid + 1;
 
                // Replace tbe answer with
                // current maximum i.e., mid
                maxVol = Math.Max(maxVol, mid);
            }
             
            // else traverse in the left half
            else
            {
                high = mid - 1;
            }
        }
 
        return maxVol;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []a = {1, 1, 1, 2, 2};
        int n = a.Length;
        int m = 3;
 
        Console.WriteLine(getMaximumVloume(a, n, m));
    }
}
 
/* This code contributed by PrinciRaj1992 */

PHP




<?php
// PHP program to implement the above approach
 
// Function to get the maximum volume that
// every person can get
function getMaximumVloume($a, $n, $m)
{
    $maxVolume = 0;
 
    // Convert the length to respective volumes
    // and find the maximum volumes
    for ($i = 0; $i < $n; $i++)
    {
        $a[$i] = $a[$i] * $a[$i] * $a[$i];
 
        $maxVolume = max($a[$i], $maxVolume);
    }
 
    // Apply binary search with initial
    // low as 0 and high as the maximum most
    $low = 0; $high = $maxVolume;
 
    // Initial answer is 0 slices
    $maxVol = 0;
 
    // Apply binary search
    while ($low <= $high)
    {
 
        // Get the mid element
        $mid = ($low + $high) >> 1;
 
        // Count the slices of volume mid
        $cnt = 0;
        for ($i = 0; $i < $n; $i++)
        {
            $cnt += (int)($a[$i] / $mid);
        }
 
        // If the slices of volume
        // exceeds the number of persons
        // then every person can get volume mid
        if ($cnt >= $m)
        {
 
            // Then check for larger in the right half
            $low = $mid + 1;
 
            // Replace tbe answer with
            // current maximum i.e., mid
            $maxVol = max($maxVol, $mid);
        }
 
        // else traverse in the left half
        else
            $high = $mid - 1;
    }
 
    return $maxVol;
}
 
// Driver code
$a = array(1, 1, 1, 2, 2);
$n = sizeof($a);
$m = 3;
 
echo getMaximumVloume($a, $n, $m);
 
// This code is contributed by Akanksha Rai
?>

Javascript




<script>
// javascript program to implement the above approach    
// Function to get the maximum volume that
    // every person can get
    function getMaximumVloume(a , n , m) {
        var maxVolume = 0;
 
        // Convert the length to respective volumes
        // and find the maximum volumes
        for (i = 0; i < n; i++) {
            a[i] = a[i] * a[i] * a[i];
 
            maxVolume = Math.max(a[i], maxVolume);
        }
 
        // Apply binary search with initial
        // low as 0 and high as the maximum most
        var low = 0, high = maxVolume;
 
        // Initial answer is 0 slices
        var maxVol = 0;
 
        // Apply binary search
        while (low <= high) {
 
            // Get the mid element
            var mid = (low + high) >> 1;
 
            // Count the slices of volume mid
            var cnt = 0;
            for (i = 0; i < n; i++) {
                cnt += parseInt(a[i] / mid);
            }
 
            // If the slices of volume
            // exceeds the number of persons
            // then every person can get volume mid
            if (cnt >= m) {
 
                // Then check for larger in the right half
                low = mid + 1;
 
                // Replace tbe answer with
                // current maximum i.e., mid
                maxVol = Math.max(maxVol, mid);
            }
 
            // else traverse in the left half
            else {
                high = mid - 1;
            }
        }
 
        return maxVol;
    }
 
    // Driver code
     
        var a = [ 1, 1, 1, 2, 2 ];
        var n = a.length;
        var m = 3;
 
        document.write(getMaximumVloume(a, n, m));
 
// This code is contributed by todaysgaurav
</script>
Output: 
4

 

Time Complexity: O(N * log (maxVolume))
 

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