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: 4
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: 8
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 getint 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 the answer with // current maximum i.e., mid maxVol = max(maxVol, mid); } // else traverse in the left half else high = mid - 1; } return maxVol;}// Driver codeint 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 the 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 getdef 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 the answer with # current maximum i.e., mid maxVol = max(maxVol, mid) # else traverse in the left half else: high = mid - 1 return maxVol# Driver codeif __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 the 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 getfunction 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 the 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 the 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)), as we are using nested loops, the outer loop traverses log(maxVolume) times as in each traversal we do right shift by 1 bit which is equivalent to floor division of 2 therefore the effective time will be 1+1/2+1/4+…..+1/2^maxVolume which is equivalent to log(maxVolume), the inner loop traverses N times.
Auxiliary Space: O(1), as we are not using any extra space.
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