Minimum boxes required to carry all gifts

Given an array containing weights of gifts, and an integer K representing maximum weight a box can contain (All boxes are uniform). Each box carries at most 2 gifts at the same time, provided the sum of the weight of those gifts is at most limit of box. The task is to find the minimum number of boxes required to carry all gifts.
Note: It is guaranteed each gift can be carried by a box.

Examples:

Input: A = [3, 2, 2, 1], K = 3
Output: 3
Explanation: 3 boxes with weights (1, 2), (2) and (3)

Input: A = [3, 5, 3, 4], K = 5
Output: 4
Explanation: 4 boxes with weights (3), (3), (4), (5)

Approach: If the heaviest gift can share a box with the lightest gift, then do so. Otherwise, the heaviest gift can’t pair with anyone, so it get an individual box.

The reason this works is because if the lightest gift can pair with anyone, it might as well pair with the heaviest gift.

Let A[i] be the currently lightest gift, and A[j] to the heaviest.

Then, if the heaviest gift can share a box with the lightest gift (if A[j] + A[i] <= K) then do so otherwise, the heaviest gift get an individual box.

Below is the implementation of above approach:

C++

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// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return number of boxes
int numBoxes(int A[], int n, int K)
{
    // Sort the boxes in ascending order
    sort(A, A + n);
  
    // Try to fit smallest box with
    // current heaviest box (from right
    // side)
    int i = 0, j = n - 1;
    int ans = 0;
    while (i <= j) {
        ans++;
        if (A[i] + A[j] <= K)
            i++;
        j--;
    }
  
    return ans;
}
  
// Driver program
int main()
{
    int A[] = { 3, 2, 2, 1 }, K = 3;
    int n = sizeof(A) / sizeof(A[0]);
    cout << numBoxes(A, n, K);
    return 0;
}
  
// This code is written by Sanjit_Prasad

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Java

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// Java implementation of above approach
import java.util.*;
  
class solution
{
  
// Function to return number of boxes
static int numBoxes(int A[], int n, int K)
{
    // Sort the boxes in ascending order
    Arrays.sort(A);
  
    // Try to fit smallest box with
    // current heaviest box (from right
    // side)
    int i = 0, j = n - 1;
    int ans = 0;
    while (i <= j) {
        ans++;
        if (A[i] + A[j] <= K)
            i++;
        j--;
    }
  
    return ans;
}
  
// Driver program
public static void main(String args[])
{
    int A[] = { 3, 2, 2, 1 }, K = 3;
    int n = A.length;
    System.out.println(numBoxes(A, n, K));
  
}
}
  
//THis code is contributed by
// Surendra_Gangwar

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Python3

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# Python3 implementation of 
# above approach 
  
# Function to return number of boxes 
def numBoxex(A,n,K):
  
    # Sort the boxes in ascending order 
    A.sort()
    # Try to fit smallest box with current 
    # heaviest box (from right side) 
    i =0
    j = n-1
    ans=0
    while i<=j:
        ans +=1
        if A[i]+A[j] <=K:
            i+=1
        j-=1
  
    return ans
  
# Driver code
if __name__=='__main__':
    A = [3, 2, 2, 1]
    K= 3
    n = len(A)
    print(numBoxex(A,n,K))
  
# This code is contributed by 
# Shrikant13

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C#

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// C# implementation of above approach
using System;
  
class GFG
{
      
// Function to return number of boxes
static int numBoxes(int []A, int n, int K)
{
    // Sort the boxes in ascending order
    Array.Sort(A);
  
    // Try to fit smallest box with
    // current heaviest box (from right
    // side)
    int i = 0, j = (n - 1);
    int ans = 0;
    while (i <= j) 
    {
        ans++;
        if (A[i] + A[j] <= K)
            i++;
        j--;
    }
  
    return ans;
}
  
// Driver Code
static public void Main ()
{
    int []A = { 3, 2, 2, 1 };
    int K = 3;
    int n = A.Length;
    Console.WriteLine(numBoxes(A, n, K));
}
}
  
// This code is contributed by ajit

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PHP

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<?php
//PHP  implementation of above approach 
// Function to return number of boxes 
function  numBoxes($A, $n, $K
    // Sort the boxes in ascending order 
    sort($A); 
  
    // Try to fit smallest box with 
    // current heaviest box (from right 
    // side) 
    $i = 0;
    $j = $n - 1; 
    $ans = 0; 
    while ($i <= $j) { 
        $ans++; 
        if ($A[$i] + $A[$j] <= $K
            $i++; 
        $j--; 
    
  
    return $ans
  
// Driver program 
    $A = array (3, 2, 2, 1 );
    $K = 3; 
    $n = sizeof($A) / sizeof($A[0]); 
    echo  numBoxes($A, $n, $K); 
      
  
// This code is written by ajit 
?>

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Output:

3

Time Complexity: O(N*log(N)), where N is the length of array.



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