Minimum boxes required to carry all gifts

10 May, 2021
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++
Java
Python3
C#
PHP
Javascript
// CPP implementation of above approach#include <bits/stdc++.h>using namespace std; // Function to return number of boxesint 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 programint 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
// Java implementation of above approachimport java.util.*; class solution{ // Function to return number of boxesstatic 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 programpublic 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
# Python3 implementation of# above approach # Function to return number of boxesdef 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 codeif __name__=='__main__':    A = [3, 2, 2, 1]    K= 3    n = len(A)    print(numBoxex(A,n,K)) # This code is contributed by# Shrikant13
// C# implementation of above approachusing System; class GFG{     // Function to return number of boxesstatic 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 Codestatic 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
<?php//PHP  implementation of above approach// Function to return number of boxesfunction  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?>
<script> // Javascript implementation of above approach // Function to return number of boxesfunction numBoxes(A, n, K){         // Sort the boxes in ascending order    A.sort(function(a, b){return a - b});     // Try to fit smallest box with    // current heaviest box (from right    // side)    let i = 0, j = (n - 1);    let ans = 0;         while (i <= j)    {        ans++;                 if (A[i] + A[j] <= K)            i++;        j--;    }    return ans;} // Driver codelet A = [ 3, 2, 2, 1 ];let K = 3;let n = A.length; document.write(numBoxes(A, n, K)); // This code is contributed by suresh07 </script>
Output: 
3
Time Complexity: O(N*log(N)), where N is the length of array. 
shrikanth13
SURENDRA_GANGWAR
jit_t
suresh07
Arrays
Greedy
Sorting
Arrays
Greedy
Sorting
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