# 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)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// CPP implementation of above approach ` `#include ` `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); ` `    ``cout << numBoxes(A, n, K); ` `    ``return` `0; ` `} ` ` `  `// This code is written by Sanjit_Prasad `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

## PHP

 ` `

Output:

`3`

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

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