# Minimum number of stacks possible using boxes of given capacities

Given N boxes with their capacities which denotes the total number of boxes that it can hold above it. You can stack up the boxes one over the other as long as the total number of boxes above each box is less than or equal to its capacity. Find the minimum number of stacks that can be made by using all the boxes.

Examples:

Input: arr[] = {0, 0, 1, 1, 2}
Output: 2
First stack (top to bottom): 0 1 2
Second stack (top to bottom): 0 1

Input: arr[] = {1, 1, 4, 4}
Output: 1
All the boxes can be put on a single stack.

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

Approach: Let’s have a map in which map[X] denotes the number of boxes with capacity X available with us. Let’s build stacks one by one. Initially the size of the stack would be 0, and then we iterate through the map greedily choosing as many boxes of current capacity as we can.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count ` `// of minimum stacks ` `int` `countPiles(``int` `n, ``int` `a[]) ` `{ ` ` `  `    ``// Keep track of occurrence ` `    ``// of each capacity ` `    ``map<``int``, ``int``> occ; ` ` `  `    ``// Fill the occurrence map ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``occ[a[i]]++; ` ` `  `    ``// Number of piles is 0 initially ` `    ``int` `pile = 0; ` ` `  `    ``// Traverse occurrences in increasing ` `    ``// order of capacities. ` `    ``while` `(occ.size()) { ` ` `  `        ``// Adding a new pile ` `        ``pile++; ` `        ``int` `size = 0; ` `        ``unordered_set<``int``> toRemove; ` ` `  `        ``// Traverse all piles in increasing ` `        ``// order of capacities ` `        ``for` `(``auto` `tm` `: occ) { ` `            ``int` `mx = ``tm``.first; ` `            ``int` `ct = ``tm``.second; ` ` `  `            ``// Number of boxes of capacity mx ` `            ``// that can be added to current pile ` `            ``int` `use = min(ct, mx - size + 1); ` ` `  `            ``// Update the occurrence ` `            ``occ[mx] -= use; ` ` `  `            ``// Update the size of the pile ` `            ``size += use; ` `            ``if` `(occ[mx] == 0) ` `                ``toRemove.insert(mx); ` `        ``} ` ` `  `        ``// Remove capacities that are ` `        ``// no longer available ` `        ``for` `(``auto` `tm` `: toRemove) ` `            ``occ.erase(``tm``); ` `    ``} ` `    ``return` `pile; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 0, 0, 1, 1, 2 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``cout << countPiles(n, a); ` ` `  `    ``return` `0; ` `} `

Output:

```2
```

Time Complexity: O(NlogN)

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Improved By : mohit kumar 29