# 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.

**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++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `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[0]); ` ` ` `cout << countPiles(n, a); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 implementation of the approach

# Function to return the count

# of minimum stacks

def countPiles(n, a):

# Keep track of occurrence

# of each capacity

occ = dict()

# Fill the occurrence map

for i in a:

if i in occ.keys():

occ[i] += 1

else:

occ[i] = 1

# Number of piles is 0 initially

pile = 0

# Traverse occurrences in increasing

# order of capacities.

while (len(occ) > 0):

# Adding a new pile

pile += 1

size = 0

toRemove = dict()

# Traverse all piles in increasing

# order of capacities

for tm in occ:

mx = tm

ct = occ[tm]

# Number of boxes of capacity mx

# that can be added to current pile

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[mx] = 1

# Remove capacities that are

# no longer available

for tm in toRemove:

del occ[tm]

return pile

# Driver code

a = [0, 0, 1, 1, 2]

n = len(a)

print(countPiles(n, a))

# This code is contributed

# by Mohit Kumar

**Output:**

2

**Time Complexity:** O(NlogN)

## Recommended Posts:

- Minimum boxes required to carry all gifts
- Minimum time required to transport all the boxes from source to the destination under the given constraints
- Find the number of boxes to be removed
- Number of visible boxes after putting one inside another
- Color N boxes using M colors such that K boxes have different color from the box on its left
- Maximum and minimum of an array using minimum number of comparisons
- Implement two stacks in an array
- Find a number which give minimum sum when XOR with every number of array of integers
- Minimum number of cubes whose sum equals to given number N
- Add minimum number to an array so that the sum becomes even
- Form minimum number from given sequence
- Minimum number of items to be delivered
- Number of subarrays whose minimum and maximum are same
- Minimum number of jumps to reach end | Set 2 (O(n) solution)
- Minimum prime number operations to convert A to B

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.