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++

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// 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;
}

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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)



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Improved By : mohit kumar 29