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 1Input: 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; } |
Java
// Java implementation of the approach import java.util.HashMap; import java.util.HashSet; class GFG { // Function to return the count // of minimum stacks static int countPiles( int n, int [] a) { // Keep track of occurrence // of each capacity HashMap<Integer, Integer> occ = new HashMap<>(); // Fill the occurrence map for ( int i = 0 ; i < n; i++) occ.put(a[i], occ.get(a[i]) == null ? 1 : occ.get(a[i]) + 1 ); // Number of piles is 0 initially int pile = 0 ; // Traverse occurrences in increasing // order of capacities. while (!occ.isEmpty()) { // Adding a new pile pile++; int size = 0 ; HashSet<Integer> toRemove = new HashSet<>(); // Traverse all piles in increasing // order of capacities for (HashMap.Entry<Integer, Integer> tm : occ.entrySet()) { int mx = tm.getKey(); int ct = tm.getValue(); // Number of boxes of capacity mx // that can be added to current pile int use = Math.min(ct, mx - size + 1 ); // Update the occurrence occ.put(mx, occ.get(mx) - use); // Update the size of the pile size += use; if (occ.get(mx) == 0 ) toRemove.add(mx); } // Remove capacities that are // no longer available for ( int tm : toRemove) occ.remove(tm); } return pile; } // Driver Code public static void main(String[] args) { int [] a = { 0 , 0 , 1 , 1 , 2 }; int n = a.length; System.out.println(countPiles(n, a)); } } // This code is contributed by // sanjeev2552 |
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 |
C#
// C# implementation of the approach using System; using System.Collections; using System.Collections.Generic; using System.Linq; class GFG { // Function to return the count // of minimum stacks static int countPiles( int n, int [] a) { // Keep track of occurrence // of each capacity Dictionary< int , int > occ = new Dictionary< int , int >(); // Fill the occurrence map for ( int i = 0; i < n; i++) { if (!occ.ContainsKey(a[i])) { occ[a[i]]=0; } occ[a[i]]++; } // Number of piles is 0 initially int pile = 0; // Traverse occurrences in increasing // order of capacities. while (occ.Count!=0) { // Adding a new pile pile++; int size = 0; HashSet< int > toRemove = new HashSet< int >(); Dictionary< int , int > tmp = occ; // Traverse all piles in increasing // order of capacities foreach ( var tm in occ.Keys.ToList()) { int mx = tm; int ct = occ[tm]; // Number of boxes of capacity mx // that can be added to current pile int use = Math.Min(ct, mx - size + 1); // Update the occurrence occ[mx]-= use; // Update the size of the pile size += use; if (occ[mx] == 0) toRemove.Add(mx); } occ = tmp; // Remove capacities that are // no longer available foreach ( int tm in toRemove.ToList()) occ.Remove(tm); } return pile; } // Driver Code public static void Main( string [] args) { int [] a = { 0, 0, 1, 1, 2 }; int n = a.Length; Console.WriteLine(countPiles(n, a)); } } // This code is contributed by rutvik_56. |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of minimum stacks function countPiles(n, a) { // Keep track of occurrence // of each capacity let occ = new Map(); // Fill the occurrence map for (let i = 0; i < n; i++) occ.set(a[i], occ.get(a[i]) == null ? 1 : occ.get(a[i]) + 1); // Number of piles is 0 initially let pile = 0; // Traverse occurrences in increasing // order of capacities. while (occ.size != 0) { // Adding a new pile pile++; let size = 0; let toRemove = new Set(); // Traverse all piles in increasing // order of capacities for (let [key, value] of occ.entries()) { let mx = key; let ct = value; // Number of boxes of capacity mx // that can be added to current pile let use = Math.min(ct, mx - size + 1); // Update the occurrence occ.set(mx, occ.get(mx) - use); // Update the size of the pile size += use; if (occ.get(mx) == 0) toRemove.add(mx); } // Remove capacities that are // no longer available for (let tm of toRemove.values()) occ. delete (tm); } return pile; } // Driver Code let a = [ 0, 0, 1, 1, 2 ]; let n = a.length; document.write(countPiles(n, a)); // This code is contributed by unknown2108 </script> |
2
Time Complexity: O(NlogN)
Auxiliary Space: O(N), where N is the size of the given input array.