Minimize number of boxes by putting small box inside bigger one
Given an array size[] of box sizes, our task is to find the number of boxes left at the end, after putting the smaller-sized box into a bigger one.
Note: Only one small box can fit inside one box.
Examples:
Input: size[] = {1, 2, 3}
Output: 1
Explanation:
Here, box of size 1 can fit inside box of size 2 and the box of size 2 can fit inside box of size 3. So at last we have only one box of size 3.Input: size[] = {1, 2, 2, 3, 7, 4, 2, 1}
Output: 3
Explanation:
Put the box of size 1, 2, 3, 4 and 7 together and for the second box put 1 and 2 together. At last 2 is left which will not fit inside anyone. So we have 3 boxes left.
Approach: The idea is to follow the steps given below:
- Sort the given array size[] in increasing order and check if the current box size is greater than the next box size. If yes then decrease the initial box number.
- Otherwise, if the current box size is equal to next box size, then check if the current box can fit inside next to next box size. If yes then move the current box pointing variable, else move next pointing variable further.
Below is the implementation of the above approach:
C++
// C++ implementation to minimize the// number of the box by putting small// box inside the bigger one#include <bits/stdc++.h>using namespace std;// Function to minimize the countvoid minBox(int arr[], int n){ // Initial number of box int box = n; // Sort array of box size // in increasing order sort(arr, arr + n); int curr_box = 0, next_box = 1; while (curr_box < n && next_box < n) { // check is current box size // is smaller than next box size if (arr[curr_box] < arr[next_box]) { // Decrement box count // Increment current box count // Increment next box count box--; curr_box++; next_box++; } // Check if both box // have same size else if (arr[curr_box] == arr[next_box]) next_box++; } // Print the result cout << box << endl;}// Driver codeint main(){ int size[] = { 1, 2, 3 }; int n = sizeof(size) / sizeof(size[0]); minBox(size, n); return 0;} |
chevron_right
filter_none
Java
// Java implementation to minimize the// number of the box by putting small// box inside the bigger oneimport java.util.Arrays; class GFG{ // Function to minimize the countpublic static void minBox(int arr[], int n){ // Initial number of box int box = n; // Sort array of box size // in increasing order Arrays.sort(arr); int curr_box = 0, next_box = 1; while (curr_box < n && next_box < n) { // Check is current box size // is smaller than next box size if (arr[curr_box] < arr[next_box]) { // Decrement box count // Increment current box count // Increment next box count box--; curr_box++; next_box++; } // Check if both box // have same size else if (arr[curr_box] == arr[next_box]) next_box++; } // Print the result System.out.println(box);}// Driver codepublic static void main(String args[]){ int []size = { 1, 2, 3 }; int n = size.length; minBox(size, n);}}// This code is contributed by SoumikMondal |
chevron_right
filter_none
Python3
# Python3 implementation to minimize the # number of the box by putting small # box inside the bigger one# Function to minimize the count def minBox(arr, n): # Initial number of box box = n # Sort array of box size # in increasing order arr.sort() curr_box, next_box = 0, 1 while (curr_box < n and next_box < n): # Check is current box size # is smaller than next box size if (arr[curr_box] < arr[next_box]): # Decrement box count # Increment current box count # Increment next box count box = box - 1 curr_box = curr_box + 1 next_box = next_box + 1 # Check if both box # have same size elif (arr[curr_box] == arr[next_box]): next_box = next_box + 1 # Print the result print(box) # Driver codesize = [ 1, 2, 3 ] n = len(size) minBox(size, n) # This code is contributed by divyeshrabadiya07 |
chevron_right
filter_none
C#
// C# implementation to minimize the// number of the box by putting small// box inside the bigger oneusing System;class GFG{ // Function to minimize the countpublic static void minBox(int []arr, int n){ // Initial number of box int box = n; // Sort array of box size // in increasing order Array.Sort(arr); int curr_box = 0, next_box = 1; while (curr_box < n && next_box < n) { // Check is current box size // is smaller than next box size if (arr[curr_box] < arr[next_box]) { // Decrement box count // Increment current box count // Increment next box count box--; curr_box++; next_box++; } // Check if both box // have same size else if (arr[curr_box] == arr[next_box]) next_box++; } // Print the result Console.WriteLine(box);}// Driver codepublic static void Main(String []args){ int []size = { 1, 2, 3 }; int n = size.Length; minBox(size, n);}}// This code is contributed by Amit Katiyar |
chevron_right
filter_none
Output
1
Time Complexity: O(N*logN) as Arrays.sort() method is used.
Auxiliary Space: O(1)
Approach: The key observation in the problem is that the minimum number of boxes is same as the maximum frequency of any element in the array. Because the rest of the elements are adjusted into one another. Below is the illustration with the help of steps:
- Create a Hash-map to store the frequency of the elements.
- Finally, after maintaining the frequency return the maximum frequency of the element.
Below is the implementation of the above approach:
Java
// Java implementation of the // above approachimport java.util.*;public class Boxes { // Function to count the boxes // at the end int countBoxes(int[] A) { // Default value if // array is empty int count = -1; int n = A.length; // Map to store frequencies Map<Integer, Integer> map = new HashMap<>(); // Loop to iterate over the // elements of the array for (int i=0; i<n; i++) { int key = A[i]; map.put(key, map.getOrDefault(key, 0) + 1); int val = map.get(key); // Condition to get the maximum // value of the key if (val > count) count = val; } return count; } // Driver Code public static void main( String[] args) { int[] a = {8, 15, 1, 10, 5, 1}; Boxes obj = new Boxes(); // Function Call int minBoxes = obj.countBoxes(a); System.out.println(minBoxes); }} |
chevron_right
filter_none
Output
2
Time Complexity: O(N)
Auxiliary Space: O(N)
