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 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. Atlast 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 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 count void 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 code int main() { int size[] = { 1, 2, 3 }; int n = sizeof(size) / sizeof(size[0]); minBox(size, n); return 0; } |
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Java
// Java implementation to minimize the // number of the box by putting small // box inside the bigger one import java.util.Arrays; class GFG{ // Function to minimize the count public 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 code public static void main(String args[]) { int []size = { 1, 2, 3 }; int n = size.length; minBox(size, n); } } // This code is contributed by SoumikMondal |
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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 code size = [ 1, 2, 3 ] n = len(size) minBox(size, n) # This code is contributed by divyeshrabadiya07 |
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C#
// C# implementation to minimize the // number of the box by putting small // box inside the bigger one using System; class GFG{ // Function to minimize the count public 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 code public static void Main(String []args) { int []size = { 1, 2, 3 }; int n = size.Length; minBox(size, n); } } // This code is contributed by Amit Katiyar |
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Output:
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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