Given N items of type-1 and M items of type-2. An object can be created from 2 items of type-1 and 1 item of type-2 or 2 items of type-2 and 1 item of type-1 pieces. The task is to find the maximum number of objects that can be created from given number of items of each type.
Examples:
Input: N = 8, M = 7
Output: 5
Explanation:
3 pairs of 2 type-1 and 1 type-2 objects.
2 pairs of 1 type-1 and 2 type-2 objects.
Input: N = 20, M = 3
Output: 3
Explanation:
3 pairs of 2 type-1 and 1 type-2 objects.
Approach:
Follow the steps below to solve the problem:
- Create two variables initial and final.
- initial = Minimum of N, M.
- final = Divide N + M by 3 ( as an object is made up of 3 components).
- Minimum of initial and final is maximum number of objects that can be created from given N type-1 and M type-2 items.
Below is the implementation of the above approach:
C++
// C++ program for the above problem#include <bits/stdc++.h> using namespace std;// Function for finding// the maximum number of// objects from N type-1 and // M type-2 itemsint numberOfObjects(int N, int M){ // storing minimum of N and M int initial = min(N, M); // storing maximum number of // objects from given items int final = (N + M) / 3; return min(initial, final);}// Driver Codeint main(){ int N = 8; int M = 7; cout << numberOfObjects(N, M) << endl; return 0;} |
Java
// Java program for the above problemclass GFG{ // Function for finding// the maximum number of// objects from N type-1 and // M type-2 itemsstatic int numberOfObjects(int N, int M){ // Storing minimum of N and M int initial = Math.min(N, M); // Storing maximum number of // objects from given items int last = (N + M) / 3; return Math.min(initial, last);}// Driver Codepublic static void main(String[] args) { int N = 8; int M = 7; System.out.println(numberOfObjects(N, M));}}// This code is contributed by rutvik_56 |
Python3
# Python3 program for the above problem# Function for finding# the maximum number of# objects from N type-1 and# M type-2 itemsdef numberOfObjects(N, M): # Storing minimum of N and M initial = min(N, M) # Storing maximum number of # objects from given items final = (N + M) // 3 return min(initial, final)# Driver Codeif __name__ == '__main__': N = 8 M = 7 print(numberOfObjects(N, M))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above problemusing System;class GFG{ // Function for finding// the maximum number of// objects from N type-1 and // M type-2 itemsstatic int numberOfObjects(int N, int M){ // Storing minimum of N and M int initial = Math.Min(N, M); // Storing maximum number of // objects from given items int last = (N + M) / 3; return Math.Min(initial, last);} // Driver Codepublic static void Main(string[] args) { int N = 8; int M = 7; Console.Write(numberOfObjects(N, M));}} // This code is contributed by rock_cool |
Javascript
<script>// JavaScript program for the above problem// Function for finding// the maximum number of// objects from N type-1 and // M type-2 itemsfunction numberOfObjects(N, M){ // storing minimum of N and M let initial = Math.min(N, M); // storing maximum number of // objects from given items let final = Math.floor((N + M) / 3); return Math.min(initial, final);}// Driver Code let N = 8; let M = 7; document.write(numberOfObjects(N, M) + "<br>");// This code is contributed by Surbhi Tyagi.</script> |
Output:
5
Time Complexity: O(1)
Auxiliary Space Complexity: O(1)
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