Given a triplet of integers (A, B, C), the task is to count the maximum number of decrements by 1 that can be performed on positive pairs of the given triplet.
Examples:
Input: A = 4, B = 3, C = 2
Output: 4
Explanation:
Operation 1: Reduce the pair (4, 3). Therefore, the triplet reduces to {3, 2, 2}.
Operation 2: Reduce the pair (3, 2). Therefore, the triplet reduces to {2, 1, 2}.
Operation 3: Reduce the pair (2, 2). Therefore, the triplet reduces to {1, 1, 1}.
Operation 3: Reduce the pair (1, 1). Therefore, the triplet reduces to {0, 0, 1}.
No further operations are possible.Input: A = 7, B = 9, C = 6
Output: 11
Approach: The idea is to use Greedy Approach to solve the problem. Follow the steps below to solve the problem:
- Store the triplet in an array.
- Initialize a variable, say count, to store the maximum possible reductions that can be performed on the triplet.
- Iterate a loop until the first two array elements are reduced to 0 and perform the following operations:
- Sort the array in increasing order.
- Pick the two maximum array elements and reduce them by 1.
- Increment count by 1.
- Print the value of count as the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the maximum// number of pair reductions// possible on a given tripletvoid maxOps(int a, int b, int c){ // Convert them into an array int arr[] = { a, b, c }; // Stores count of operations int count = 0; while (1) { // Sort the array sort(arr, arr + 3); // If the first two array // elements reduce to 0 if (!arr[0] && !arr[1]) break; // Apply the operations arr[1] -= 1; arr[2] -= 1; // Increment count count += 1; } // Print the maximum count cout << count;}// Driver Codeint main(){ // Given triplet int a = 4, b = 3, c = 2; maxOps(a, b, c); return 0;}// This code is contributed by subhammahato348. |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to count the maximum // number of pair reductions // possible on a given triplet static void maxOps(int a, int b, int c) { // Convert them into an array int arr[] = { a, b, c }; // Stores count of operations int count = 0; while (1 != 0) { // Sort the array Arrays.sort(arr); // If the first two array // elements reduce to 0 if (arr[0] == 0 && arr[1] == 0) break; // Apply the operations arr[1] -= 1; arr[2] -= 1; // Increment count count += 1; } // Print the maximum count System.out.print(count); } // Driver Code public static void main(String[] args) { // Given triplet int a = 4, b = 3, c = 2; maxOps(a, b, c); }}// This code is contributed by code_hunt. |
Python3
# Python3 program for the above approach# Function to count the maximum# number of pair reductions# possible on a given tripletdef maxOps(a, b, c): # Convert them into an array arr = [a, b, c] # Stores count of operations count = 0 while True: # Sort the array arr.sort() # If the first two array # elements reduce to 0 if not arr[0] and not arr[1]: break # Apply the operations arr[1] -= 1 arr[2] -= 1 # Increment count count += 1 # Print the maximum count print(count)# Given tripleta, b, c = 4, 3, 2maxOps(a, b, c) |
C#
// C# program for the above approachusing System;class GFG{ // Function to count the maximum // number of pair reductions // possible on a given triplet static void maxOps(int a, int b, int c) { // Convert them into an array int[] arr = { a, b, c }; // Stores count of operations int count = 0; while (1 != 0) { // Sort the array Array.Sort(arr); // If the first two array // elements reduce to 0 if (arr[0] == 0 && arr[1] == 0) break; // Apply the operations arr[1] -= 1; arr[2] -= 1; // Increment count count += 1; } // Print the maximum count Console.WriteLine(count); }// Driver Codepublic static void Main(String[] args){ // Given triplet int a = 4, b = 3, c = 2; maxOps(a, b, c);}}// This code is contributed by susmitakundugoaldanga. |
4
Time Complexity: O(max(arr[i])*(3 * log(3)))
Auxiliary Space: O(1)
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