Given three integers **A**, **B**, and **C**. In one operation, choose any two of the three integers, subject to the condition that both of them should be greater than 0, and reduce them by **1**. The task is to find the maximum number of operations that can be performed until at least two of them becomes **0**.**Examples:**

Input:A = 1, B = 3, C = 1Output:2Explanation:

Operation 1: Choose A and B, reduce both by 1. Current values: A = 0, B = 2, C = 1

Operation 2: Choose B and C, reduce both by 1. Current Values: A = 0, B = 1, C = 0

No more opeartions are possible as any pair chosen will have at least one 0 in it.Input:A = 8, B = 1, C = 4Output:5

**Approach:** The idea is to arrange the given numbers in decreasing order to find the maximum number of operations on the basis of the following condition: **Case 1: When A ≥ (B + C)**

- Choose pair A and B for B times. As a result, after B operations the current status will be A = (A – B) and B = 0.
- As A ≥ (B + C) which implies (A – B) ≥ C. So the pair A and C can be chosen for C operations, and the current status will be A = (A – B – C), B = 0, and C = 0.
- Total operations performed =
**(B + C)**

**Case 2: When A < (B + C)**

- Try to make A, B, C equal after performing some operations.
- First, make A and B equal. For this choose A and C for performing (A – B) operations. Let the updated values be named
**A**, and_{1}, B_{1}**C**. The values A_{1}_{1}, B_{1}, and C_{1}will be:

A_{1}= A – (A – B)B_{1}= BC_{1}= C – (A – B)

- The number of operations performed =
**(A – B)**. - A
_{1}and B_{1}are equal. So, choose the pair A_{1}and B_{1}for (A_{1}– C_{1}) operations. - Let A
_{2}, B_{2}, and C_{2}be the updated values of A, B, and C after the above operation. The values of A_{2}, B_{2, }and C_{2}Will be the same and that will be:

A= A_{2}_{1}– (A_{1}– C_{1}) = C_{1}=(C – A + B)B_{2}= C – A + BC_{2}= C – A + B

- Let the total number of operations performed as of now be
**Z**. So the value of Z will be:

Z= (A – B) + (A_{1}– C_{1}) = (A – B) + (B – C + A – B)=2A – B – C

- As A2 = B2 = C2, then there arises two cases:
**A2, B2, C2 are even:**For every set of 3 operations on the pairs (A2, B2), (B2, C2), and (C2, A2) the count of the**A2, B2**,**and C2 decreases by 2**.

Let A2 = B2 = C2 = 4. Let the operations that can be performed be X. So, X = (4 + 4 + 4) / 2 = 6. Thus, the value of X can be generalized as:

**A2, B2, C2 are odd:**For every set of 3 operations on the pairs (A2, B2), (B2, C2), and (C2, A2) the count of the**A2, B2**,**and C2 decreases by 2**, finally, the values A2, B2, and C2 reach 1, 1 and 1 respectively. Here one additional operation can be performed.

Let A2 = B2 = C2 = 5. After performing 6 operations, A2 = B2 = C2 = 1. Here one more operation can be performed. Therefore, total operations that can be performed are 7 (6+1). Let the operations that can be performed be Y. So, Y = floor((5 + 5 + 5) / 2) = 7. Thus, the value of Y can be generalized as:

- Since from the above steps X = Y, therefore a total number of possible causes can be given by:

Total number of possible cases = (Z + X) = (Z + Y) = (A + B + C) / 2.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the minimum number` `// operations` `int` `solution(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` `int` `arr[3];` ` ` `// Insert the three numbers in array` ` ` `arr[0] = A, arr[1] = B, arr[2] = C;` ` ` `// Sort the array` ` ` `sort(arr, arr + 3);` ` ` `// Case 2` ` ` `if` `(arr[2] < arr[0] + arr[1])` ` ` `return` `((arr[0] + arr[1]` ` ` `+ arr[2])` ` ` `/ 2);` ` ` `// Case 1` ` ` `else` ` ` `return` `(arr[0] + arr[1]);` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given A, B, C` ` ` `int` `A = 8, B = 1, C = 5;` ` ` `// Function Call` ` ` `cout << solution(A, B, C);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to find the minimum number` `// operations` `public` `static` `int` `solution(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` `int` `arr[] = ` `new` `int` `[` `3` `];` ` ` `// Insert the three numbers in array` ` ` `arr[` `0` `] = A; arr[` `1` `] = B; arr[` `2` `] = C;` ` ` `// Sort the array` ` ` `Arrays.sort(arr);` ` ` `// Case 2` ` ` `if` `(arr[` `2` `] < arr[` `0` `] + arr[` `1` `])` ` ` `return` `((arr[` `0` `] + arr[` `1` `] + arr[` `2` `]) / ` `2` `);` ` ` ` ` `// Case 1` ` ` `else` ` ` `return` `(arr[` `0` `] + arr[` `1` `]);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given A, B, C` ` ` `int` `A = ` `8` `, B = ` `1` `, C = ` `5` `;` ` ` `// Function call` ` ` `System.out.println(solution(A, B, C));` `}` `}` `// This code is contributed by jrishabh99` |

## Python

`#Python3 program for the above approach` `#Function to find the minimum number` `#operations` `def` `solution(A, B, C):` ` ` `arr` `=` `[` `0` `] ` `*` `3` ` ` `#Insert the three numbers in array` ` ` `arr[` `0` `] ` `=` `A` ` ` `arr[` `1` `] ` `=` `B` ` ` `arr[` `2` `] ` `=` `C` ` ` `#Sort the array` ` ` `arr ` `=` `sorted` `(arr)` ` ` `#Case 2` ` ` `if` `(arr[` `2` `] < arr[` `0` `] ` `+` `arr[` `1` `]):` ` ` `return` `((arr[` `0` `] ` `+` `arr[` `1` `] ` `+` `arr[` `2` `]) ` `/` `/` `2` `)` ` ` ` ` `#Case 1` ` ` `else` `:` ` ` `return` `(arr[` `0` `] ` `+` `arr[` `1` `])` `#Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `#Given A, B, C` ` ` `A ` `=` `8` ` ` `B ` `=` `1` ` ` `C ` `=` `5` ` ` `#Function Call` ` ` `print` `(solution(A, B, C))` `# This code is contributed by Mohit Kumar 29` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the minimum number` `// operations` `public` `static` `int` `solution(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` `int` `[]arr = ` `new` `int` `[3];` ` ` `// Insert the three numbers in array` ` ` `arr[0] = A; arr[1] = B; arr[2] = C;` ` ` `// Sort the array` ` ` `Array.Sort(arr);` ` ` `// Case 2` ` ` `if` `(arr[2] < arr[0] + arr[1])` ` ` `return` `((arr[0] + arr[1] + arr[2]) / 2);` ` ` ` ` `// Case 1` ` ` `else` ` ` `return` `(arr[0] + arr[1]);` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ` `// Given A, B, C` ` ` `int` `A = 8, B = 1, C = 5;` ` ` `// Function call` ` ` `Console.WriteLine(solution(A, B, C));` `}` `}` `// This code is contributed by Rajput-Ji` |

**Output:**

6

**Time Complexity:** O(1) **Space Complexity:** O(1)

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