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Let us suppose we have three numbers. The task is to count the total number of optimal steps to make all these numbers '0'.

**For Example**

**Input-1:**

a = 4 b = 4c = 6

**Output:**

7

**Explanation: **Total number of optimal steps to make all the numbers '0' is,

(4, 4, 6)

Removing '1' from 1st and 2nd number = (3, 3, 6)

Removing '1' from 1st and 3rd number = (2, 3, 5)

Removing '1' from 1st and 3rd number = (1 ,3, 4)

Removing '1' from 1st and 3rd number = (0 ,3 ,3)

Removing '1' from 2nd and 3rd number = (0 ,2, 2)

Removing '1' from 2nd and 3rd number = (0, 1, 1)

Removing '1' from 2nd and 3rd number = (0, 0, 0)

Thus, the total number of steps to make all the numbers zero is '7'

To solve this particular problem, we will remove '1' from any two numbers such that the sum of these two numbers is greater than the last one. To find the minimum steps to make it zero, we will calculate the minimum number of steps.

- Take three numbers as the input.
- Check if the sum of two numbers, let's say 'a' and 'b' is greater than 'c', and a > 0, b > 0, then remove '1' from 'a' and 'b'.
- Find the minimum from the answers and return the result.

#include <bits/stdc++.h> using namespace std; int maxSteps(int a, int b, int c) { int res = 0; while (a + b > c and a > 0 and b > 0) { a--; b--; res++; } res += min(c, a + b); return res; } int main() { int a = 4; int b = 4; int c = 6; cout << maxSteps(a, b, c) << endl; return 0; }

Running the above code will generate the output as,

7

In the given input a = 4, b = 4 and c = 6, it will take seven steps to make all the numbers zero, thus the program returns the output as 7.

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