Integer Moves codeforces solution

Problem Statement

The "Integer Moves" problem on Codeforces is as follows:

You are given two positive integers a and b. In one move, you can choose any positive integer x and subtract it from a or add it to b. You can perform as many moves as you want (possibly zero).

Your task is to find the minimum number of moves required to make a and b equal. In other words, you need to find the minimum value of |a - b|.

Implement a C++ function int minMoves(int a, int b) that returns the minimum number of moves required to make a and b equal.

Approach

To solve this problem, we can observe that the difference between a and b can only be reduced by either subtracting a positive integer from a or adding a positive integer to b. Therefore, we need to find the minimum number of moves required to make a and b equal.

One way to do this is to iterate through all possible values of x from 0 to the maximum possible value of a or b. For each value of x, we calculate the absolute difference between a - x and b + x. We keep track of the minimum absolute difference encountered so far and return it as the result.

Here is the step-by-step implementation of the minMoves function:

  1. Initialize a variable minDiff to store the minimum absolute difference encountered so far. Set it to a large value initially.
  2. Iterate through all possible values of x from 0 to the maximum of a and b.
  3. For each value of x, calculate the absolute difference between a - x and b + x.
  4. If the absolute difference is less than minDiff, update minDiff with the new value.
  5. After the loop ends, return the value of minDiff as the minimum number of moves required to make a and b equal.

C++ Implementation

Here is the C++ implementation of the minMoves function:

int minMoves(int a, int b) {
    int maxVal = max(a, b);
    int minDiff = INT_MAX;

    for (int x = 0; x <= maxVal; x++) {
        int diff = abs(a - x) + abs(b + x);
        minDiff = min(minDiff, diff);
    }

    return minDiff;
}

This implementation calculates the minimum number of moves required to make a and b equal by iterating through all possible values of x and keeping track of the minimum absolute difference encountered so far.

Complexity Analysis

The time complexity of this solution is O(max(a, b)) since we iterate through all possible values of x from 0 to the maximum of a and b.

The space complexity is O(1) since we only use a constant amount of additional space to store the minimum absolute difference encountered so far.