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:
- Initialize a variable
minDiff
to store the minimum absolute difference encountered so far. Set it to a large value initially. - Iterate through all possible values of x from 0 to the maximum of a and b.
- For each value of x, calculate the absolute difference between a - x and b + x.
- If the absolute difference is less than
minDiff
, updateminDiff
with the new value. - 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.