Solution 1 for Scaler Topics Fortnightly Contest - 26
Make GCD Problem
This article is part of the Scaler Topics Fortnightly Contest - 26
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Solution Approach
The solution is to find the nearest multiple of B for each element. If doing this gives the GCD as a multiple of B, we should change an element to either A[i] + B or A[i] - B to make the GCD equal to B. So, we would see which element was closest to the other multiple of B. Why does this work? Consider that every element was divisible by KB after the first set of operations. If we decrease B or increase B to an element and now take the gcd, GCD(KBF, KB*F - B). The GCD is going to be B.
Time Complexity: O(|A|) Space Complexity: O(|A|)
C++ Implementation
Java Implementation
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