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This article is part of the Scaler Topics Fortnightly Contest - 24

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Solution Approach

We can greedily operate on the numbers. Note that the final number is always fixed and doesn’t change according to the operations. It is basically the GCD of the array.
We need to subtract a higher number in order to reach the final number earlier. So we select the highest 2 non-equal numbers say x and y and operate on them. But this would take a lot of operations.
Rather than subtracting again and again, we can just see how many times we need to subtract that number from the larger number and we can just change the larger number to y%x or x if y%x is equal to 0.
This would be the most optimal way because subtracting by a lower number would take more moves. So, we can use a map that also

Time Complexity: $O(|A| * log(A))$