BinarySearch 0829 Reverse Equivalent Pairs

Problem statement

https://binarysearch.com/problems/Reverse-Equivalent-Pairs/

Solution

We need to find number of paris where x - f(x) = y - f(y). So, we can precalculate all of values and then use counter.

Complexity

It is O(n) for time and space.

Code

class Solution:
    def solve(self, nums):
        A = [x - int(str(x)[::-1]) for x in nums]
        cnt = Counter(A)
        ans = 2 * len(A)
        for x, f in cnt.items():
            ans += f*(f-1)
            
        return (ans//2) % (10**9 + 7)