Problem URL: Count nice pairs in an array

The problem statement says nums[i]+rev(nums[j]) == nums[j]+rev(nums[i]) is the defination of nice pairs. This can also be restated as nums[i]-rev(nums[i]) == nums[j]-rev(nums[j]). We will use this formula, and add the difference in a lookup table. And then just like the two sum problem, we will look for a matching pair on lookup hashmap, and add it to our result. Finally, we will return that in the result by moduling the given value 10^9+7.

class Solution:
    def countNicePairs(self, nums: List[int]) -> int:
        MOD = 10 ** 9 + 7
        res = 0
        lookup = collections.defaultdict(int)
        for num in nums:
            diff = num - int(str(num)[::-1])
            res += lookup[diff]
            lookup[diff] += 1
        return res % MOD

Time Complexity: O(n)
Space Complexity: O(n)