Count nice pairs in an array

class Solution:
    def countNicePairs(self, nums: List[int]) -> int:
        n = len(nums)
        arr = [0] * n
        def rev(num):
            res = 0
            while num:
                res = res * 10 + num % 10
                num //= 10
            return res
        for i in range(n):
            arr[i] = nums[i] - rev(nums[i])
        count = {}
        MOD = 10 ** 9 + 7
        res = 0
        for i in arr:
            res = (res + count.get(i, 0)) % MOD
            count[i] = count.get(i, 0) + 1

        return res

Count Nice Pairs in an Array

You are given an array nums that consists of non-negative integers. Let us define rev(x) as the reverse of the non-negative integer x. For example, rev(123) = 321, and rev(120) = 21. A pair of indices (i, j) is nice if it satisfies all of the following conditions:

0 <= i < j < nums.length
nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])

Return the number of nice pairs of indices. Since that number can be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [42,11,1,97]
Output: 2
Explanation: The two pairs are:
 - (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.
 - (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.

Example 2:

Input: nums = [13,10,35,24,76]
Output: 4

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹