CountNicePairs.py

'''
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
'''

class Solution:
    def countNicePairs(self, nums: List[int]) -> int:
        
        def rev(num):
            return int(str(num)[::-1])
        
        for i in range(len(nums)):
            nums[i] = nums[i] - rev(nums[i])
            
        count = 0
        freq = {}
        
        for i in nums:
            if i in freq.keys(): freq[i] += 1
            else: freq[i] = 1

        for k, v in freq.items():
            count += ((v * (v -1 )) // 2)
            
        return count % (10 ** 9 + 7)