Count Nice Pairs in an Array - LeetCode javascript solutions

Problem

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

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var countNicePairs = function(nums) {
    var diffArray = nums.map(num => {
        var rev = Number(String(num).split('').reverse().join(''));
        return num - rev;
    }).sort((a, b) => a - b);
    var tmp = 0;
    var res = 0;
    var mod = Math.pow(10, 9) + 7;
    for (var i = 0; i < diffArray.length; i++) {
        tmp += 1;
        if (diffArray[i + 1] !== diffArray[i]) {
            res += tmp * (tmp - 1) / 2;
            res %= mod;
            tmp = 0;
        }
    }
    return res;
};

Explain:

nope.

Complexity:

Time complexity : O(n * log(n)).
Space complexity : O(n).