# LeetCode 1814. 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 <= 10^5`
* `0 <= nums[i] <= 10^9`

## Solution

```
class Solution {
public:
    int reverse(int x) {
        int result = 0;
        while (x > 0) {
            result = result * 10 + (x % 10);
            x /= 10;
        }
        return result;
    }
    
    int countNicePairs(vector<int>& nums) {
        unordered_map<int, int> mymap;
        int mod = 1e9+7;
        int res = 0;
        for (int num : nums) {
            int rnum = reverse(num);
            int diff = num - rnum;
            res += mymap[diff];
            res %= mod;
            mymap[diff]++;
        }
        
        return res;
    }
};
```
