1814. Count Nice Pairs in an Array

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 10^9 + 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 (Java)

class Solution {
    private int rev(int n) {
        int r = 0;
        while (n > 0) {
            r = r * 10 + n % 10;
            n = n / 10;
        }
        return r;
    }

    public int countNicePairs(int[] nums) {
        HashMap<Integer, Integer> revMap = new HashMap<>();
        int cnt = 0;
        for (int num : nums) {
            int lhs = num - rev(num);
            int prevCnt = revMap.getOrDefault(lhs, 0);
            cnt += prevCnt;
            int mod = 1000000007;
            cnt = cnt % mod;
            revMap.put(lhs, prevCnt + 1);
        }
        return cnt;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).