1814

1814. Count Nice Pairs in an Array (Medium)

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

Companies: Uber, Navi, Square

Related Topics:
Array, Hash Table, Math, Counting

Similar Questions:

• Number of Pairs of Interchangeable Rectangles (Medium)
• Count Number of Bad Pairs (Medium)
• Number of Pairs Satisfying Inequality (Hard)

Hints:

• The condition can be rearranged to (nums[i] - rev(nums[i])) == (nums[j] - rev(nums[j])).
• Transform each nums[i] into (nums[i] - rev(nums[i])). Then, count the number of (i, j) pairs that have equal values.
• Keep a map storing the frequencies of values that you have seen so far. For each i, check if nums[i] is in the map. If it is, then add that count to the overall count. Then, increment the frequency of nums[i].

Solution 1.

// OJ: https://leetcode.com/problems/count-nice-pairs-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
    long rev(long x) {
        long ans = 0;
        while (x) {
            ans = ans * 10 + x % 10;
            x /= 10;
        }
        return ans;
    }
public:
    int countNicePairs(vector<int>& A) {
        unordered_map<int, int> m;
        long ans = 0, mod = 1e9 + 7;
        for (int n : A) {
            long x = n - rev(n);
            ans = (ans + m[x]++) % mod;
        }
        return ans;
    }
};