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 <= 105

• 0 <= nums[i] <= 109

hint 1：

The condition can be rearranged to (nums[i] - rev(nums[i])) == (nums[j] - rev(nums[j])).

hint 2:

Transform each nums[i] into (nums[i] - rev(nums[i])). Then, count the number of (i, j) pairs that have equal values.

hint 3:

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].

Runtime: 98 ms, faster than 10.67% of Java online submissions for Count Nice Pairs in an Array.

Memory Usage: 55.7 MB, less than 59.67% of Java online submissions for Count Nice Pairs in an Array.