Formatted question description: https://leetcode.ca/all/1814.html

# 1814. Count Nice Pairs in an Array

Medium

## Description

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

If a pair of indices (i, j) (where i < j) is nice, then there is nums[i] + rev(nums[j]) == nums[j] + rev(nums[i]), which is equivalent to nums[i] - rev(nums[i]) == nums[j] - rev(nums[j]). Therefore, for each 0 <= i < nums.length, calculate nums[i] - rev(nums[i]) and use an array differences to store the calculated differences. Then sort differences. For each difference, if there are k values, then the number of nice pairs with difference is k * (k - 1) / 2. Loop over differences to calculate the number of nice pairs.

class Solution {
public int countNicePairs(int[] nums) {
final int MODULO = 1000000007;
int length = nums.length;
int[] differences = new int[length];
for (int i = 0; i < length; i++) {
int reverseNum = reverse(nums[i]);
differences[i] = nums[i] - reverseNum;
}
Arrays.sort(differences);
long pairs = 0;
int prev = differences[0];
long consecutive = 1;
for (int i = 1; i < length; i++) {
int curr = differences[i];
if (curr == prev)
consecutive++;
else {
pairs = (pairs + consecutive * (consecutive - 1) / 2) % MODULO;
prev = curr;
consecutive = 1;
}
}
pairs = (pairs + consecutive * (consecutive - 1) / 2) % MODULO;
return (int) pairs;
}

public int reverse(int x) {
int max = Integer.MAX_VALUE / 10, min = Integer.MIN_VALUE / 10;
int reversed = 0;
while (x != 0) {
int lastDigit = x % 10;
if (reversed > max || reversed == max && lastDigit > 7)
return 0;
if (reversed < min || reversed == min && lastDigit < -8)
return 0;
x /= 10;
reversed = reversed * 10 + lastDigit;
}
return reversed;
}
}