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

1814. Count Nice Pairs in an Array
struct Solution;

use std::collections::HashMap;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn count_nice_pairs(nums: Vec<i32>) -> i32 {
        let mut hm: HashMap<i32, i32> = HashMap::new();
        let mut res: i64 = 0;
        for x in nums {
            let diff = x - rev(x);
            let count = hm.entry(diff).or_default();
            res += *count as i64;
            *count += 1;
        }
        (res % MOD) as i32
    }
}

fn rev(mut x: i32) -> i32 {
    let mut digits: Vec<i32> = vec![];
    while x > 0 {
        digits.push(x % 10);
        x /= 10;
    }
    let mut res = 0;
    for x in digits {
        res *= 10;
        res += x;
    }
    res
}

#[test]
fn test() {
    let nums = vec![42, 11, 1, 97];
    let res = 2;
    assert_eq!(Solution::count_nice_pairs(nums), res);
    let nums = vec![13, 10, 35, 24, 76];
    let res = 4;
    assert_eq!(Solution::count_nice_pairs(nums), res);
}