Count nice pairs in an array

Array Hash Table Math Counting

48.4% acceptance

Feb 25, 2026

2041

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.

Solution

Rust

Time O(2^n)

Space O(n)

LeetCode

solution.rs

use std::collections::HashMap;


impl Solution {
  pub fn count_nice_pairs(nums: Vec<i32>) -> i32 {
    const MOD: i64 = 1_000_000_007;

    fn rev(mut x: i32) -> i32 {
      let mut r = 0i32;
      while x > 0 {
        r = r * 10 + x % 10;
        x /= 10;
      }
      r
    }

    let mut freq: HashMap<i32, i64> = HashMap::new();
    for &n in &nums {
      *freq.entry(n - rev(n)).or_insert(0) += 1;
    }
    let mut result: i64 = 0;
    for &c in freq.values() {
      result = (result + c * (c - 1) / 2) % MOD;
    }
    result as i32
  }
}