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1814. Count Nice Pairs in an Array

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 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

 

Solutions

Solution: Hash Map

  • Time complexity: O(n)
  • Space complexity: O(n)

 

JavaScript

js
/**
 * @param {number[]} nums
 * @return {number}
 */
const countNicePairs = function (nums) {
  const MODULO = 10 ** 9 + 7;
  const uniqueNums = [...new Set(nums)];
  const countMap = new Map();
  const reverseMap = uniqueNums.reduce((map, num) => {
    return (map[num] = +`${num}`.split('').reverse().join('')), map;
  }, {});
  let result = 0;

  for (const num of nums) {
    const diff = num - reverseMap[num];
    const count = countMap.get(diff) ?? 0;

    if (count) {
      result += count;
      result %= MODULO;
    }
    countMap.set(diff, count + 1);
  }
  return result;
};

Released under the MIT license