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

1814. Count Nice Pairs in an Array

Level

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.

• Java
• C++
• Python

• 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;
      }
  }
  

• // OJ: https://leetcode.com/problems/count-nice-pairs-in-an-array/
  // Time: O(N)
  // Space: O(N)
  class Solution {
      long rev(long x) {
          long ans = 0;
          while (x) {
              ans = ans * 10 + x % 10;
              x /= 10;
          }
          return ans;
      }
  public:
      int countNicePairs(vector<int>& A) {
          unordered_map<int, int> m;
          long ans = 0, mod = 1e9+7;
          for (int n : A) {
              long x = n - rev(n);
              ans = (ans + m[x]) % mod;
              m[x]++;
          }
          return ans;
      }
  };
  

• class Solution:
      def countNicePairs(self, nums: List[int]) -> int:
          def rev(x):
              y = 0
              while x:
                  y = y * 10 + x % 10
                  x //= 10
              return y
  
          cnt = Counter(x - rev(x) for x in nums)
          mod = 10**9 + 7
          return sum(v * (v - 1) // 2 for v in cnt.values()) % mod
  
  ############
  
  # 1814. Count Nice Pairs in an Array
  # https://leetcode.com/problems/count-nice-pairs-in-an-array/
  
  class Solution:
      def countNicePairs(self, nums: List[int]) -> int:
          count = Counter(num - int(str(num)[::-1]) for num in nums)
          
          return sum((c * (c-1) // 2) for c in count.values()) % (10 ** 9 + 7)
  
  

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