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.

  • 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;
        }
    };
    
  • # 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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