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)

1814 - Count Nice Pairs in an Array

1814. Count Nice Pairs in an Array

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