1803. Count Pairs With XOR in a Range

Problem

Given a (0-indexed) integer array nums and two integers low and high, return **the number of *nice pairs***.

A nice pair is a pair (i, j) where 0 <= i < j < nums.length and low <= (nums[i] XOR nums[j]) <= high.

  Example 1:

Input: nums = [1,4,2,7], low = 2, high = 6
Output: 6
Explanation: All nice pairs (i, j) are as follows:
    - (0, 1): nums[0] XOR nums[1] = 5 
    - (0, 2): nums[0] XOR nums[2] = 3
    - (0, 3): nums[0] XOR nums[3] = 6
    - (1, 2): nums[1] XOR nums[2] = 6
    - (1, 3): nums[1] XOR nums[3] = 3
    - (2, 3): nums[2] XOR nums[3] = 5

Example 2:

Input: nums = [9,8,4,2,1], low = 5, high = 14
Output: 8
Explanation: All nice pairs (i, j) are as follows:
​​​​​    - (0, 2): nums[0] XOR nums[2] = 13
    - (0, 3): nums[0] XOR nums[3] = 11
    - (0, 4): nums[0] XOR nums[4] = 8
    - (1, 2): nums[1] XOR nums[2] = 12
    - (1, 3): nums[1] XOR nums[3] = 10
    - (1, 4): nums[1] XOR nums[4] = 9
    - (2, 3): nums[2] XOR nums[3] = 6
    - (2, 4): nums[2] XOR nums[4] = 5

  Constraints:

• 1 <= nums.length <= 2 * 10^4

• 1 <= nums[i] <= 2 * 10^4

• 1 <= low <= high <= 2 * 10^4

Solution

class Solution {
    public int countPairs(int[] nums, int low, int high) {
        Trie root = new Trie();
        int pairsCount = 0;
        for (int num : nums) {
            int pairsCountHigh = countPairsWhoseXorLessThanX(num, root, high + 1);
            int pairsCountLow = countPairsWhoseXorLessThanX(num, root, low);
            pairsCount += (pairsCountHigh - pairsCountLow);
            root.insertNumber(num);
        }
        return pairsCount;
    }

    private int countPairsWhoseXorLessThanX(int num, Trie root, int x) {
        int pairs = 0;
        Trie curr = root;
        for (int i = 14; i >= 0 && curr != null; i--) {
            int numIthBit = (num >> i) & 1;
            int xIthBit = (x >> i) & 1;
            if (xIthBit == 1) {
                if (curr.child[numIthBit] != null) {
                    pairs += curr.child[numIthBit].count;
                }
                curr = curr.child[1 - numIthBit];
            } else {
                curr = curr.child[numIthBit];
            }
        }
        return pairs;
    }

    private static class Trie {
        Trie[] child;
        int count;

        public Trie() {
            child = new Trie[2];
            count = 0;
        }

        public void insertNumber(int num) {
            Trie curr = this;
            for (int i = 14; i >= 0; i--) {
                int ithBit = (num >> i) & 1;
                if (curr.child[ithBit] == null) {
                    curr.child[ithBit] = new Trie();
                }
                curr.child[ithBit].count++;
                curr = curr.child[ithBit];
            }
        }
    }
}

Explain:

nope.

Complexity:

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