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Formatted question description: https://leetcode.ca/all/1803.html

1803. Count Pairs With XOR in a Range

Level

Hard

Description

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

Calculate the highest bit of high, which is highBit. Sort the array nums and count the pairs using nested loops with pruning.

Let 0 <= i < j < nums.length. If nums[i] < highBit * 2 && nums[j] >= highBit * 2, then do not search for more indices j for the current index i. This is the pruning operation.

Finally, return the number of nice pairs.

• Java
• C++
• Python

• class Solution {
      public int countPairs(int[] nums, int low, int high) {
          int highBit = 1 << ((int) (Math.log(high) / Math.log(2)));
          int pairs = 0;
          Arrays.sort(nums);
          int length = nums.length;
          for (int i = 0; i < length; i++) {
              for (int j = i + 1; j < length; j++) {
                  int xor = nums[i] ^ nums[j];
                  if (xor >= low && xor <= high)
                      pairs++;
                  else if (nums[i] < highBit * 2 && nums[j] >= highBit * 2)
                      break;
              }
          }
          return pairs;
      }
  }
  

• // OJ: https://leetcode.com/problems/count-pairs-with-xor-in-a-range/
  // Time: O(N)
  // Space: O(1)
  struct TrieNode {
      TrieNode *next[2] = {};
      int cnt = 0;
  };
  const int BIT = 15;
  class Solution {
      void addNode(TrieNode *node, int n) {
          for (int i = BIT; i >= 0; --i) {
              int b = n >> i & 1;
              if (!node->next[b]) node->next[b] = new TrieNode();
              node = node->next[b];
              node->cnt++;
          }
      }
      int count(TrieNode *node, int n, int low, int high, int i = BIT, int rangeMin = 0, int rangeMax = (1 << (BIT + 1)) - 1) {
          if (rangeMin >= low && rangeMax <= high) return node->cnt;
          if (rangeMax < low || rangeMin > high) return 0;
          int ans = 0, b = n >> i & 1, r = 1 - b, mask = 1 << i;
          if (node->next[b]) ans += count(node->next[b], n, low, high, i - 1, rangeMin & ~mask, rangeMax & ~mask);
          if (node->next[r]) ans += count(node->next[r], n, low, high, i - 1, rangeMin | mask, rangeMax | mask);
          return ans;
      }
  public:
      int countPairs(vector<int>& A, int low, int high) {
          TrieNode root;
          int ans = 0;
          for (int n : A) {
              ans += count(&root, n, low, high);
              addNode(&root, n);
          }
          return ans;
      }
  };
  

• class Trie:
      def __init__(self):
          self.children = [None] * 2
          self.cnt = 0
  
      def insert(self, x):
          node = self
          for i in range(15, -1, -1):
              v = x >> i & 1
              if node.children[v] is None:
                  node.children[v] = Trie()
              node = node.children[v]
              node.cnt += 1
  
      def search(self, x, limit):
          node = self
          ans = 0
          for i in range(15, -1, -1):
              if node is None:
                  return ans
              v = x >> i & 1
              if limit >> i & 1:
                  if node.children[v]:
                      ans += node.children[v].cnt
                  node = node.children[v ^ 1]
              else:
                  node = node.children[v]
          return ans
  
  
  class Solution:
      def countPairs(self, nums: List[int], low: int, high: int) -> int:
          ans = 0
          tree = Trie()
          for x in nums:
              ans += tree.search(x, high + 1) - tree.search(x, low)
              tree.insert(x)
          return ans
  
  
  

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