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Problem solution · Java

Count Pairs With XOR in a Range

Count Pairs With XOR in a Range: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
52 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Count Pairs With XOR in a Range, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 52 lines of Java from the credited upstream file 1803.java.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Pairs With XOR in a Range · JavaJava
Use this to learn the idea, then write your own version.
class TrieNode {  public TrieNode[] children = new TrieNode[2];  public int count = 0;} class Solution {  public int countPairs(int[] nums, int low, int high) {    int ans = 0;     for (final int num : nums) {      ans += getCount(num, high + 1) - getCount(num, low);      insert(num);    }     return ans;  }   private final int HEIGHT = 14;  private TrieNode root = new TrieNode();   private void insert(int num) {    TrieNode node = root;    for (int i = HEIGHT; i >= 0; --i) {      final int bit = num >> i & 1;      if (node.children[bit] == null)        node.children[bit] = new TrieNode();      node = node.children[bit];      ++node.count;    }  }   // Returns the number of numbers < limit.  private int getCount(int num, int limit) {    int count = 0;    TrieNode node = root;    for (int i = HEIGHT; i >= 0; --i) {      final int bit = num >> i & 1;      final int bitLimit = ((limit >> i) & 1);      if (bitLimit == 1) {        if (node.children[bit] != null)          count += node.children[bit].count;        node = node.children[bit ^ 1];      } else {        node = node.children[bit];      }      if (node == null)        break;    }    return count;  }}

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