LeetCode #1803 — HARD

Count Pairs With XOR in a Range

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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 * 10⁴
1 <= nums[i] <= 2 * 10⁴
1 <= low <= high <= 2 * 10⁴

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation · Trie

Example 1

[1,4,2,7]
2
6

Example 2

[9,8,4,2,1]
5
14

Core Insight

What unlocks the optimal approach

Let's note that we can count all pairs with XOR ≤ K, so the answer would be to subtract the number of pairs withs XOR < low from the number of pairs with XOR ≤ high.
For each value, find out the number of values when you XOR it with the result is ≤ K using a trie.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1803: Count Pairs With XOR in a Range
class Trie {
    private Trie[] children = new Trie[2];
    private int cnt;

    public void insert(int x) {
        Trie node = this;
        for (int i = 15; i >= 0; --i) {
            int v = (x >> i) & 1;
            if (node.children[v] == null) {
                node.children[v] = new Trie();
            }
            node = node.children[v];
            ++node.cnt;
        }
    }

    public int search(int x, int limit) {
        Trie node = this;
        int ans = 0;
        for (int i = 15; i >= 0 && node != null; --i) {
            int v = (x >> i) & 1;
            if (((limit >> i) & 1) == 1) {
                if (node.children[v] != null) {
                    ans += node.children[v].cnt;
                }
                node = node.children[v ^ 1];
            } else {
                node = node.children[v];
            }
        }
        return ans;
    }
}

class Solution {
    public int countPairs(int[] nums, int low, int high) {
        Trie trie = new Trie();
        int ans = 0;
        for (int x : nums) {
            ans += trie.search(x, high + 1) - trie.search(x, low);
            trie.insert(x);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1803: Count Pairs With XOR in a Range
type Trie struct {
	children [2]*Trie
	cnt      int
}

func newTrie() *Trie {
	return &Trie{}
}

func (this *Trie) insert(x int) {
	node := this
	for i := 15; i >= 0; i-- {
		v := (x >> i) & 1
		if node.children[v] == nil {
			node.children[v] = newTrie()
		}
		node = node.children[v]
		node.cnt++
	}
}

func (this *Trie) search(x, limit int) (ans int) {
	node := this
	for i := 15; i >= 0 && node != nil; i-- {
		v := (x >> i) & 1
		if (limit >> i & 1) == 1 {
			if node.children[v] != nil {
				ans += node.children[v].cnt
			}
			node = node.children[v^1]
		} else {
			node = node.children[v]
		}
	}
	return
}

func countPairs(nums []int, low int, high int) (ans int) {
	tree := newTrie()
	for _, x := range nums {
		ans += tree.search(x, high+1) - tree.search(x, low)
		tree.insert(x)
	}
	return
}

# Accepted solution for LeetCode #1803: Count Pairs With XOR in a Range
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

// Accepted solution for LeetCode #1803: Count Pairs With XOR in a Range
/**
 * [1803] Count Pairs With XOR in a Range
 *
 * 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
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-pairs-with-xor-in-a-range/
// discuss: https://leetcode.com/problems/count-pairs-with-xor-in-a-range/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_pairs(nums: Vec<i32>, low: i32, high: i32) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1803_example_1() {
        let nums = vec![1, 4, 2, 7];
        let low = 2;
        let high = 6;

        let result = 6;

        assert_eq!(Solution::count_pairs(nums, low, high), result);
    }

    #[test]
    #[ignore]
    fn test_1803_example_2() {
        let nums = vec![9, 8, 4, 2, 1];
        let low = 5;
        let high = 14;

        let result = 8;

        assert_eq!(Solution::count_pairs(nums, low, high), result);
    }
}

// Accepted solution for LeetCode #1803: Count Pairs With XOR in a Range
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1803: Count Pairs With XOR in a Range
// class Trie {
//     private Trie[] children = new Trie[2];
//     private int cnt;
// 
//     public void insert(int x) {
//         Trie node = this;
//         for (int i = 15; i >= 0; --i) {
//             int v = (x >> i) & 1;
//             if (node.children[v] == null) {
//                 node.children[v] = new Trie();
//             }
//             node = node.children[v];
//             ++node.cnt;
//         }
//     }
// 
//     public int search(int x, int limit) {
//         Trie node = this;
//         int ans = 0;
//         for (int i = 15; i >= 0 && node != null; --i) {
//             int v = (x >> i) & 1;
//             if (((limit >> i) & 1) == 1) {
//                 if (node.children[v] != null) {
//                     ans += node.children[v].cnt;
//                 }
//                 node = node.children[v ^ 1];
//             } else {
//                 node = node.children[v];
//             }
//         }
//         return ans;
//     }
// }
// 
// class Solution {
//     public int countPairs(int[] nums, int low, int high) {
//         Trie trie = new Trie();
//         int ans = 0;
//         for (int x : nums) {
//             ans += trie.search(x, high + 1) - trie.search(x, low);
//             trie.insert(x);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(n × log M)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.