LeetCode #3209 — HARD

Number of Subarrays With AND Value of K

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

The Problem

Problem Statement

Given an array of integers nums and an integer k, return the number of subarrays of nums where the bitwise AND of the elements of the subarray equals k.

Example 1:

Input: nums = [1,1,1], k = 1

Output: 6

Explanation:

All subarrays contain only 1's.

Example 2:

Input: nums = [1,1,2], k = 1

Output: 3

Explanation:

Subarrays having an AND value of 1 are: [1,1,2], [1,1,2], [1,1,2].

Example 3:

Input: nums = [1,2,3], k = 2

Output: 2

Explanation:

Subarrays having an AND value of 2 are: [1,2,3], [1,2,3].

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i], k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given an array of integers nums and an integer k, return the number of subarrays of nums where the bitwise AND of the elements of the subarray equals k.

Baseline thinking

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

Pattern signal: Array · Binary Search · Bit Manipulation · Segment Tree

Example 1

[1,1,1]
1

Example 2

[1,1,2]
1

Example 3

[1,2,3]
2

Step 02

Core Insight

What unlocks the optimal approach

Let’s say we want to count the number of pairs <code>(l, r)</code> such that <code>nums[l] & nums[l + 1] & … & nums[r] == k</code>.
Fix the left index <code>l</code>.
Note that if you increase <code>r</code> for a fixed <code>l</code>, then the AND value of the subarray either decreases or remains unchanged.
Therefore, consider using binary search.
To calculate the AND value of a subarray, use sparse tables.

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 #3209: Number of Subarrays With AND Value of K
class Solution {
    public long countSubarrays(int[] nums, int k) {
        long ans = 0;
        Map<Integer, Integer> pre = new HashMap<>();
        for (int x : nums) {
            Map<Integer, Integer> cur = new HashMap<>();
            for (var e : pre.entrySet()) {
                int y = e.getKey(), v = e.getValue();
                cur.merge(x & y, v, Integer::sum);
            }
            cur.merge(x, 1, Integer::sum);
            ans += cur.getOrDefault(k, 0);
            pre = cur;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3209: Number of Subarrays With AND Value of K
func countSubarrays(nums []int, k int) (ans int64) {
	pre := map[int]int{}
	for _, x := range nums {
		cur := map[int]int{}
		for y, v := range pre {
			cur[x&y] += v
		}
		cur[x]++
		ans += int64(cur[k])
		pre = cur
	}
	return
}

# Accepted solution for LeetCode #3209: Number of Subarrays With AND Value of K
class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        ans = 0
        pre = Counter()
        for x in nums:
            cur = Counter()
            for y, v in pre.items():
                cur[x & y] += v
            cur[x] += 1
            ans += cur[k]
            pre = cur
        return ans

// Accepted solution for LeetCode #3209: Number of Subarrays With AND Value of K
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3209: Number of Subarrays With AND Value of K
// class Solution {
//     public long countSubarrays(int[] nums, int k) {
//         long ans = 0;
//         Map<Integer, Integer> pre = new HashMap<>();
//         for (int x : nums) {
//             Map<Integer, Integer> cur = new HashMap<>();
//             for (var e : pre.entrySet()) {
//                 int y = e.getKey(), v = e.getValue();
//                 cur.merge(x & y, v, Integer::sum);
//             }
//             cur.merge(x, 1, Integer::sum);
//             ans += cur.getOrDefault(k, 0);
//             pre = cur;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3209: Number of Subarrays With AND Value of K
function countSubarrays(nums: number[], k: number): number {
    let ans = 0;
    let pre = new Map<number, number>();
    for (const x of nums) {
        const cur = new Map<number, number>();
        for (const [y, v] of pre) {
            const z = x & y;
            cur.set(z, (cur.get(z) || 0) + v);
        }
        cur.set(x, (cur.get(x) || 0) + 1);
        ans += cur.get(k) || 0;
        pre = cur;
    }
    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(log M)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.