LeetCode #2275 — MEDIUM

Largest Combination With Bitwise AND Greater Than Zero

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

The bitwise AND of an array nums is the bitwise AND of all integers in nums.

  • For example, for nums = [1, 5, 3], the bitwise AND is equal to 1 & 5 & 3 = 1.
  • Also, for nums = [7], the bitwise AND is 7.

You are given an array of positive integers candidates. Compute the bitwise AND for all possible combinations of elements in the candidates array.

Return the size of the largest combination of candidates with a bitwise AND greater than 0.

Example 1:

Input: candidates = [16,17,71,62,12,24,14]
Output: 4
Explanation: The combination [16,17,62,24] has a bitwise AND of 16 & 17 & 62 & 24 = 16 > 0.
The size of the combination is 4.
It can be shown that no combination with a size greater than 4 has a bitwise AND greater than 0.
Note that more than one combination may have the largest size.
For example, the combination [62,12,24,14] has a bitwise AND of 62 & 12 & 24 & 14 = 8 > 0.

Example 2:

Input: candidates = [8,8]
Output: 2
Explanation: The largest combination [8,8] has a bitwise AND of 8 & 8 = 8 > 0.
The size of the combination is 2, so we return 2.

Constraints:

  • 1 <= candidates.length <= 105
  • 1 <= candidates[i] <= 107
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: The bitwise AND of an array nums is the bitwise AND of all integers in nums. For example, for nums = [1, 5, 3], the bitwise AND is equal to 1 & 5 & 3 = 1. Also, for nums = [7], the bitwise AND is 7. You are given an array of positive integers candidates. Compute the bitwise AND for all possible combinations of elements in the candidates array. Return the size of the largest combination of candidates with a bitwise AND greater than 0.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

[16,17,71,62,12,24,14]

Example 2

[8,8]

Related Problems

  • Count Number of Maximum Bitwise-OR Subsets (count-number-of-maximum-bitwise-or-subsets)
Step 02

Core Insight

What unlocks the optimal approach

  • For the bitwise AND to be greater than zero, at least one bit should be 1 for every number in the combination.
  • The candidates are 24 bits long, so for every bit position, we can calculate the size of the largest combination such that the bitwise AND will have a 1 at that bit position.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Upper-end input sizes
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 #2275: Largest Combination With Bitwise AND Greater Than Zero
class Solution {
    public int largestCombination(int[] candidates) {
        int mx = Arrays.stream(candidates).max().getAsInt();
        int m = Integer.SIZE - Integer.numberOfLeadingZeros(mx);
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            int cnt = 0;
            for (int x : candidates) {
                cnt += x >> i & 1;
            }
            ans = Math.max(ans, cnt);
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.