LeetCode #2568 — MEDIUM

Minimum Impossible OR

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

The Problem

Problem Statement

You are given a 0-indexed integer array nums.

We say that an integer x is expressible from nums if there exist some integers 0 <= index₁ < index₂ < ... < index_k < nums.length for which nums[index₁] | nums[index₂] | ... | nums[index_k] = x. In other words, an integer is expressible if it can be written as the bitwise OR of some subsequence of nums.

Return the minimum positive non-zero integer that is not expressible from nums.

Example 1:

Input: nums = [2,1]
Output: 4
Explanation: 1 and 2 are already present in the array. We know that 3 is expressible, since nums[0] | nums[1] = 2 | 1 = 3. Since 4 is not expressible, we return 4.

Example 2:

Input: nums = [5,3,2]
Output: 1
Explanation: We can show that 1 is the smallest number that is not expressible.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. We say that an integer x is expressible from nums if there exist some integers 0 <= index1 < index2 < ... < indexk < nums.length for which nums[index1] | nums[index2] | ... | nums[indexk] = x. In other words, an integer is expressible if it can be written as the bitwise OR of some subsequence of nums. Return the minimum positive non-zero integer that is not expressible from nums.

Baseline thinking

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

Pattern signal: Array · Bit Manipulation

Example 1

[2,1]

Example 2

[5,3,2]

Step 02

Core Insight

What unlocks the optimal approach

Think about forming numbers in the powers of 2 using their bit representation.
The minimum power of 2 not present in the array will be the first number that could not be expressed using the given operation.

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

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 #2568: Minimum Impossible OR
class Solution {
    public int minImpossibleOR(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int x : nums) {
            s.add(x);
        }
        for (int i = 0;; ++i) {
            if (!s.contains(1 << i)) {
                return 1 << i;
            }
        }
    }
}

// Accepted solution for LeetCode #2568: Minimum Impossible OR
func minImpossibleOR(nums []int) int {
	s := map[int]bool{}
	for _, x := range nums {
		s[x] = true
	}
	for i := 0; ; i++ {
		if !s[1<<i] {
			return 1 << i
		}
	}
}

# Accepted solution for LeetCode #2568: Minimum Impossible OR
class Solution:
    def minImpossibleOR(self, nums: List[int]) -> int:
        s = set(nums)
        return next(1 << i for i in range(32) if 1 << i not in s)

// Accepted solution for LeetCode #2568: Minimum Impossible OR
// 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 #2568: Minimum Impossible OR
// class Solution {
//     public int minImpossibleOR(int[] nums) {
//         Set<Integer> s = new HashSet<>();
//         for (int x : nums) {
//             s.add(x);
//         }
//         for (int i = 0;; ++i) {
//             if (!s.contains(1 << i)) {
//                 return 1 << i;
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #2568: Minimum Impossible OR
function minImpossibleOR(nums: number[]): number {
    const s: Set<number> = new Set();
    for (const x of nums) {
        s.add(x);
    }
    for (let i = 0; ; ++i) {
        if (!s.has(1 << i)) {
            return 1 << i;
        }
    }
}

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)

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.