What is the time complexity of Shortest Subarray With OR at Least K I?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Shortest Subarray With OR at Least K I?

This problem uses Array, Bit Manipulation, Sliding Window. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Shortest Subarray With OR at Least K I?

The problem can be efficiently solved using a sliding window approach, although a brute-force approach is also feasible given the constraints. Bitwise OR operations are crucial; understanding their properties (associativity, commutativity) is essential.

Which companies ask LeetCode 3095. Shortest Subarray With OR at Least K I in interviews?

This problem is frequently asked at Mitsogo.

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Shortest Subarray With OR at Least K I - LeetCode 3095 Solution

Shortest Subarray With OR at Least K I - Complete Solution Guide

Shortest Subarray With OR at Least K I is LeetCode problem 3095, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array nums of non-negative integers and an integer k . An array is called special if the bitwise OR of all of its elements is at least k . Return the length of the shortest special non-empty subarray of nums , or return -1 if no special subarray exists . Example 1: Input: nums = [1,2,3], k = 2 Output: 1 Explanation: The subarray [3] has OR value of 3 . Hence, we return 1 . Note that [2] is also a special subarray. Example 2: Input: nums = [2,1,8], k = 10 Output: 3 Explanation: T

Detailed Explanation

The problem asks you to find the length of the shortest subarray within a given array `nums` such that the bitwise OR of all elements in that subarray is at least `k`. The input is an array of non-negative integers (`nums`) and an integer (`k`). The output is the length of the shortest subarray satisfying the condition, or -1 if no such subarray exists. Subarrays must be non-empty. The constraints limit the array size and element values to relatively small ranges.

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible subarrays of `nums`. For each subarray, it calculates the bitwise OR of its elements. If the OR result is greater than or equal to `k`, it updates the minimum length seen so far. The algorithm then returns the minimum length or -1 if no subarray meets the criteria.

Step-by-Step Algorithm

Initialize `min_len` to infinity (or a large value).
Iterate through all possible starting indices `i` of subarrays (outer loop).
For each starting index `i`, iterate through all possible ending indices `j` (inner loop).
Calculate the bitwise OR (`current_or`) of elements from `nums[i]` to `nums[j]` (inclusive).
If `current_or` is greater than or equal to `k`, update `min_len` to the minimum of `min_len` and `(j - i + 1)`.
After checking all subarrays, return `min_len` if it's less than infinity; otherwise, return -1.

Key Insights

The problem can be efficiently solved using a sliding window approach, although a brute-force approach is also feasible given the constraints.
Bitwise OR operations are crucial; understanding their properties (associativity, commutativity) is essential.
The solution needs to handle the case where no subarray satisfies the condition (returning -1).

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Bit Manipulation, Sliding Window.

Companies

Asked at: Mitsogo.

Shortest Subarray With OR at Least K I | LeetCode 3095

Easy

Array Bit Manipulation Sliding Window

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimumSubarrayLength(int[] nums, int k) {
        int minLen = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; i++) {
            int currentOr = 0;
            for (int j = i; j < nums.length; j++) {
                currentOr |= nums[j];
                if (currentOr >= k) {
                    minLen = Math.min(minLen, j - i + 1);
                    break;
                }
            }
        }
        return minLen == Integer.MAX_VALUE ? -1 : minLen;
    }
}

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Problem Info

Description

Problem 3095: Shortest Subarray With OR at Least K I - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Sliding Window

Companies

Mitsogo

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3095

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Initialize `min_len` to infinity (or a large value).
2
Iterate through all possible starting indices `i` of subarrays (outer loop).
3
For each starting index `i`, iterate through all possible ending indices `j` (inner loop).
4
Calculate the bitwise OR (`current_or`) of elements from `nums[i]` to `nums[j]` (inclusive).
5
If `current_or` is greater than or equal to `k`, update `min_len` to the minimum of `min_len` and `(j - i + 1)`.
6
After checking all subarrays, return `min_len` if it's less than infinity; otherwise, return -1.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through all possible subarrays, resulting in a time complexity of O(n^2), where n is the length of the input array. The space complexity is O(1) because only a few variables are used regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Optimized Sliding Window (Potentially)

While not directly implemented in the given code, a sliding window approach *could* be more efficient. This would involve maintaining a window and updating the bitwise OR as the window slides. If the OR falls below k, the window expands; otherwise, the minimum length is updated.

Time: Potentially O(n)Space: O(1)

Trade-offs: A sliding window would be more efficient for large input arrays, but the provided brute-force approach is sufficient given the constraints in the problem description.

✨

Key Insights

1The problem can be efficiently solved using a sliding window approach, although a brute-force approach is also feasible given the constraints.
2Bitwise OR operations are crucial; understanding their properties (associativity, commutativity) is essential.
3The solution needs to handle the case where no subarray satisfies the condition (returning -1).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Empty input array: The code correctly handles this by implicitly returning -1 because the outer loop won't execute.
No subarray satisfying the condition: The code correctly handles this by initializing `min_len` to a large value and returning -1 if `min_len` remains unchanged.

⚠️

Common Mistakes to Avoid

✗Incorrectly handling the bitwise OR operation: Beginners might mistakenly use other logical or arithmetic operators instead of the bitwise OR operator (`|`).

✗Forgetting to handle the case where no subarray satisfies the condition: The code needs a mechanism to return -1 when no subarray has a bitwise OR at least `k`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Network routing: Finding the shortest path between nodes where the 'OR' represents a combination of network properties (e.g., bandwidth, latency).

🏦Data compression: Identifying the smallest data segment with a specific characteristic based on a bitwise OR of properties.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Shortest Subarray With OR at Least K I"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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