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

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

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

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

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

Insight 1: The problem can be efficiently solved using the sliding window technique due to the contiguous subarray requirement. Insight 2: Bitwise OR operation is monotonic, i.e., adding more elements to a subarray can only increase its OR value (or keep it the same).

Which companies ask LeetCode 3097. Shortest Subarray With OR at Least K II in interviews?

This problem is frequently asked at Mitsogo.

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Shortest Subarray With OR at Least K II - LeetCode 3097 Solution

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

Shortest Subarray With OR at Least K II is LeetCode problem 3097, a Medium 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 . Example 2: Input: nums = [2,1,8], k = 10 Output: 3 Explanation: The subarray [2,1,8] has OR value of 11 . H

Detailed Explanation

The problem requires us to find the shortest subarray within a given array `nums` of non-negative integers whose bitwise OR is at least `k`. We need to return the length of this shortest subarray, or -1 if no such subarray exists. A subarray is considered 'special' if the bitwise OR of all its elements is greater than or equal to `k`. The constraints specify the size of `nums` and the range of values it contains.

Solution Approach

The solution uses a sliding window approach with two pointers, `left` and `right`, to define the current subarray. It maintains an array `bit_counts` to store the count of each bit (0 to 30) present in the current window. The `current_or` variable tracks the bitwise OR of the elements within the current window. As the right pointer moves through the array, the corresponding bits of the new element are added to the `bit_counts`. If a bit count becomes 1 from 0, it means this bit is added to the `current_or`. Then, the left pointer moves to shrink the window as long as the `current_or` remains greater than or equal to `k`, updating `bit_counts` and `current_or` accordingly. During the window shrinking process, the minimum length is tracked.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0, `min_length` to infinity, `current_or` to 0, and `bit_counts` to an array of 31 zeros.
Step 2: Iterate through the array `nums` using the `right` pointer (sliding window's right boundary).
Step 3: For each element at index `right`, update the `bit_counts` array by incrementing the count of each bit present in the element. If adding the element sets a bit to 1 for the first time in the window, update the `current_or` accordingly.
Step 4: While the `current_or` is greater than or equal to `k` and `left` is less than or equal to `right`, update `min_length` with the minimum between the current `min_length` and the length of the current window (`right - left + 1`).
Step 5: Shrink the window from the left by decrementing the `bit_counts` for the element at index `left`. If removing the element causes a bit to become 0 from 1 in the window, clear this bit in the `current_or` as well.
Step 6: Increment the `left` pointer to move the window's left boundary.
Step 7: After the loop finishes, return `min_length` if it was updated (i.e., not infinity), otherwise return -1.

Key Insights

Insight 1: The problem can be efficiently solved using the sliding window technique due to the contiguous subarray requirement.
Insight 2: Bitwise OR operation is monotonic, i.e., adding more elements to a subarray can only increase its OR value (or keep it the same).
Insight 3: Maintaining a count of the occurrences of each bit within the sliding window helps track the current OR value and determine when to shrink the window.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Mitsogo.

Shortest Subarray With OR at Least K II | LeetCode 3097

Medium

Array Bit Manipulation Sliding Window

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

Solution Code

Available in:

class Solution {
    public int minimumSubarrayLength(int[] nums, int k) {
        int n = nums.length;
        int min_length = Integer.MAX_VALUE;
        int left = 0;
        int[] bit_counts = new int[31];
        int current_or = 0;

        for (int right = 0; right < n; right++) {
            int num_r = nums[right];
            for (int i = 0; i < 31; i++) {
                if ((num_r >> i) % 2 == 1) {
                    if (bit_counts[i] == 0) {
                        current_or |= (1 << i);
                    }
                    bit_counts[i] += 1;
                }
            }

            while (left <= right && current_or >= k) {
                min_length = Math.min(min_length, right - left + 1);

                int num_l = nums[left];
                for (int i = 0; i < 31; i++) {
                    if ((num_l >> i) % 2 == 1) {
                        bit_counts[i] -= 1;
                        if (bit_counts[i] == 0) {
                            current_or &= ~(1 << i);
                        }
                    }
                }
                left += 1;
            }
        }

        return min_length == Integer.MAX_VALUE ? -1 : min_length;
    }
}

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Problem Info

Description

Problem 3097: Shortest Subarray With OR at Least K II - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Sliding Window

Companies

Mitsogo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3097

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

7 steps

1
Step 1: Initialize `left` to 0, `min_length` to infinity, `current_or` to 0, and `bit_counts` to an array of 31 zeros.
2
Step 2: Iterate through the array `nums` using the `right` pointer (sliding window's right boundary).
3
Step 3: For each element at index `right`, update the `bit_counts` array by incrementing the count of each bit present in the element. If adding the element sets a bit to 1 for the first time in the window, update the `current_or` accordingly.
4
Step 4: While the `current_or` is greater than or equal to `k` and `left` is less than or equal to `right`, update `min_length` with the minimum between the current `min_length` and the length of the current window (`right - left + 1`).
5
Step 5: Shrink the window from the left by decrementing the `bit_counts` for the element at index `left`. If removing the element causes a bit to become 0 from 1 in the window, clear this bit in the `current_or` as well.
6
Step 6: Increment the `left` pointer to move the window's left boundary.
7
Step 7: After the loop finishes, return `min_length` if it was updated (i.e., not 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because each element of the array is visited at most twice (once by the `right` pointer and once by the `left` pointer). The space complexity is O(1) because the `bit_counts` array's size depends on the number of bits in an integer, not the size of the input array. All other used variables take constant space.

📖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

Brute-force approach

Iterate through all possible subarrays and compute the bitwise OR of each subarray. If the OR is at least `k`, update the minimum length found so far.

Time: O(n^2), where n is the length of the input array `nums`. There are O(n^2) possible subarrays, and calculating the OR of each takes O(n) in worst case.Space: O(1). Constant extra space.

Trade-offs: The brute-force approach is simple to understand but has a higher time complexity compared to the sliding window approach. It is suitable for small input sizes but becomes inefficient for larger inputs.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using the sliding window technique due to the contiguous subarray requirement.
2Insight 2: Bitwise OR operation is monotonic, i.e., adding more elements to a subarray can only increase its OR value (or keep it the same).
3Insight 3: Maintaining a count of the occurrences of each bit within the sliding window helps track the current OR value and determine when to shrink the window.

📚

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

Edge case 1: If the input array is empty, or no subarray with OR >= k is found, the algorithm correctly returns -1. This is because `min_length` remains `float('inf')` or `Integer.MAX_VALUE` or `INT_MAX` and the final check handles this scenario.
Edge case 2: If k is 0, the algorithm will find the shortest subarray of length 1 because any non-negative number has an OR >= 0. It iterates and sets min_length to 1 immediately.
Edge case 3: If the nums contains large numbers that can cause integer overflow during OR operation the algorithm will still work fine, since we do not store the accumulated OR of all elements. Instead we track the counts of each bit, which are safe.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where no special subarray exists, leading to an incorrect return value (not -1).

✗Common mistake 2: Incorrectly updating the `current_or` variable when adding or removing elements from the window, leading to inaccurate OR values and incorrect results.

✗Common mistake 3: Failing to update `bit_counts` correctly, leading to incorrect updates of the 'current_or' variable and incorrect results.

✗Common mistake 4: Integer overflow in 'current_or' if not handled carefully.

✅

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?

💼Real-world scenario 1: Analyzing network traffic to detect anomalies or security threats. You could use this type of algorithm to identify bursts of specific traffic patterns that exceed a predefined threshold (k).

🏦Real-world scenario 2: Finding the shortest period of time in sensor data where a certain combination of sensor readings exceeds a safety threshold. The bitwise OR can represent the combination of active sensors.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

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 8, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 II"?

For this medium-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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