What is the time complexity of Median of Two Sorted Arrays?

The optimal solution has a time complexity of O(log(min(m,n))).

What data structures are used to solve Median of Two Sorted Arrays?

This problem uses Array, Binary Search, Divide and Conquer. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Median of Two Sorted Arrays?

Insight 1: Direct merging and sorting would result in O(m+n) time complexity, violating the logarithmic time constraint. Binary search is essential. Insight 2: The problem can be reframed as finding the correct partition points in both arrays such that all elements to the left of the partitions are smaller than all elements to the right.

Which companies ask LeetCode 4. Median of Two Sorted Arrays in interviews?

This problem is frequently asked at Accenture, Adobe, Akamai, Amazon, Apple and 37 more companies.

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Median of Two Sorted Arrays - LeetCode 4 Solution

Q: How to solve Median of Two Sorted Arrays?

The solution employs a binary search approach to find the optimal partition points in the two sorted arrays. By partitioning the arrays, the algorithm seeks to satisfy the condition where the maximum element on the left side of the partition in both arrays is less than or equal to the minimum element on the right side. This indicates that the elements are correctly 'merged' such that the median can be calculated. We perform binary search on the smaller array to optimize the search space.

Median of Two Sorted Arrays - Complete Solution Guide

Median of Two Sorted Arrays is LeetCode problem 4, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)) . Example 1: Input: nums1 = [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2. Example 2: Input: nums1 = [1,2], nums2 = [3,4] Output: 2.50000 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. Constraints: nums1.length == m nums2.length == n 0 <= m <= 1000 0 <= n <= 1000 1

Detailed Explanation

The problem requires finding the median of two sorted arrays, `nums1` and `nums2`. The median is the middle element (or the average of the two middle elements if the total number of elements is even) when the combined array is sorted. The crucial constraint is to achieve this with a time complexity of O(log(m+n)), where m and n are the lengths of the input arrays.

Solution Approach

The solution employs a binary search approach to find the optimal partition points in the two sorted arrays. By partitioning the arrays, the algorithm seeks to satisfy the condition where the maximum element on the left side of the partition in both arrays is less than or equal to the minimum element on the right side. This indicates that the elements are correctly 'merged' such that the median can be calculated. We perform binary search on the smaller array to optimize the search space.

Step-by-Step Algorithm

Step 1: Ensure that `nums1` is the shorter array. If not, swap `nums1` and `nums2`.
Step 2: Initialize `low` to 0 and `high` to the length of `nums1` (`m`). These represent the possible number of elements to take from `nums1` in the left partition.
Step 3: Calculate `half_len` as `(m + n + 1) // 2`. This represents the number of elements that should be in the left partition of the merged array.
Step 4: Perform binary search within the range `[low, high]`. In each iteration:
Step 5: Calculate `partition1` as `(low + high) // 2`. This is the number of elements taken from `nums1` in the left partition.
Step 6: Calculate `partition2` as `half_len - partition1`. This ensures that the total number of elements in the left partitions of both arrays is `half_len`.
Step 7: Determine `maxLeft1`, `minRight1`, `maxLeft2`, and `minRight2`. These are the maximum element to the left and the minimum element to the right of the partitions in each array. Handle edge cases where partitions are at the beginning or end of the arrays by assigning negative or positive infinity respectively.
Step 8: Check if `maxLeft1 <= minRight2` and `maxLeft2 <= minRight1`. If this condition is met, the correct partition is found.
Step 9: If the condition in Step 8 is met, check if the total length of both arrays is even or odd.
Step 10: If the total length is even, the median is the average of the maximum of `maxLeft1` and `maxLeft2` and the minimum of `minRight1` and `minRight2`.
Step 11: If the total length is odd, the median is the maximum of `maxLeft1` and `maxLeft2`.
Step 12: If `maxLeft1 > minRight2`, move the `high` pointer to `partition1 - 1` to search in the left half.
Step 13: Otherwise, move the `low` pointer to `partition1 + 1` to search in the right half.
Step 14: Repeat steps 4-13 until the correct partition is found.

Key Insights

Insight 1: Direct merging and sorting would result in O(m+n) time complexity, violating the logarithmic time constraint. Binary search is essential.
Insight 2: The problem can be reframed as finding the correct partition points in both arrays such that all elements to the left of the partitions are smaller than all elements to the right.
Insight 3: Swapping arrays to ensure that binary search is performed on the smaller array guarantees a tighter upper bound for the search, improving efficiency.

Complexity Analysis

Time Complexity: O(log(min(m,n)))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Divide and Conquer.

Companies

Asked at: Accenture, Adobe, Akamai, Amazon, Apple, Autodesk, Bloomberg, Capgemini, Citadel, Cognizant, DE Shaw, Dropbox, Flipkart, Goldman Sachs, Google, IBM, Infosys, Intuit, LinkedIn, Meta, Microsoft, Oracle, Palo Alto Networks, PayPal, PornHub, Pwc, Rippling, ServiceNow, Snap, Swiggy, TikTok, Uber, VMware, Visa, Walmart Labs, Wipro, Wix, Yahoo, Yandex, Zenefits, Zoho, eBay.

Median of Two Sorted Arrays | LeetCode 4

Hard

Array Binary Search Divide and Conquer

Time: O(log(min(m,n)))Space: O(1)

Solution Code

Available in:

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if (nums1.length > nums2.length) {
            int[] temp = nums1;
            nums1 = nums2;
            nums2 = temp;
        }

        int m = nums1.length;
        int n = nums2.length;
        int low = 0;
        int high = m;

        int half_len = (m + n + 1) / 2;

        while (low <= high) {
            int partition1 = (low + high) / 2;
            int partition2 = half_len - partition1;

            int maxLeft1 = (partition1 > 0) ? nums1[partition1 - 1] : Integer.MIN_VALUE;
            int minRight1 = (partition1 < m) ? nums1[partition1] : Integer.MAX_VALUE;

            int maxLeft2 = (partition2 > 0) ? nums2[partition2 - 1] : Integer.MIN_VALUE;
            int minRight2 = (partition2 < n) ? nums2[partition2] : Integer.MAX_VALUE;

            if (maxLeft1 <= minRight2 && maxLeft2 <= minRight1) {
                if ((m + n) % 2 == 0) {
                    return (double)(Math.max(maxLeft1, maxLeft2) + Math.min(minRight1, minRight2)) / 2.0;
                } else {
                    return (double)Math.max(maxLeft1, maxLeft2);
                }
            } else if (maxLeft1 > minRight2) {
                high = partition1 - 1;
            } else {
                low = partition1 + 1;
            }
        }
        return 0.0; // should never reach here, but required for compilation
    }
}

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Problem Info

Description

Problem 4: Median of Two Sorted Arrays - Detailed explanation and solution approaches.

Topics

Array Binary Search Divide and Conquer

Companies

AccentureAdobeAkamaiAmazonAppleAutodeskBloombergCapgemini

Stats

Difficulty:Hard

Languages:4

Companies:42

Problem #:4

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

14 steps

1
Step 1: Ensure that `nums1` is the shorter array. If not, swap `nums1` and `nums2`.
2
Step 2: Initialize `low` to 0 and `high` to the length of `nums1` (`m`). These represent the possible number of elements to take from `nums1` in the left partition.
3
Step 3: Calculate `half_len` as `(m + n + 1) // 2`. This represents the number of elements that should be in the left partition of the merged array.
4
Step 4: Perform binary search within the range `[low, high]`. In each iteration:
5
Step 5: Calculate `partition1` as `(low + high) // 2`. This is the number of elements taken from `nums1` in the left partition.
6
Step 6: Calculate `partition2` as `half_len - partition1`. This ensures that the total number of elements in the left partitions of both arrays is `half_len`.
7
Step 7: Determine `maxLeft1`, `minRight1`, `maxLeft2`, and `minRight2`. These are the maximum element to the left and the minimum element to the right of the partitions in each array. Handle edge cases where partitions are at the beginning or end of the arrays by assigning negative or positive infinity respectively.
8
Step 8: Check if `maxLeft1 <= minRight2` and `maxLeft2 <= minRight1`. If this condition is met, the correct partition is found.
9
Step 9: If the condition in Step 8 is met, check if the total length of both arrays is even or odd.
10
Step 10: If the total length is even, the median is the average of the maximum of `maxLeft1` and `maxLeft2` and the minimum of `minRight1` and `minRight2`.
11
Step 11: If the total length is odd, the median is the maximum of `maxLeft1` and `maxLeft2`.
12
Step 12: If `maxLeft1 > minRight2`, move the `high` pointer to `partition1 - 1` to search in the left half.
13
Step 13: Otherwise, move the `low` pointer to `partition1 + 1` to search in the right half.
14
Step 14: Repeat steps 4-13 until the correct partition is found.

💡

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(log(min(m,n)))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the binary search. Since the binary search is performed on the smaller of the two arrays, the maximum number of iterations is logarithmic with respect to the size of the smaller array. The other operations (comparisons, calculations) take constant time.

📖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

Merging and Sorting

Merge the two sorted arrays into a single sorted array and then find the median. This approach is straightforward but less efficient.

Time: O(m+n log(m+n)) for sorting or O(m+n) if we merge in sorted order and find the median simultaneously. The time complexity comes from the sorting or merging procedure.Space: O(m+n) to store the merged array.

Trade-offs: This approach is simpler to implement but has a higher time complexity than the binary search approach. It is suitable for small arrays but inefficient for large arrays.

Iterative Merging

Iteratively merge the arrays and keep track of the indices necessary to calculate the median. This avoids creating a new array.

Time: O(m+n). In the worst case, we need to iterate through all elements of both arrays to find the median.Space: O(1). We only use a constant amount of extra space.

Trade-offs: This is simpler than binary search but not as efficient for large arrays when the target time complexity is O(log(n)).

✨

Key Insights

1Insight 1: Direct merging and sorting would result in O(m+n) time complexity, violating the logarithmic time constraint. Binary search is essential.
2Insight 2: The problem can be reframed as finding the correct partition points in both arrays such that all elements to the left of the partitions are smaller than all elements to the right.
3Insight 3: Swapping arrays to ensure that binary search is performed on the smaller array guarantees a tighter upper bound for the search, improving efficiency.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: One of the arrays is empty. The algorithm handles this correctly because the binary search is performed on the smaller array. When one array is empty, binary search essentially finds the median in the non-empty array.
Edge case 2: `partition1` or `partition2` is 0 or equal to the array's length. This is handled by assigning `-infinity` to `maxLeft1` or `maxLeft2` and `infinity` to `minRight1` or `minRight2`, respectively, when the partition is at the beginning or the end of the array.
Edge case 3: All elements in one array are smaller than all elements in the other array. The binary search will eventually find the correct partition points to determine the median.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling edge cases where `partition1` or `partition2` is 0 or equal to the length of the array. This can lead to out-of-bounds array access errors.

✗Common mistake 2: Failing to ensure that the binary search is performed on the smaller array. This does not affect the correctness but increases the number of iterations and might be considered less efficient.

✗Common mistake 3: Incorrectly calculating `half_len`. It should be `(m + n + 1) // 2` to handle both even and odd length cases properly.

✗Common mistake 4: Not correctly handling integer overflow when calculating the median of two integers. Casting to double before averaging is essential.

✅

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: Database systems often need to perform operations on sorted data. Finding the median of data distributed across multiple sorted partitions is a direct application.

🏦Real-world scenario 2: Statistical analysis often involves finding percentiles (including the median) of large datasets. This algorithm can be adapted to efficiently find other percentiles as well.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: June 5, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Median of Two Sorted Arrays"?

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

🤖 Need help with this problem?