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Leetcode #4: Median of Two Sorted Arrays

In this guide, we solve Leetcode #4 Median of Two Sorted Arrays in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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)).

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Array, Binary Search, Divide and Conquer

Intuition

The problem structure suggests a monotonic decision, which makes binary search a natural fit.

By halving the search space each step, we reach the answer efficiently.

Approach

Search either directly on a sorted array or on the answer space using a check function.

Each check is fast, and the logarithmic search keeps the overall runtime low.

Steps:

  • Define the search bounds.
  • Check the mid point condition.
  • Narrow the bounds until convergence.

Example

Input: nums1 = [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2.

Python Solution

class Solution: def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float: def f(i: int, j: int, k: int) -> int: if i >= m: return nums2[j + k - 1] if j >= n: return nums1[i + k - 1] if k == 1: return min(nums1[i], nums2[j]) p = k // 2 x = nums1[i + p - 1] if i + p - 1 < m else inf y = nums2[j + p - 1] if j + p - 1 < n else inf return f(i + p, j, k - p) if x < y else f(i, j + p, k - p) m, n = len(nums1), len(nums2) a = f(0, 0, (m + n + 1) // 2) b = f(0, 0, (m + n + 2) // 2) return (a + b) / 2

Complexity

The time complexity is O(log⁡(m+n))O(\log(m + n))O(log(m+n)), and the space complexity is O(log⁡(m+n))O(\log(m + n))O(log(m+n)). The space complexity is O(log⁡(m+n))O(\log(m + n))O(log(m+n)).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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