Apple interviews focus on the algorithm optimization process. This article demonstrates how to progressively optimize from a brute force solution to an optimal solution through a comprehensive problem. oavoservice helps you master optimization thinking.
📋 Problem: Longest Increasing Subsequence (LIS)
Given an array, find the length of the longest strictly increasing subsequence.
🎯 Solution Evolution
Method 1: Brute Force Recursion
def lengthOfLIS_v1(nums):
def dfs(i, prev):
if i == len(nums):
return 0
# Do not select current element
skip = dfs(i + 1, prev)
# Select current element
take = 0
if prev == -1 or nums[i] > nums[prev]:
take = 1 + dfs(i + 1, i)
return max(skip, take)
return dfs(0, -1)
Time: O(2^n)
Method 2: Memoization
def lengthOfLIS_v2(nums):
memo = {}
def dfs(i, prev):
if i == len(nums):
return 0
if (i, prev) in memo:
return memo[(i, prev)]
skip = dfs(i + 1, prev)
take = 0
if prev == -1 or nums[i] > nums[prev]:
take = 1 + dfs(i + 1, i)
memo[(i, prev)] = max(skip, take)
return memo[(i, prev)]
return dfs(0, -1)
Time: O(n²)
Method 3: Dynamic Programming
def lengthOfLIS_v3(nums):
if not nums:
return 0
dp = [1] * len(nums)
for i in range(1, len(nums)):
for j in range(i):
if nums[i] > nums[j]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)
Time: O(n²)
Method 4: Binary Search (Optimal)
def lengthOfLIS_v4(nums):
tails = []
for num in nums:
left, right = 0, len(tails)
while left < right:
mid = (left + right) // 2
if tails[mid] < num:
left = mid + 1
else:
right = mid
if left == len(tails):
tails.append(num)
else:
tails[left] = num
return len(tails)
Time: O(n log n)
💼 How oavoservice Helps
Optimization Path - Complete process from brute force to optimal. Complexity Analysis - Improvements at each step. Interview Skills - How to demonstrate your thought process.
Contact oavoservice for professional algorithm optimization interview assistance!
Tags: #Apple #AlgorithmOptimization #DynamicProgramming #BinarySearch #VOHelp #InterviewPrep #1point3acres
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