LeetCode

Subtitle / Summary Maximum Subarray is the classic 1D DP / greedy template. This ACERS guide explains Kadane’s idea, engineering use cases, and runnable multi-language solutions. Reading time: 10–12 min Tags: Hot100, dynamic programming, greedy SEO keywords: Maximum Subarray, Kadane, dynamic programming, O(n), Hot100 Meta description: Kadane O(n) maximum subarray sum with engineering scenarios and multi-language code. A — Algorithm Problem Restatement Given an integer array nums, find the contiguous subarray with the largest sum (must contain at least one element) and return the sum. ...

Subtitle / Summary A classic event-spacing validation model. This ACERS guide explains the one-pass logic, engineering use cases, and runnable multi-language solutions. Reading time: 10–12 min Tags: array, two pointers, event spacing SEO keywords: LeetCode 1437, event spacing, O(n) Meta description: One-pass validation for minimum spacing between 1s, with engineering use cases and multi-language code. A — Algorithm Problem Restatement Given an integer array nums and integer k, return true if every pair of 1s is at least k apart; otherwise return false. ...

Subtitle / Summary A classic bit-manipulation template: determine if a number is a power of two in O(1). This ACERS guide covers the core insight, practical uses, and runnable multi-language implementations. Reading time: 8–12 min Tags: bit manipulation, binary, math SEO keywords: Power of Two, bit manipulation, binary, O(1), LeetCode 231 Meta description: O(1) power-of-two check using bit tricks, with engineering scenarios and multi-language code. A — Algorithm Problem Restatement Given an integer n, determine whether it is a power of two. Return true if it is; otherwise, return false. ...

Subtitle / Summary A standard fixed-window counting problem. This ACERS guide explains the sliding-window model, engineering use cases, and runnable multi-language solutions. Reading time: 10–12 min Tags: sliding window, string, fixed window SEO keywords: Maximum Number of Vowels, Sliding Window, Fixed Window Meta description: Fixed-window sliding count for maximum vowels with engineering applications. A — Algorithm Problem Restatement Given a string s and an integer k, return the maximum number of vowels in any substring of length k. ...

Title LeetCode 239: Sliding Window Maximum (Monotonic Queue ACERS Guide) Subtitle / Summary Sliding Window Maximum is the classic combo of sliding window + monotonic queue. This article follows the ACERS template with reusable engineering patterns and multi-language implementations. Estimated reading time: 12–15 minutes Tags: sliding window, monotonic queue, array SEO keywords: Sliding Window Maximum, monotonic queue, deque, O(n) Meta description: Monotonic-queue solution for Sliding Window Maximum with engineering practice and multi-language implementations. A — Algorithm (Problem & Algorithm) Problem Restatement Given an integer array nums and a window size k, a sliding window moves from left to right. Each move shifts the window by one. Return the maximum for each window. ...

Subtitle / Abstract A basic counting problem: use frequency + combinations to drop O(n^2) to O(n). Includes engineering use cases and portable implementations. Reading time: 8-10 minutes Tags: hash-table, counting, array SEO keywords: Good Pairs, hash map, frequency Meta description: Hash counting solution for Good Pairs with complexity and code. Target readers Beginners learning hash tables and counting Engineers who want to map interview patterns to real stats tasks Interview prep for basic counting models Background / Motivation Counting equal pairs is a classic problem. A double loop is O(n^2). With frequency counting, you can solve it in linear time and scale to large data. ...

LeetCode 1: Two Sum Summary Find two indices such that nums[i] + nums[j] = target. Use a hash map for O(n) time. Approach Iterate and store value -> index. For each number x, check if target - x exists. Complexity Time: O(n) Space: O(n) Python reference implementation def two_sum(nums, target): seen = {} for i, x in enumerate(nums): y = target - x if y in seen: return [seen[y], i] seen[x] = i

LeetCode 2300: Successful Pairs of Spells and Potions Summary For each spell, count how many potions make spell * potion >= success. Sort potions and binary search the threshold. Approach Sort potions. For each spell, compute need = ceil(success / spell). Use binary search to find the first potion >= need. Complexity Time: O(n log n) Space: O(1) extra (or O(n) if sorting a copy) Python reference implementation import bisect import math def successful_pairs(spells, potions, success): potions = sorted(potions) n = len(potions) res = [] for s in spells: need = (success + s - 1) // s idx = bisect.bisect_left(potions, need) res.append(n - idx) return res

LeetCode 2379: Minimum Recolors to Get K Consecutive Black Blocks Summary Given a string of ‘B’ and ‘W’, find the minimum recolors to make a substring of length k all black. Approach Use a sliding window of length k and count the number of whites in the window. The minimum whites across all windows is the answer. Complexity Time: O(n) Space: O(1) Python reference implementation def minimum_recolors(blocks, k): whites = sum(1 for c in blocks[:k] if c == 'W') ans = whites for i in range(k, len(blocks)): if blocks[i-k] == 'W': whites -= 1 if blocks[i] == 'W': whites += 1 ans = min(ans, whites) return ans

LeetCode 2841: Maximum Sum of Almost Unique Subarray Summary Given an array, window size k, and threshold m, find the maximum sum of any length-k subarray that contains at least m distinct elements. Approach Use a sliding window with a frequency map, track window sum and number of distinct values. Complexity Time: O(n) Space: O(n) for frequency map Python reference implementation def max_sum_almost_unique(nums, m, k): from collections import defaultdict count = defaultdict(int) distinct = 0 window_sum = 0 ans = 0 for i, x in enumerate(nums): window_sum += x if count[x] == 0: distinct += 1 count[x] += 1 if i >= k: y = nums[i - k] window_sum -= y count[y] -= 1 if count[y] == 0: distinct -= 1 if i >= k - 1 and distinct >= m: ans = max(ans, window_sum) return ans