LeetCode

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