Redddy's Devlog

🔈 문제You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer $k$. Each minute, you may take either the leftmost character of s, or the rightmost character of s.Return the minimum number of minutes needed for you to take at least $k$ of each character, or return -1 if it is not possible to take $k$ of each character. 💫 예시Input: s = "aabaaaacaabc", k = 2..

🔈 문제Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs.A pair (i, j) is fair if:0  💫 예시Input: nums = [0,1,7,4,4,5], lower = 3, upper = 6 Output: 6 Explanation: There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5). 🎱 제한1 nums.length == n$-10^9$ $-10^9$  📚 문제 풀이 문제를 요약하자면 nums에서 i, j 를 골라(i != j)  더했을 때 lower 보다 같거나 ..

🔈 문제You are given an array of people, people, which are the attributes of some people in a queue (not necessarily in order). Each people[i] = [$h_i$, $k_i$] represents the $i^{th}$ person of height $h_i$ with exactly $k_i$ other people in front who have a height greater than or equal to $h_i$.Reconstruct and return the queue that is represented by the input array people. The returned queue shou..

🔈 문제 You are given a 0-indexed integer array nums of size n, and a 0-indexed integer array pattern of size m consisting of integers -1, 0, and 1. A subarray nums[i..j] of size m + 1 is said to match the pattern if the following conditions hold for each element pattern[k]: nums[i + k + 1] > nums[i + k] if pattern[k] == 1. nums[i + k + 1] == nums[i + k] if pattern[k] == 0. nums[i + k + 1] < nums[i..

🔈 문제 Implement the RandomizedSet class: RandomizedSet() Initializes the RandomizedSet object. bool insert(int val) Inserts an item val into the set if not present. Returns true if the item was not present, false otherwise. bool remove(int val) Removes an item val from the set if present. Returns true if the item was present, false otherwise. int getRandom() Returns a random element from the curr..

🔈 문제 Two strings are considered close if you can attain one from the other using the following operations: Operation 1: Swap any two existing characters.For example, abcde -> aecdb Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character. For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into ..