We have a wooden plank of the length n units. Some ants are walking on the plank, each ant moves with speed 1 unit per second. Some of the ants move to the left, the other move to the right. When two ants moving in two different directions meet at some point, they change their directions and continue moving again. Assume changing directions doesn’t take any additional time. ...
Given an array nums and an integer target. Return the maximum number of non-empty non-overlapping subarrays such that the sum of values in each subarray is equal to target. Example 1: Input: nums = [1,1,1,1,1], target = 2 Output: 2 Explanation: There are 2 non-overlapping subarrays [1,1,1,1,1] with sum equals to target(2). Example 2: Input: nums = [-1,3,5,1,4,2,-9], target = 6 Output: 2 Explanation: There are 3 subarrays with sum equal to 6. ([5,1], [4,2], [3,5,1,4,2,-9]) but only the first 2 are non-overlapping. Example 3: Input: nums = [-2,6,6,3,5,4,1,2,8], target = 10 Output: 3 Example 4: Input: nums = [0,0,0], target = 0 Output: 3 Constraints: ...
We can represent a sentence as an array of words, for example, the sentence “I am happy with leetcode” can be represented as arr = [“I”,“am”,happy",“with”,“leetcode”]. Given two sentences sentence1 and sentence2 each represented as a string array and given an array of string pairs similarPairs where similarPairs[i] = [xi, yi] indicates that the two words xi and yi are similar. Return true if sentence1 and sentence2 are similar, or false if they are not similar. ...
Given an array nums of integers, return the length of the longest arithmetic subsequence in nums. Recall that a subsequence of an array nums is a list nums[i1], nums[i2], …, nums[ik] with 0 <= i1 < i2 < … < ik <= nums.length - 1, and that a sequence seq is arithmetic if seq[i+1] - seq[i] are all the same value (for 0 <= i < seq.length - 1). ...
Given the root of a binary tree, determine if it is a complete binary tree. In a complete binary tree, every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h. Example 1: Input: root = [1,2,3,4,5,6] Output: true Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible. ...
You have a graph of n nodes labeled from 0 to n - 1. You are given an integer n and a list of edges where edges[i] = [ai, bi] indicates that there is an undirected edge between nodes ai and bi in the graph. Return true if the edges of the given graph make up a valid tree, and false otherwise. Example 1: Input: n = 5, edges = [[0,1],[0,2],[0,3],[1,4]] Output: true ...
You are given a 0-indexed array of positive integers w where w[i] describes the weight of the ith index. You need to implement the function pickIndex(), which randomly picks an index in the range [0, w.length - 1] (inclusive) and returns it. The probability of picking an index i is w[i] / sum(w). For example, if w = [1, 3], the probability of picking index 0 is 1 / (1 + 3) = 0.25 (i.e., 25%), and the probability of picking index 1 is 3 / (1 + 3) = 0.75 (i.e., 75%). Example 1: Input ["Solution","pickIndex"] [[[1]],[]] Output [null,0] Explanation Solution solution = new Solution([1]); solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w. Example 2: Input ["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"] [[[1,3]],[],[],[],[],[]] Output [null,1,1,1,1,0] Explanation Solution solution = new Solution([1, 3]); solution.pickIndex(); // return 1. It is returning the second element (index = 1) that has a probability of 3/4. solution.pickIndex(); // return 1 solution.pickIndex(); // return 1 solution.pickIndex(); // return 1 solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4. Since this is a randomization problem, multiple answers are allowed. All of the following outputs can be considered correct: [null,1,1,1,1,0] [null,1,1,1,1,1] [null,1,1,1,0,0] [null,1,1,1,0,1] [null,1,0,1,0,0] ...... and so on. Constraints: ...
Implement a thread-safe bounded blocking queue that has the following methods: BoundedBlockingQueue(int capacity) The constructor initializes the queue with a maximum capacity. void enqueue(int element) Adds an element to the front of the queue. If the queue is full, the calling thread is blocked until the queue is no longer full. int dequeue() Returns the element at the rear of the queue and removes it. If the queue is empty, the calling thread is blocked until the queue is no longer empty. int size() Returns the number of elements currently in the queue. Your implementation will be tested using multiple threads at the same time. Each thread will either be a producer thread that only makes calls to the enqueue method or a consumer thread that only makes calls to the dequeue method. The size method will be called after every test case. ...
You are given an m x n binary matrix grid. An island is a group of 1’s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other. ...
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time. Example 1: Input: nums = [100,4,200,1,3,2] Output: 4 Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. Example 2: Input: nums = [0,3,7,2,5,8,4,6,0,1] Output: 9 Constraints: 0 <= nums.length <= 105 -10^9 <= nums[i] <= 10^9 DFS Solution class Solution { public int longestConsecutive(int[] nums) { int n = nums.length; if (n == 0) return 0; Arrays.sort(nums); int max = 1; int currentMax = 1; for (int i = 1; i < n ; i++) { if (nums[i] - nums[i - 1] == 0) { continue; } else if (nums[i] - nums[i - 1] == 1) { currentMax++; max = Math.max(max, currentMax); } else { currentMax = 1; } } return max; } }