Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false. Each letter in the magazine string can only be used once in your ransom note. Note: You may assume that both strings contain only lowercase letters. Example canConstruct("a", "b") -> false canConstruct("aa", "ab") -> false canConstruct("aa", "aab") -> true Solution Java
You’re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels. The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".
Given a binary tree where each path going from the root to any leaf form a valid sequence, check if a given string is a valid sequence in such binary tree. We get the given string from the concatenation of an array of integers arr and the concatenation of all values of the nodes along a path results in a sequence in the given binary tree. Example 1: Input: root = [0,1,0,0,1,0,null,null,1,0,0], arr = [0,1,0,1] Output: true Explanation: The path 0 -> 1 -> 0 -> 1 is a valid sequence (green color in the figure).
Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. The length of path between two nodes is represented by the number of edges between them. Example 1: Input: 5 / \ 4 5 / \ \ 1 1 5 Output: 2 Example 2: Input: 1 / \ 4 5 / \ \ 4 4 5 Output: 2 Note: The given binary tree has not more than 10000 nodes.
Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Your goal is to reach the last index in the minimum number of jumps. Example: Input: [2,3,1,1,4] Output: 2 Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
You have a queue of integers, you need to retrieve the first unique integer in the queue. Implement the FirstUnique class: FirstUnique(int nums) Initializes the object with the numbers in the queue. int showFirstUnique() returns the value of the first unique integer of the queue, and returns -1 if there is no such integer. void add(int value) insert value to the queue. Example 1 Input: ["FirstUnique","showFirstUnique","add","showFirstUnique","add","showFirstUnique","add","showFirstUnique"] [[[2,3,5]],,,,,,,] Output: [null,2,null,2,null,3,null,-1] Explanation: FirstUnique firstUnique = new FirstUnique([2,3,5]); firstUnique.
Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Determine if you are able to reach the last index. Example 1 Input: [2,3,1,1,4] Output: true Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index. Example 2 Input: [3,2,1,0,4] Output: false Explanation: You will always arrive at index 3 no matter what.
Return the root node of a binary search tree that matches the given preorder traversal. (Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val. Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.) Example 1 Input: [8,5,1,7,10,12] Output: [8,5,10,1,7,null,12] Note: 1 <= preorder.
Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules: Any left parenthesis '(' must have a corresponding right parenthesis ')'. Any right parenthesis ')' must have a corresponding left parenthesis '('. Left parenthesis '(' must go before the corresponding right parenthesis ')'. '*' could be treated as a single right parenthesis ‘)’ or a single left parenthesis '(' or an empty string.
You are given a string s containing lowercase English letters, and a matrix shift, where shift[i] = [direction, amount]: direction can be 0 (for left shift) or 1 (for right shift). amount is the amount by which string s is to be shifted. A left shift by 1 means remove the first character of s and append it to the end. Similarly, a right shift by 1 means remove the last character of s and add it to the beginning.