Medium

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1]. You must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4] Example 2: Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1] Example 3: Input: nums = [], target = 0 Output: [-1,-1] Constraints: ...

Given two strings s and t of length N, find the maximum number of possible matching pairs in strings s and t after swapping exactly two characters within s. A swap is switching s[i] and s[j], where s[i] and s[j] denotes the character that is present at the ith and jth index of s, respectively. The matching pairs of the two strings are defined as the number of indices for which s[i] and t[i] are equal. Note: This means you must swap two characters at different indices. Signature int matchingPairs(String s, String t) Input ...

Pair Sums Given a list of n integers arr[0..(n-1)], determine the number of different pairs of elements within it which sum to k. If an integer appears in the list multiple times, each copy is considered to be different; that is, two pairs are considered different if one pair includes at least one array index which the other doesn’t, even if they include the same values. Signature int numberOfWays(int[] arr, int k) Input n is in the range [1, 100,000]. Each value arr[i] is in the range [1, 1,000,000,000]. k is in the range [1, 1,000,000,000]. Output Return the number of different pairs of elements which sum to k. ...

Android devices have a special lock screen with a 3 x 3 grid of dots. Users can set an “unlock pattern” by connecting the dots in a specific sequence, forming a series of joined line segments where each segment’s endpoints are two consecutive dots in the sequence. A sequence of k dots is a valid unlock pattern if both of the following are true: All the dots in the sequence are distinct. If the line segment connecting two consecutive dots in the sequence passes through the center of any other dot, the other dot must have previously appeared in the sequence. No jumps through the center non-selected dots are allowed. For example, connecting dots 2 and 9 without dots 5 or 6 appearing beforehand is valid because the line from dot 2 to dot 9 does not pass through the center of either dot 5 or 6. However, connecting dots 1 and 3 without dot 2 appearing beforehand is invalid because the line from dot 1 to dot 3 passes through the center of dot 2. ...

You are given an array arr of N integers. For each index i, you are required to determine the number of contiguous subarrays that fulfill the following conditions: The value at index i must be the maximum element in the contiguous subarrays, and These contiguous subarrays must either start from or end on index i. Signature int[] countSubarrays(int[] arr) Input Array arr is a non-empty list of unique integers that range between 1 to 1,000,000,000 Size N is between 1 and 1,000,000 Output An array where each index i contains an integer denoting the maximum number of contiguous subarrays of arr[i] ...

Given two binary search trees root1 and root2. Return a list containing all the integers from both trees sorted in ascending order. Example 1: Input: root1 = [2,1,4], root2 = [1,0,3] Output: [0,1,1,2,3,4] Example 2: Input: root1 = [0,-10,10], root2 = [5,1,7,0,2] Output: [-10,0,0,1,2,5,7,10] Example 3: Input: root1 = [], root2 = [5,1,7,0,2] Output: [0,1,2,5,7] Example 4: Input: root1 = [0,-10,10], root2 = [] Output: [-10,0,10] Example 5: Input: root1 = [1,null,8], root2 = [8,1] Output: [1,1,8,8] ...

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST): BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST. boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false. int next() Moves the pointer to the right, then returns the number at the pointer. Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST. ...

An integer array is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, [1,3,5,7,9], [7,7,7,7], and [3,-1,-5,-9] are arithmetic sequences. Given an integer array nums, return the number of arithmetic subarrays of nums. A subarray is a contiguous subsequence of the array. Example 1: Input: nums = [1,2,3,4] Output: 3 Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself. Example 2: Input: nums = [1] Output: 0 Constraints: ...

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key’s value at a certain timestamp. Implement the TimeMap class: TimeMap() Initializes the object of the data structure. void set(String key, String value, int timestamp) Stores the key key with the value value at the given time timestamp. String get(String key, int timestamp) Returns a value such that set was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the value associated with the largest timestamp_prev. If there are no values, it returns “”. Example 1: Input ["TimeMap", "set", "get", "get", "set", "get", "get"] [[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]] Output [null, null, "bar", "bar", null, "bar2", "bar2"] Explanation TimeMap timeMap = new TimeMap(); timeMap.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1. timeMap.get("foo", 1); // return "bar" timeMap.get("foo", 3); // return "bar", since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar". timeMap.set("foo", "bar2", 4); // store the key "foo" and value "ba2r" along with timestamp = 4. timeMap.get("foo", 4); // return "bar2" timeMap.get("foo", 5); // return "bar2" Constraints: ...

Given the root of a binary tree, return the lowest common ancestor of its deepest leaves. Recall that: The node of a binary tree is a leaf if and only if it has no children The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1. The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A. Example 1: Input: root = [3,5,1,6,2,0,8,null,null,7,4] Output: [2,7,4] Explanation: We return the node with value 2, colored in yellow in the diagram. The nodes coloured in blue are the deepest leaf-nodes of the tree. Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3. ...