Leetcode

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers. Given a string s, return true if it is a palindrome, or false otherwise. Example 1: Input: s = "A man, a plan, a canal: Panama" Output: true Explanation: "amanaplanacanalpanama" is a palindrome. Example 2: Input: s = "race a car" Output: false Explanation: "raceacar" is not a palindrome. Example 3: Input: s = " " Output: true Explanation: s is an empty string "" after removing non-alphanumeric characters. Since an empty string reads the same forward and backward, it is a palindrome. Constraints: ...

Given the root of a binary search tree, return a balanced binary search tree with the same node values. If there is more than one answer, return any of them. A binary search tree is balanced if the depth of the two subtrees of every node never differs by more than 1. Example 1: Input: root = [1,null,2,null,3,null,4,null,null] Output: [2,1,3,null,null,null,4] Explanation: This is not the only correct answer, [3,1,4,null,2] is also correct. ...

Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators ‘+’, ‘-’, and/or ‘*’ between the digits of num so that the resultant expression evaluates to the target value. Note that operands in the returned expressions should not contain leading zeros. Example 1: Input: num = "123", target = 6 Output: ["1*2*3","1+2+3"] Explanation: Both "1*2*3" and "1+2+3" evaluate to 6. Example 2: Input: num = "232", target = 8 Output: ["2*3+2","2+3*2"] Explanation: Both "2*3+2" and "2+3*2" evaluate to 8. Example 3: Input: num = "105", target = 5 Output: ["1*0+5","10-5"] Explanation: Both "1*0+5" and "10-5" evaluate to 5. Note that "1-05" is not a valid expression because the 5 has a leading zero. Example 4: Input: num = "00", target = 0 Output: ["0*0","0+0","0-0"] Explanation: "0*0", "0+0", and "0-0" all evaluate to 0. Note that "00" is not a valid expression because the 0 has a leading zero. Example 5: Input: num = "3456237490", target = 9191 Output: [] Explanation: There are no expressions that can be created from "3456237490" to evaluate to 9191. Constraints: ...

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively. Merge nums1 and nums2 into a single array sorted in non-decreasing order. The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n. ...

Given two sparse vectors, compute their dot product. Implement class SparseVector: SparseVector(nums) Initializes the object with the vector nums dotProduct(vec) Compute the dot product between the instance of SparseVector and vec A sparse vector is a vector that has mostly zero values, you should store the sparse vector efficiently and compute the dot product between two SparseVector. Follow up: What if only one of the vectors is sparse? ...

Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window. Implement the MovingAverage class: MovingAverage(int size) Initializes the object with the size of the window size. double next(int val) Returns the moving average of the last size values of the stream. Example 1: Input ["MovingAverage", "next", "next", "next", "next"] [[3], [1], [10], [3], [5]] Output [null, 1.0, 5.5, 4.66667, 6.0] Explanation MovingAverage movingAverage = new MovingAverage(3); movingAverage.next(1); // return 1.0 = 1 / 1 movingAverage.next(10); // return 5.5 = (1 + 10) / 2 movingAverage.next(3); // return 4.66667 = (1 + 10 + 3) / 3 movingAverage.next(5); // return 6.0 = (10 + 3 + 5) / 3 Constraints: ...

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements. ...

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. Note: You must not use any built-in BigInteger library or convert the inputs to integer directly. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123", num2 = "456" Output: "56088" Constraints: 1 <= num1.length, num2.length <= 200 num1 and num2 consist of digits only. Both num1 and num2 do not contain any leading zero, except the number 0 itself. Solution class Solution { public String multiply(String num1, String num2) { int[] products = new int[num1.length() + num2.length()]; for (int i = 0; i < num1.length(); i++) { for (int j = 0; j < num2.length(); j++) { int d1 = num1.charAt(i) - '0'; int d2 = num2.charAt(j) - '0'; products[i + j + 1] += d1 * d2; } } int carry = 0; for (int i = products.length - 1; i >= 0; i--) { int tmp = (products[i] + carry) % 10; carry = (products[i] + carry) / 10; products[i] = tmp; } StringBuilder sb = new StringBuilder(); for (int product: products) { if (product == 0 && sb.length() == 0) { continue; } sb.append(product); } if (sb.length() == 0) { sb.append("0"); } return sb.toString(); } }

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++’s atoi function). The algorithm for myAtoi(string s) is as follows: Read in and ignore any leading whitespace. Check if the next character (if not already at the end of the string) is ‘-’ or ‘+’. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored. Convert these digits into an integer (i.e. “123” -> 123, “0032” -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2). If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1. Return the integer as the final result. Note: ...

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: ...