Leetcode

![https://leetcode.com/problems/k-closest-points-to-origin/] We have a list of points on the plane. Find the K closest points to the origin (0, 0). (Here, the distance between two points on a plane is the Euclidean distance.) You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.) Example 1: Input: points = [[1,3],[-2,2]], K = 1 Output: [[-2,2]] Explanation: The distance between (1, 3) and the origin is sqrt(10). The distance between (-2, 2) and the origin is sqrt(8). Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin. We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]]. Example 2: Input: points = [[3,3],[5,-1],[-2,4]], K = 2 Output: [[3,3],[-2,4]] (The answer [[-2,4],[3,3]] would also be accepted.) Note: ...

![https://leetcode.com/problems/majority-element/] Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array. Example 1: Input: nums = [3,2,3] Output: 3 Example 2: Input: nums = [2,2,1,1,1,2,2] Output: 2 Constraints: n == nums.length 1 <= n <= 5 * 104 -231 <= nums[i] <= 231 - 1 Follow-up: Could you solve the problem in linear time and in O(1) space? ...

![https://leetcode.com/problems/kth-largest-element-in-an-array/] Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element. Example 1: Input: [3,2,1,5,6,4] and k = 2 Output: 5 Example 2: Input: [3,2,3,1,2,4,5,5,6] and k = 4 Output: 4 Note: You may assume k is always valid, 1 ≤ k ≤ array’s length. class Solution { public int findKthLargest(int[] nums, int k) { int lo = 0; int hi = nums.length - 1; k = nums.length - k; while (lo < hi) { int j = partition(nums, lo, hi); if (j < k) { lo = j + 1; } else if (j > k) { hi = j - 1; } else { return nums[k]; } } return nums[k]; } int partition(int[] a, int lo, int hi) { int i = lo; int j = hi + 1; while (true) { while (a[++i] < a[lo]) { if (i == hi) break; } while (a[lo] < a[--j]) { if (j == lo) break; } if (i >= j) break; swap(a, i, j); } swap(a, lo, j); return j; } void swap(int[] a, int i, int j) { int temp = a[i]; a[i] = a[j]; a[j] = temp; } }

![https://leetcode.com/problems/shuffle-string/] Given a string s and an integer array indices of the same length. The string s will be shuffled such that the character at the ith position moves to indices[i] in the shuffled string. Return the shuffled string. Example 1: Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3] Output: "leetcode" Explanation: As shown, "codeleet" becomes "leetcode" after shuffling. Example 2: Input: s = "abc", indices = [0,1,2] Output: "abc" Explanation: After shuffling, each character remains in its position. Example 3: Input: s = "aiohn", indices = [3,1,4,2,0] Output: "nihao" Example 4: Input: s = "aaiougrt", indices = [4,0,2,6,7,3,1,5] Output: "arigatou" Example 5: Input: s = "art", indices = [1,0,2] Output: "rat" Constraints: ...

![https://leetcode.com/problems/remove-outermost-parentheses/] There are n computers numbered from 0 to n-1 connected by ethernet cables connections forming a network where connections[i] = [a, b] represents a connection between computers a and b. Any computer can reach any other computer directly or indirectly through the network. Given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the minimum number of times you need to do this in order to make all the computers connected. If it’s not possible, return -1. ...

![https://leetcode.com/problems/evaluate-reverse-polish-notation/] Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note: Division between two integers should truncate toward zero. The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation. Example 1: Input: ["2", "1", "+", "3", "*"] Output: 9 Explanation: ((2 + 1) * 3) = 9 Example 2: Input: ["4", "13", "5", "/", "+"] Output: 6 Explanation: (4 + (13 / 5)) = 6 Example 3: Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] Output: 22 Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 = ((10 * (6 / (12 * -11))) + 17) + 5 = ((10 * (6 / -132)) + 17) + 5 = ((10 * 0) + 17) + 5 = (0 + 17) + 5 = 17 + 5 = 22 class Solution { public int evalRPN(String[] tokens) { int[] stack = new int[tokens.length]; int size = 0; for (String token: tokens) { if (token.equals("+") || token.equals("-") || token.equals("*") || token.equals("/") ) { int b = stack[--size]; int a = stack[--size]; int c; switch (token) { case "+": c = a + b; break; case "-": c = a - b; break; case "*": c = a * b; break; case "/": c = a / b; break; default: throw new RuntimeException(); } stack[size++] = c; } else { stack[size++] = Integer.valueOf(token); } } return stack[0]; } }

![https://leetcode.com/problems/crawler-log-folder/] The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: “../” : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). “./” : Remain in the same folder. “x/” : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. ...

![https://leetcode.com/problems/crawler-log-folder/] The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: “../” : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). “./” : Remain in the same folder. “x/” : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. ...

![https://leetcode.com/problems/remove-outermost-parentheses/] A valid parentheses string is either empty (""), “(” + A + “)”, or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, “”, “()”, “(())()”, and “(()(()))” are all valid parentheses strings. A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings. ...

Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them. We repeatedly make duplicate removals on S until we no longer can. Return the final string after all such duplicate removals have been made. It is guaranteed the answer is unique. Example 1: Input: "abbaca" Output: "ca" Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca". Note: ...