Implement the following operations of a stack using queues. push(x) -- Push element x onto stack.pop() -- Removes the element on top of the stack.top() -- Get the top element.empty() -- Return whether the stack is empty. Example: MyStack stack = new MyStack(); stack.push(1); stack.push(2); stack.top(); // returns 2 stack.pop(); // returns 2 stack.empty(); // …

# Day: March 22, 2020

# Implement Queue using Stacks

Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of queue.pop() -- Removes the element from in front of queue.peek() -- Get the front element.empty() -- Return whether the queue is empty. Example: MyQueue queue = new MyQueue(); queue.push(1); queue.push(2); queue.peek(); // returns 1 queue.pop(); // returns …

# First Missing Positive

Given an unsorted integer array, find the smallest missing positive integer. Example 1: Input: [1,2,0] Output: 3 Example 2: Input: [3,4,-1,1] Output: 2 Example 3: Input: [7,8,9,11,12] Output: 1 Note: Your algorithm should run in O(n) time and uses constant extra space. Solution With comments public class Solution { public int firstMissingPositive(int[] nums) { int n = …

# Find All Numbers Disappeared in an Array

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count …

# Find All Duplicates in an Array

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,3,2,7,8,2,3,1] Output: [2,3] Approach When find a number i, flip the number at position i-1 to negative.If …

# How Many Numbers Are Smaller Than the Current Number

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i]. Return the answer in an array. Example 1: Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums[0]=8 there exist four smaller numbers than …

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