abhinavjonnada82

import React from 'react';
import './SearchBar.css';


constructor(props){
  super(props);
  this.handleTermChange.handleLocationChange.handleSearch = this.handleTermChange.handleLocationChange.handleSearch.bind(this);
}
const sortByOptions = {
       "Best Match": "best_match",

abhinavjonnada82

Studying for a Tech Interview Sucks, so Here's a Cheat Sheet to Help

This list is meant to be a both a quick guide and reference for further research into these topics. It's basically a summary of that comp sci course you never took or forgot about, so there's no way it can cover everything in depth. It also will be available as a gist on Github for everyone to edit and add to.

Data Structure Basics

###Array ####Definition:

• Stores data elements based on an sequential, most commonly 0 based, index.
• Based on tuples from set theory.

abhinavjonnada82

   # cycle in a inked lists
    def hasCycle(self, head: ListNode) -> bool:
        first = head
        second = head
        while first and second and second.next: # Important takes care of edge cases
            first = first.next
            second = second.next.next
            if first == second:
                return True
        return False

abhinavjonnada82

# Balance Parantheses
def isValid(string):
    balLis = [i for i in string ]
    stack = []
    mapping = {")": "(", "}": "{", "]": "["} # Dictionary
    for i in balLis:
        if i == "(" or i == "[" or i == "{":
            stack.append(i)
        elif i ==")" or i == "]" or i == "}" and len(stack) > 0:
            if stack[-1] == mapping[i]:

abhinavjonnada82

# containerMostWater
def containerMostWater(arr):
    # O(N) runtime
    # O(1) space
    start = 0 
    end = len(arr)-1
    maxDist = 0
    if (len(arr) < 2):
        return -1
    while start < end:

abhinavjonnada82

# Finonacci 
def getNthFib(n):
    if n == 1:
        return 0
    if n == 2:
        return 1
    else:
        return getNthFib(n-1) + getNthFib(n-2)
        
# Powerset  

abhinavjonnada82

# Binary Search (Edge case: i/p: [1] o/p: [0,0])
class Solution:
    def searchRange(nums, target):
        left = searchLeft(nums, target, 0, len(nums)-1)
        if left < 0:
            return [-1, -1]
        else:
            right = searchRight(nums, target, left, len(nums)-1)
            return [left, right]
        

abhinavjonnada82

# Need to work on this!
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

class BST:
    def __init__(self):
        self.root = None

abhinavjonnada82

# O(n) time | O(d) space
def invertBinaryTree(tree):
    if tree is None:
         return 0
    (tree.left, tree.right) = (tree.right, tree.left)
    if tree.left:
        invertBinaryTree(tree.left)
    if tree.right:
        invertBinaryTree(tree.right)

abhinavjonnada82

# isPalindrome
# Runtime: O(N)
# Space: O(N)

def isPalindrome(strings):
    mid = len(strings)//2
    if len(strings) <= 1:
        return False
    if len(strings) % 2 == 0:
        leftPtr = mid - 1