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  Big O Cheat Sheet In this post, we will look at the Big O Notation both time and space complexity! For all our examples we will be using Ruby. What is Big O Notation? Big O notation is just a fancy way of describing how your code’s performance depends on the amount of data its processing. It allows us to calculate the worst possible runtime of an algorithm, or how long it takes the algorithm to complete. In other words, it informs us of how much it will slow down based on the input size. It’s usually talked about in two ways the  time complexity ( how long an algorithm takes )and  space complexity ( how much space an algorithm uses ) Let me put this graph here we can refer to it as we go on, it’s a graph showing the complexities compare against each other. Big O complexity chart Time Complexity Before we get try wrapping our heads around the different big O notations here is a nice table that I am borrowing to show how speed will slow down based on the size of the dataset. I found th

Game Theory with examples Introduction In this article I will be covering problems related to two players game in which it is assumed that both players play optimally and we have to find out the winner of the game. First we will look at the  basic   division of positions to winning and losing . Then we will see the  Game of Nim  and then see how it will be used to solve the  Composite games . Basic Division of positions to winning and losing Problem Statement:  Consider a simple game played by two players A and B . There are n stones on the table. Each player can pick 1 , 2 or 5 stones in each turn. Both players pick the stones alternately until the total number of stones left on the table is 0. The player unable to make the move will lose. Assuming that both the players play optimally, output the winner of the game. Solution:  As you can see positions 1 , 2 and 5 are winning positions since the player can pick up all the stones and other player will not be able to make a move . 0 is a

 1. Sum of Odd numbers: 1+3+5+7+9+....+n =? where n is an odd number Solve: let, n = 3 list = 1+3+5 = 9 ; 3^2=9 or 3*3 = 9 let, n = 5 list = 1+3+5+7+9 = 25 ; 5^2 =25 or 5*5 =25 • so, summation of n odd numbers = n^2 = n*n 2.Sum of Even numbers: 2+4+6+8+10+....+n =? where n is an even number Solve: n=4, list = 2+4+6+8 = 20; sum = 4^2+4 = 20 or 4*5 =20 n=7, list = 2+4+6+8+10+12+14 = 56; sum = 7^2+7 = 56 or 7*8 = 56 • so, summation of n even numbers = n^2+n = n*(n+1)

  Suppose you are given an array of numbers and you need to find whether the number X is in the array. The straight forward approach to do that would be to run a loop and compare each number in the array with X. This is what the pseudo code would look like: function findNumber ( A , x ) for i = 0 to length of array A if A [ i ] == x return true return false For example, let’s assume you are given this array: [7, 14, 17, 21, 35, 60, 92, 121, 155] And, you were looking for the number 92. The loop would start checking each of the number in the array against 92 and would find after iterating through the array 7 times. The worst case time complexity of this approach would be O(n). However, whenever the array itself is in sorted order (like in this case where the numbers are increasing in value from left to right of the array), you can perform something that is much more efficient: binary search. The prerequisite of binary search is that the array should be sorted.