Objective In this challenge, we're going to use loops to help us do some simple math. Task Given an integer,  , print its first   multiples. Each multiple   (where  ) should be printed on a new line in the form:  N x i = result . Input Format A single integer,  . Constraints Output Format Print   lines of output; each line   (where  ) contains the   of   in the form: N x i = result . Sample Input 2 Sample Output 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 2 x 6 = 12 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 2 x 10 = 20 Explanation:           Here, we just need to use for loops to achieve the result Solution : import  java.io.* ;  import  java.math.* ;  import  java.security.* ;  import  java.text.* ;  import  java.util.* ;  import  java.util.concurrent.* ;  import  java.util.regex.* ;     public  class  Solution  {      public  static  void  main ( String []  args )  t...
Dynamic Programming:Memoization| Program to find Fibonaaci number in that postion using Memoization Technique
Idea Behind Memoize technique:
In the Recursion there are so many repetitive calls in that,in order to
overcome that we need to declare an array to store the intermediate
results and use that.
Logic:
    Declare the Memo array and store the result and call them byposition later.
Code Logic:
- input to find fibonacci number at the position
 - function declaration fib(n,memo)
 - if memo[n]!=null:return memo[n]
 - base condition n==1 or n==2 return 1
 - else return fib(n-1)+fib(n-2)
 
Program:
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