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 ) throws IOException { BufferedReader bufferedReader = new BufferedReader ( new InputStreamReader ( Syste
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 by
position 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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