** Day 13**
Minimum falling Path
using hashmap
Day 10
Airtmetic Slices
Logic and other Solution: LeetCode
package Day10; import java.util.Scanner; public class AirthmeticSlice { public static int numberOfArithmeticSlices(int[] nums) { int[] DP= new int[nums.length]; int count=0; for(int i=2; i<nums.length; i++){ int diff=nums[i]-nums[i-1]; if(diff==nums[i-1]- nums[i-2]){ //previous series grown +1 new ap DP[i]=DP[i-1]+1; count=DP[i]+count; } ; } return count; } public static void main(String[] args) { Scanner sc= new Scanner(System.in); System.out.println("Enter the length of your array"); int n= sc.nextInt(); int[] nums= new int[n]; for(int i=0; i<n; ++i){ nums[i]=sc.nextInt(); sc.close(); } System.out.println(numberOfArithmeticSlices(nums)); } } Decode Ways :
LeetCode Solution :
we can use subString Intger.Parseint(s.substring(i-1, i+1), a point to be noticed i-1 to i+1, will cover range of i-1 and 1. eg:(2,5)-> 2,3,4
Day 5
Day 4
public class jumpGame { //I will store my maximum value in Dp array and compare the current maximum dp[i] to //the previous maximum Dp[i-1], so the currrent maximum or the maximum value will depend on the value of dp[i-1], so we can store dp[i-1] in variable MaxJump //maxJump= Math.max(maxJump, nums[i]+i); public static boolean canJump(int[] nums) { if (nums.length==1) return true; int[] dp = new int[nums.length]; dp[0] = nums[0]; for(int i = 1; i < nums.length; i++){ if(dp[i - 1] < i){ return false; } //if my capcity to jump is smaller than to capacity to reach last index,i, then we will return false dp[i] = Math.max(dp[i - 1], i + nums[i]); //maxJump= Math.max(maxJump, i+nums[i]); //whether we can reach to end index or not from current: i+nums[i] } return true; } public static void main(String[] args ){ int[] nums= {2,1,4,3,0,1}; System.out.println(canJump(nums)); } } Day 3
Day 2:
Day 1
Problem 1: Fibonacci Series
class Solution { public int fib(int n) { if(n==0 || n==1) return n; int[] dp = new int[n+1]; dp[0]=0; dp[1] =1; for(int i=2;i<=n;i++){ dp[i] = dp[i-1]+dp[i-2]; } return dp[n]; } } Problem 2: n-th tribonacci Series
class Solution { public int tribonacci(int n) { if(n==0 ){ return 0; } else if(n==1 || n==2){ return 1; } int dp[]= new int[n+1]; dp[0]=0; dp[1]=1; dp[2]=1; for(int i =3;i<=n;i++){ dp[i] = dp[i-1]+dp[i-2]+dp[i-3]; } return dp[n]; 
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