DEV Community 👩‍💻👨‍💻

Cover image for LEET Code 21 Days of Dynamic Programming
Swapnil Gupta
Swapnil Gupta

Posted on • Updated on

LEET Code 21 Days of Dynamic Programming

** Day 13**
Minimum falling Path
using hashmap
Image description

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
            } ;

           return count;
 public static void main(String[] args) {
     Scanner sc= new Scanner(;
     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){


Enter fullscreen mode Exit fullscreen mode

Decode Ways :
LeetCode Solution :
Image description
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

Enter fullscreen mode Exit fullscreen mode

Day 5

Day 4

Image description

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};

Enter fullscreen mode Exit fullscreen mode

Day 3

Image description

Day 2:

Image description

Day 1
Image description

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[1] =1;
        for(int i=2;i<=n;i++){
            dp[i] = dp[i-1]+dp[i-2];

        return dp[n];

Enter fullscreen mode Exit fullscreen mode

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];
        for(int i =3;i<=n;i++){
        dp[i] = dp[i-1]+dp[i-2]+dp[i-3];
        return dp[n];

Enter fullscreen mode Exit fullscreen mode

Top comments (0)


Thank you.

Thanks for visiting DEV, we’ve worked really hard to cultivate this great community and would love to have you join us. If you’d like to create an account, you can sign up here.