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Swapnil Gupta
Swapnil Gupta

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LEET Code 21 Days of Dynamic Programming

** Day 13**
Minimum falling Path
using hashmap
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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){


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Decode Ways :
LeetCode Solution :
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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

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Day 5

Day 4

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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};

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Day 3

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Day 2:

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Day 1
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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];

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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];

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