# Count ways to reach end from start stone with at most K jumps at each step

Given N stones in a row from left to right. From each stone, you can jump to at most K stones. The task is to find the total number of ways to reach from sth stone to Nth stone.
Examples:

Input: N = 5, s = 2, K = 2
Output: Total Ways = 3
Explanation:
Assume s1, s2, s3, s4, s5 be the stones. The possible paths from 2nd stone to 5th stone:
s2 -> s3 -> s4 -> s5
s2 -> s4 -> s5
s2 -> s3 -> s5
Hence total number of ways = 3
Input: N = 8, s = 1, K = 3
Output: Total Ways = 44

Approach:

1. Let assume dp[i] be the number of ways to reach ith stone.
2. Since there are atmost K jumps, So the ith stone can be reach by all it’s previous K stones.
3. Iterate for all possible K jumps and keep adding this possible combination to the array dp[].
4. Then the total number of possible ways to reach Nth node from sth stone will be dp[N-1].
5. For Example:

Let N = 5, s = 2, K = 2, then we have to reach Nth stone from sth stone.
Let dp[N+1] is the array that stores the number of paths to reach the Nth Node from sth stone.
Initially, dp[] = { 0, 0, 0, 0, 0, 0 } and dp[s] = 1, then
dp[] = { 0, 0, 1, 0, 0, 0 }
To reach the 3rd,
There is only 1 way with at most 2 jumps i.e., from stone 2(with jump = 1). Update dp[3] = dp[2]
dp[] = { 0, 0, 1, 1, 0, 0 }
To reach the 4th stone,
The two ways with at most 2 jumps i.e., from stone 2(with jump = 2) and stone 3(jump = 1). Update dp[4] = dp[3] + dp[2]
dp[] = { 0, 0, 1, 1, 2, 0 }
To reach the 5th stone,
The two ways with at most 2 jumps i.e., from stone 3(with jump = 2) and stone 4(with jump = 1). Update dp[5] = dp[4] + dp[3]
dp[] = { 0, 0, 1, 1, 2, 3 }
Now dp[N] = 3 is the number of ways to reach Nth stone from sth stone.

1.

Below is the implementation of the above approach:

## C++

 // C++ program to find total no.of ways // to reach nth step #include "bits/stdc++.h" using namespace std;   // Function which returns total no.of ways // to reach nth step from sth steps int TotalWays(int n, int s, int k) {     // Initialize dp array with 0s.     vector dp(n,0);         // Initialize (s-1)th index to 1     dp[s - 1] = 1;       // Iterate a loop from s to n     for (int i = s; i < n; i++) {           // starting range for counting ranges         int idx = max(s - 1, i - k);           // Calculate Maximum moves to         // Reach ith step         for (int j = idx; j < i; j++) {             dp[i] += dp[j];         }     }       // For nth step return dp[n-1]     return dp[n - 1]; }   // Driver Code int main() {     // no of steps     int n = 5;       // Atmost steps allowed     int k = 2;       // starting range     int s = 2;     cout << "Total Ways = "          << TotalWays(n, s, k); }

## Java

 // Java program to find total no.of ways // to reach nth step class GFG{    // Function which returns total no.of ways // to reach nth step from sth steps static int TotalWays(int n, int s, int k) {     // Initialize dp array     int []dp = new int[n];        // Initialize (s-1)th index to 1     dp[s - 1] = 1;        // Iterate a loop from s to n     for (int i = s; i < n; i++) {            // starting range for counting ranges         int idx = Math.max(s - 1, i - k);            // Calculate Maximum moves to         // Reach ith step         for (int j = idx; j < i; j++) {             dp[i] += dp[j];         }     }        // For nth step return dp[n-1]     return dp[n - 1]; }    // Driver Code public static void main(String[] args) {     // no of steps     int n = 5;        // Atmost steps allowed     int k = 2;        // starting range     int s = 2;     System.out.print("Total Ways = "          + TotalWays(n, s, k)); } }   // This code is contributed by sapnasingh4991

## Python3

 # Python 3 program to find total no.of ways # to reach nth step   # Function which returns total no.of ways # to reach nth step from sth steps def TotalWays(n, s, k):       # Initialize dp array     dp = [0]*n       # Initialize (s-1)th index to 1     dp[s - 1] = 1       # Iterate a loop from s to n     for i in range(s, n):           # starting range for counting ranges         idx = max(s - 1, i - k)           # Calculate Maximum moves to         # Reach ith step         for j in range( idx, i) :             dp[i] += dp[j]       # For nth step return dp[n-1]     return dp[n - 1]   # Driver Code if __name__ == "__main__":     # no of steps     n = 5       # Atmost steps allowed     k = 2       # starting range     s = 2     print("Total Ways = ", TotalWays(n, s, k))       # This code is contributed by chitranayal

## C#

 // C# program to find total no.of ways // to reach nth step using System;   class GFG{            // Function which returns total no.of ways     // to reach nth step from sth steps     static int TotalWays(int n, int s, int k)     {         // Initialize dp array         int []dp = new int[n];                // Initialize (s-1)th index to 1         dp[s - 1] = 1;                // Iterate a loop from s to n         for (int i = s; i < n; i++) {                    // starting range for counting ranges             int idx = Math.Max(s - 1, i - k);                    // Calculate Maximum moves to             // Reach ith step             for (int j = idx; j < i; j++) {                 dp[i] += dp[j];             }         }                // For nth step return dp[n-1]         return dp[n - 1];     }            // Driver Code     public static void Main(string[] args)     {         // no of steps         int n = 5;                // Atmost steps allowed         int k = 2;                // starting range         int s = 2;         Console.Write("Total Ways = "+ TotalWays(n, s, k));     } }   // This code is contributed by Yash_R

## Javascript



Output:

Total Ways = 3

Time Complexity: O(N*K), where N is the number of stones, k is maximum jump range.
Auxiliary Space: O(N)

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