# 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 = dp
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 = dp + dp
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 = dp + dp
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<``int``> 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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