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:**

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

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 ` ` ` `int` `dp[n]; ` ` ` ` ` `// filling all the elements with 0 ` ` ` `memset` `(dp, 0, ` `sizeof` `(dp)); ` ` ` ` ` `// 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); ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

Total Ways = 3

**Time Complexity:** O(N^{2}), where N is the number of stones.

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