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

• Difficulty Level : Easy
• Last Updated : 24 Nov, 2021

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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(N2), where N is the number of stones.
Auxiliary Space: O(N)

My Personal Notes arrow_drop_up