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:
- 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] (assume 1-indexed) is the array that stores the number of paths to reach the Nth Node from sth stone.
Initially, dp[] = { 0, 0, 0, 0, 0} and dp[s] = 1, then
dp[] = { 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, 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, 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, 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++ 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 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
# 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# 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
<script>
// Javascript 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
function TotalWays(n, s, k)
{
// Initialize dp array
let dp = new Array(n);
// filling all the elements with 0
dp.fill(0);
// Initialize (s-1)th index to 1
dp[s - 1] = 1;
// Iterate a loop from s to n
for (let i = s; i < n; i++) {
// starting range for counting ranges
let idx = Math.max(s - 1, i - k);
// Calculate Maximum moves to
// Reach ith step
for (let j = idx; j < i; j++) {
dp[i] += dp[j];
}
}
// For nth step return dp[n-1]
return dp[n - 1];
}
// Driver Code
// no of steps
let n = 5;
// Atmost steps allowed
let k = 2;
// starting range
let s = 2;
document.write("Total Ways = " + TotalWays(n, s, k));
// This code is contributed by _saurabh_jaiswal
</script>
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)