计算数组中的路径

给定一个由正整数组成的数组A ，大小为N 。如果数组中索引i处的元素是K，那么您可以在索引范围(i + 1)到(i + K)之间跳转。
任务是通过模块10 ⁹ + 7找出到达终点的可能方式的数量。
起始位置被视为索引0 。
例子：

Input: A = {5, 3, 1, 4, 3} 
Output: 6
Input: A = {2, 3, 1, 1, 2} 
Output: 4 
 

朴素的方法：我们可以形成一个递归结构来解决问题。
让 F[i] 表示从索引i开始的路径数，如果元素A[i]是K，则在每个索引i处，可以执行跳转的方式总数为：

F(i) = F(i+1) + F(i+2) +...+ F(i+k), where i + k <= n, 
where F(n) = 1

通过使用这个递归公式，我们可以解决这个问题：

C++

// C++ implementation of
// the above approach
#include 
using namespace std;
 
const int mod = 1e9 + 7;
 
// Find the number of ways
// to reach the end
int ways(int i, int arr[], int n)
{
    // Base case
    if (i == n - 1)
        return 1;
 
    int sum = 0;
 
    // Recursive structure
    for (int j = 1;
         j + i < n && j <= arr[i];
         j++) {
        sum += (ways(i + j,
                     arr, n))
               % mod;
        sum %= mod;
    }
 
    return sum % mod;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 3, 1, 4, 3 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << ways(0, arr, n) << endl;
 
    return 0;
}

Java

// Java implementation of
// the above approach
import java.io.*;
 
class GFG
{
static int mod = 1000000000;
 
// Find the number of ways
// to reach the end
static int ways(int i,
                int arr[], int n)
{
    // Base case
    if (i == n - 1)
        return 1;
 
    int sum = 0;
 
    // Recursive structure
    for (int j = 1; j + i < n &&
                    j <= arr[i]; j++)
    {
        sum += (ways(i + j,
                     arr, n)) % mod;
        sum %= mod;
    }
    return sum % mod;
}
 
// Driver code
public static void main (String[] args)
{
    int arr[] = { 5, 3, 1, 4, 3 };
     
    int n = arr.length;
 
    System.out.println (ways(0, arr, n));
}
}
 
// This code is contributed by ajit

Python3

# Python3 implementation of
# the above approach
 
mod = 1e9 + 7;
 
# Find the number of ways
# to reach the end
def ways(i, arr, n):
     
    # Base case
    if (i == n - 1):
        return 1;
 
    sum = 0;
 
    # Recursive structure
    for j in range(1, arr[i] + 1):
        if(i + j < n):
            sum += (ways(i + j, arr, n)) % mod;
            sum %= mod;
 
    return int(sum % mod);
 
# Driver code
if __name__ == '__main__':
    arr = [5, 3, 1, 4, 3];
 
    n = len(arr);
 
    print(ways(0, arr, n));
 
# This code is contributed by PrinciRaj1992

C#

// C# implementation of
// the above approach
using System;
     
class GFG
{
static int mod = 1000000000;
 
// Find the number of ways
// to reach the end
static int ways(int i,
                int []arr, int n)
{
    // Base case
    if (i == n - 1)
        return 1;
 
    int sum = 0;
 
    // Recursive structure
    for (int j = 1; j + i < n &&
                    j <= arr[i]; j++)
    {
        sum += (ways(i + j,
                     arr, n)) % mod;
        sum %= mod;
    }
    return sum % mod;
}
 
// Driver code
public static void Main (String[] args)
{
    int []arr = { 5, 3, 1, 4, 3 };
     
    int n = arr.Length;
 
    Console.WriteLine(ways(0, arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

    // Javascript implementation of the above approach
     
    let mod = 1000000000;
   
    // Find the number of ways
    // to reach the end
    function ways(i, arr, n)
    {
        // Base case
        if (i == n - 1)
            return 1;
 
        let sum = 0;
 
        // Recursive structure
        for (let j = 1; j + i < n && j <= arr[i]; j++)
        {
            sum += (ways(i + j, arr, n)) % mod;
            sum %= mod;
        }
        return sum % mod;
    }
     
    let arr = [ 5, 3, 1, 4, 3 ];
       
    let n = arr.length;
   
    document.write(ways(0, arr, n));
     

输出：

6

高效的方法：在之前的方法中，有一些计算被多次执行。最好将这些值存储在 dp 数组中，dp[i] 将存储从索引i开始到数组末尾结束的路径数。
因此 dp[0] 将是问题的解决方案。
下面是该方法的实现：

C++

// C++ implementation
#include 
using namespace std;
 
const int mod = 1e9 + 7;
 
// find the number of ways to reach the end
int ways(int arr[], int n)
{
    // dp to store value
    int dp[n + 1];
 
    // base case
    dp[n - 1] = 1;
 
    // Bottom up dp structure
    for (int i = n - 2; i >= 0; i--) {
        dp[i] = 0;
 
        // F[i] is dependent of
        // F[i+1] to F[i+k]
        for (int j = 1; ((j + i) < n
                         && j <= arr[i]);
             j++) {
            dp[i] += dp[i + j];
            dp[i] %= mod;
        }
    }
 
    // Return value of dp[0]
    return dp[0] % mod;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 3, 1, 4, 3 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << ways(arr, n) % mod << endl;
    return 0;
}

Java

// Java implementation of above approach
class GFG
{
    static final int mod = (int)(1e9 + 7);
     
    // find the number of ways to reach the end
    static int ways(int arr[], int n)
    {
        // dp to store value
        int dp[] = new int[n + 1];
     
        // base case
        dp[n - 1] = 1;
     
        // Bottom up dp structure
        for (int i = n - 2; i >= 0; i--)
        {
            dp[i] = 0;
     
            // F[i] is dependent of
            // F[i+1] to F[i+k]
            for (int j = 1; ((j + i) < n &&
                              j <= arr[i]); j++)
            {
                dp[i] += dp[i + j];
                dp[i] %= mod;
            }
        }
     
        // Return value of dp[0]
        return dp[0] % mod;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 5, 3, 1, 4, 3 };
     
        int n = arr.length;
     
        System.out.println(ways(arr, n) % mod);
    }
}
 
// This code is contributed by AnkitRai01

蟒蛇3

# Python3 implementation of above approach
mod = 10**9 + 7
 
# find the number of ways to reach the end
def ways(arr, n):
     
    # dp to store value
    dp = [0] * (n + 1)
     
    # base case
    dp[n - 1] = 1
     
    # Bottom up dp structure
    for i in range(n - 2, -1, -1):
        dp[i] = 0
         
        # F[i] is dependent of
        # F[i + 1] to F[i + k]
        j = 1
        while((j + i) < n and j <= arr[i]):
            dp[i] += dp[i + j]
            dp[i] %= mod
            j += 1
     
    # Return value of dp[0]
    return dp[0] % mod
 
# Driver code
arr = [5, 3, 1, 4, 3 ]
n = len(arr)
print(ways(arr, n) % mod)
 
# This code is contributed by SHUBHAMSINGH10

C＃

// C# implementation of above approach
using System;
     
class GFG
{
    static readonly int mod = (int)(1e9 + 7);
     
    // find the number of ways to reach the end
    static int ways(int []arr, int n)
    {
        // dp to store value
        int []dp = new int[n + 1];
     
        // base case
        dp[n - 1] = 1;
     
        // Bottom up dp structure
        for (int i = n - 2; i >= 0; i--)
        {
            dp[i] = 0;
     
            // F[i] is dependent of
            // F[i+1] to F[i+k]
            for (int j = 1; ((j + i) < n &&
                              j <= arr[i]); j++)
            {
                dp[i] += dp[i + j];
                dp[i] %= mod;
            }
        }
     
        // Return value of dp[0]
        return dp[0] % mod;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int []arr = { 5, 3, 1, 4, 3 };
     
        int n = arr.Length;
     
        Console.WriteLine(ways(arr, n) % mod);
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
    // Javascript implementation
    // of above approach
     
    let mod = (1e9 + 7);
       
    // find the number of ways
    // to reach the end
    function ways(arr, n)
    {
        // dp to store value
        let dp = new Array(n + 1);
        dp.fill(0);
       
        // base case
        dp[n - 1] = 1;
       
        // Bottom up dp structure
        for (let i = n - 2; i >= 0; i--)
        {
            dp[i] = 0;
       
            // F[i] is dependent of
            // F[i+1] to F[i+k]
            for (let j = 1; ((j + i)
            < n && j <= arr[i]); j++)
            {
                dp[i] += dp[i + j];
                dp[i] %= mod;
            }
        }
       
        // Return value of dp[0]
        return dp[0] % mod;
    }
     
    let arr = [ 5, 3, 1, 4, 3 ];
       
    let n = arr.length;
 
    document.write(ways(arr, n) % mod);
     

输出：

6

时间复杂度： O(K)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。