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# Counts Path in an Array

• Difficulty Level : Medium
• Last Updated : 30 Apr, 2021

Given an array A consisting of positive integer, of size N. If the element in the array at index i is K then you can jump between index ranges (i + 1) to (i + K)
The task is to find the number of possible ways to reach the end with module 109 + 7.
The starting position is considered as index 0.
Examples:

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

Naive Approach: We can form a recursive structure to solve the problem.
Let F[i] denotes the number of paths starting at index i, at every index i if the element A[i] is K then the total number of ways the jump can be performed is:

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

By using this recursive formula we can solve the problem:

## 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);` `    ``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

 ``
Output:
`6`

Efficient Approach: In the previous approach, there are some calculations that are being done more than once. It will be better to store these values in a dp array and dp[i] will store the number of paths starting at index i and ending at the end of the array.
Hence dp will be the solution to the problem.
Below is the implementation of the approach:

## 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``    ``return` `dp % mod;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 3, 1, 4, 3 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``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``        ``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`

## Python3

 `# 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``    ``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``        ``return` `dp % 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

 ``
Output:
`6`

Time Complexity: O(K)

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