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: 4
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 <bits/stdc++.h>using namespace std;const int mod = 1e9 + 7;// Find the number of ways// to reach the endint 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 codeint 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 approachimport java.io.*;class GFG{static int mod = 1000000000;// Find the number of ways// to reach the endstatic 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 codepublic 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 approachmod = 1e9 + 7;# Find the number of ways# to reach the enddef 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 codeif __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 approachusing System; class GFG{static int mod = 1000000000;// Find the number of ways// to reach the endstatic 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 codepublic 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
<script> // 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)); </script> |
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[0] will be the solution to the problem.
Below is the implementation of the approach:
C++
// C++ implementation#include <bits/stdc++.h>using namespace std;const int mod = 1e9 + 7;// find the number of ways to reach the endint 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 codeint 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 approachclass 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 |
Python3
# Python3 implementation of above approachmod = 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 approachusing 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
<script> // 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); </script> |
Output:
6
Time Complexity: O(K)
Auxiliary Space: O(n)
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