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Given an array of numbers where each element represents the max number of jumps that can be made forward from that element. For each array element, count the number of ways jumps can be made from that element to reach the end of the array. If an element is 0, then a move cannot be made through that element. The element that cannot reach the end should have a count “-1“.

Examples:

Input: {3, 2, 0, 1}
Output: 2 1 -1 0
Explanation:

  • For 3 number of steps or jumps that can be taken are 1, 2 or 3. The different ways are:
    3 -> 2 -> 1
    3 -> 1
  • For 2 number of steps or jumps that can be taken are 1, or 2. The different ways are:
    2 -> 1
  • For 0 number of steps or jumps that can be taken are 0. One cannot move forward from this point.
  • For 1 number of steps or jumps that can be taken are 1. But the element is at the end so no jump is required.

Input: {1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9}
Output: 52 52 28 16 8 -1 -1 4 2 1 0

This problem is a variation of the Minimum number of jumps to reach end(Method 3). Here we need to count all the ways to reach the end from every cell.
The solution is a modified version of the solution to the problem of the Minimum number of jumps to reach end(Method 3)
This problem aims to count the different ways to jump from each element to reach the end. For each element, the count is being calculated by adding the counts of all those forward elements that can reach the end and to which the current element could reach + 1(if the element can directly reach the end).

Algorithm: 

countWays(arr, n)
    Initialize array count_jump[n] = {0}

    count_jump[n-1] = 0
    for i = n-2 to 0
        if arr[i] >= (n-i-1)
         count_jump[i]++
        for j=i+1; j < n-1 && j <= arr[i]+i; i++
          if count_jump[j] != -1
             count_jump[i] += count_jump[j]
        if count_jump[i] == 0
         count_jump[i] = -1

    for i = 0 to n-1
        print count_jump[i]

Below is the implementation of the above approach:

C++




// C++ implementation to count number
// of ways to jump to reach end
#include <bits/stdc++.h>
using namespace std;
 
// function to count ways to jump to
// reach end for each array element
void countWaysToJump(int arr[], int n)
{
    // count_jump[i] store number of ways
    // arr[i] can reach to the end
    int count_jump[n];
    memset(count_jump, 0, sizeof(count_jump));
 
    // Last element does not require to jump.
    // Count ways to jump for remaining
    // elements
    for (int i=n-2; i>=0; i--)
    {
        // if the element can directly
        // jump to the end
        if (arr[i] >= n - i - 1)
            count_jump[i]++;
 
        // add the count of all the elements
        // that can reach to end and arr[i] can
        // reach to them
        for (int j=i+1; j < n-1 && j <= arr[i] + i; j++)
 
            // if element can reach to end then add
            // its count to count_jump[i]
            if (count_jump[j] != -1)
                 count_jump[i] += count_jump[j];
 
        // if arr[i] cannot reach to the end
        if (count_jump[i] == 0)
            count_jump[i] = -1;
    }
 
    // print count_jump for each
    // array element
    for (int i=0; i<n; i++)
        cout << count_jump[i] << " ";
}
 
// Driver program to test above
int main()
{
    int arr[] = {1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9};
    int n = sizeof(arr) / sizeof(arr[0]);
    countWaysToJump(arr, n);
    return 0;
}


Java




// Java implementation to count number
// of ways to jump to reach end
import java.util.Arrays;
 
class GFG {
     
    // function to count ways to jump to
    // reach end for each array element
    static void countWaysToJump(int arr[], int n)
    {
         
        // count_jump[i] store number of ways
        // arr[i] can reach to the end
        int count_jump[] = new int[n];
        Arrays.fill(count_jump, 0);
     
        // Last element does not require to jump.
        // Count ways to jump for remaining
        // elements
        for (int i = n-2; i >= 0; i--)
        {
             
            // if the element can directly
            // jump to the end
            if (arr[i] >= n - i - 1)
                count_jump[i]++;
     
            // add the count of all the elements
            // that can reach to end and arr[i] can
            // reach to them
            for (int j = i+1; j < n-1 && j <= arr[i] + i; j++)
     
                // if element can reach to end then add
                // its count to count_jump[i]
                if (count_jump[j] != -1)
                    count_jump[i] += count_jump[j];
     
            // if arr[i] cannot reach to the end
            if (count_jump[i] == 0)
                count_jump[i] = -1;
        }
     
        // print count_jump for each
        // array element
        for (int i = 0; i < n; i++)
            System.out.print(count_jump[i] + " ");
    }
     
    //driver code
    public static void main (String[] args)
    {
         
        int arr[] = {1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9};
        int n = arr.length;
         
        countWaysToJump(arr, n);
    }
}
 
// This code is contributed by Anant Agarwal.


Python3




# Python3 implementation to count
# number of ways to jump to reach end
 
# Function to count ways to jump to
# reach end for each array element
def countWaysToJump(arr, n):
 
    # count_jump[i] store number of ways
    # arr[i] can reach to the end
    count_jump = [0 for i in range(n)]
 
    # Last element does not require
    # to jump. Count ways to jump for
    # remaining elements
    for i in range(n - 2, -1, -1):
     
        # if the element can directly
        # jump to the end
        if (arr[i] >= n - i - 1):
            count_jump[i] += 1
 
        # Add the count of all the elements
        # that can reach to end and arr[i]
        # can reach to them
        j = i + 1
        while(j < n-1 and j <= arr[i] + i):
 
            # if element can reach to end then
            # add its count to count_jump[i]
            if (count_jump[j] != -1):
                count_jump[i] += count_jump[j]
            j += 1
             
        # if arr[i] cannot reach to the end
        if (count_jump[i] == 0):
            count_jump[i] = -1
     
 
    # print count_jump for each
    # array element
    for i in range(n):
        print(count_jump[i], end = " ")
 
# Driver code
arr = [1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9]
n = len(arr)
countWaysToJump(arr, n)
 
# This code is contributed by Anant Agarwal.


Javascript




<script>
 
// Javascript implementation to count number
// of ways to jump to reach end
     
    // function to count ways to jump to
    // reach end for each array element
    function countWaysToJump(arr,n)
    {
        // count_jump[i] store number of ways
        // arr[i] can reach to the end
        let count_jump = new Array(n);
        for(let i=0;i<n;i++)
        {
            count_jump[i]=0;
        }
       
        // Last element does not require to jump.
        // Count ways to jump for remaining
        // elements
        for (let i = n-2; i >= 0; i--)
        {
               
            // if the element can directly
            // jump to the end
            if (arr[i] >= n - i - 1)
                count_jump[i]++;
       
            // add the count of all the elements
            // that can reach to end and arr[i] can
            // reach to them
            for (let j = i+1; j < n-1 && j <= arr[i] + i; j++)
       
                // if element can reach to end then add
                // its count to count_jump[i]
                if (count_jump[j] != -1)
                    count_jump[i] += count_jump[j];
       
            // if arr[i] cannot reach to the end
            if (count_jump[i] == 0)
                count_jump[i] = -1;
        }
       
        // print count_jump for each
        // array element
        for (let i = 0; i < n; i++)
            document.write(count_jump[i] + " ");
    }
     
    //driver code
    let arr=[1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9];
    let n = arr.length;
    countWaysToJump(arr, n);
     
     
     
    // This code is contributed by avanitrachhadiya2155
</script>


C#




// C# implementation to count number
// of ways to jump to reach end
using System;
 
class GFG {
     
    // function to count ways to jump to
    // reach end for each array element
    static void countWaysToJump(int[] arr, int n)
    {
         
        // count_jump[i] store number of ways
        // arr[i] can reach to the end
        int[] count_jump = new int[n];
         
        for(int i = 0; i < n; i++)
            count_jump[i] = 0;
         
     
        // Last element does not require to jump.
        // Count ways to jump for remaining
        // elements
        for (int i = n-2; i >= 0; i--)
        {
             
            // if the element can directly
            // jump to the end
            if (arr[i] >= n - i - 1)
                count_jump[i]++;
     
            // add the count of all the elements
            // that can reach to end and arr[i] can
            // reach to them
            for (int j = i+1; j < n-1 && j <= arr[i] + i; j++)
     
                // if element can reach to end then add
                // its count to count_jump[i]
                if (count_jump[j] != -1)
                    count_jump[i] += count_jump[j];
     
            // if arr[i] cannot reach to the end
            if (count_jump[i] == 0)
                count_jump[i] = -1;
        }
     
        // print count_jump for each
        // array element
        for (int i = 0; i < n; i++)
            Console.Write(count_jump[i] + " ");
    }
     
    // Driver code
    public static void Main ()
    {
        int[] arr = {1, 3, 5, 8, 9,
                 1, 0, 7, 6, 8, 9};
        int n = arr.Length;
        countWaysToJump(arr, n);
    }
}
 
// This code is contributed by ChitraNayal


PHP




<?php
// PHP implementation to count number
// of ways to jump to reach end
 
// function to count ways to jump to
// reach end for each array element
function countWaysToJump($arr, $n)
{
    // count_jump[i] store number of ways
    // arr[i] can reach to the end
    $count_jump;
    for($i = 0; $i < $n; $i++)
        $count_jump[$i] = 0;
 
    // Last element does not require to jump.
    // Count ways to jump for remaining
    // elements
    for ($i = $n - 2; $i >= 0; $i--)
    {
        // if the element can directly
        // jump to the end
        if ($arr[$i] >= $n - $i - 1)
            $count_jump[$i]++;
 
        // add the count of all the elements
        // that can reach to end and arr[i]
        // can reach to them
        for ($j = $i + 1; $j < $n - 1 &&
             $j <= $arr[$i] + $i; $j++)
 
            // if element can reach to end then
            // add its count to count_jump[i]
            if ($count_jump[$j] != -1)
                $count_jump[$i] += $count_jump[$j];
 
        // if arr[i] cannot reach to the end
        if ($count_jump[$i] == 0)
            $count_jump[$i] = -1;
    }
 
    // print count_jump for each
    // array element
    for ($i = 0; $i < $n; $i++)
        echo $count_jump[$i] . " ";
}
 
// Driver Code
$arr = array(1, 3, 5, 8, 9, 1,
                0, 7, 6, 8, 9);
$n = count($arr);
countWaysToJump($arr, $n);
 
// This code is contributed by Rajput-Ji
?>


Output

52 52 28 16 8 -1 -1 4 2 1 0 

Time Complexity: O(n2), Where n is the size of the given array.
Auxiliary Space: O(n), As we are creating count_jump array of size n.

Approach 2: Recursion with Memoization

  •    Create a memoization table of size n+1, where n is the number of cells in the given array. Initialize all values in the table to -1.
  •    Define a recursive function, countWays(arr, i, n, memo) where arr is the input array, i is the current index, n is the total number of cells, and memo is the memoization table.
  •    Check if the value of memo[i] is not -1. If it is not -1, return memo[i].
  •    If i is equal to n-1, return 1, as there is only one way to reach the end from the last cell.
  •    If arr[i] is 0, return 0 as we cannot jump from a cell with value 0.
  •    Otherwise, for each valid jump length j from the current cell, call the countWays function recursively with the new index i+j.
  •    Add up all the values returned by the recursive calls, and store the result in memo[i].
  •    Return memo[i].
  •    Call the countWays function with initial parameters arr, 0, n, memo.
  •    Return the result obtained from the function call.

C++




#include <bits/stdc++.h>
using namespace std;
 
int jumpWays(int arr[], int n, int pos, int dp[]) {
    // Base case
    if(pos == n-1) {
        return 1;
    }
    if(pos >= n) {
        return 0;
    }
    // Memoization
    if(dp[pos] != -1) {
        return dp[pos];
    }
    int ans = 0;
    for(int i = 1; i <= arr[pos]; i++) {
        ans += jumpWays(arr, n, pos + i, dp);
    }
    dp[pos] = ans;
    return ans;
}
 
int main() {
    int arr[] = {1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9};
    int n = sizeof(arr) / sizeof(int);
    int dp[n];
    memset(dp, -1, sizeof(dp));
    cout << jumpWays(arr, n, 0, dp) << endl;
    return 0;
}


Java




import java.util.Arrays;
 
public class Main {
 
  public static int jumpWays(int[] arr, int n, int pos, int[] dp)
  {
 
    // Base case
    if(pos == n-1) {
      return 1;
    }
    if(pos >= n) {
      return 0;
    }
 
    // Memoization
    if(dp[pos] != -1) {
      return dp[pos];
    }
    int ans = 0;
    for(int i = 1; i <= arr[pos]; i++) {
      ans += jumpWays(arr, n, pos + i, dp);
    }
    dp[pos] = ans;
    return ans;
  }
 
  public static void main(String[] args) {
    int[] arr = {1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9};
    int n = arr.length;
    int[] dp = new int[n];
    Arrays.fill(dp, -1);
    System.out.println(jumpWays(arr, n, 0, dp));
  }
}


Python3




def jumpWays(arr, n, pos, dp):
    # Base case
    if pos == n-1:
        return 1
    if pos >= n:
        return 0
    # Memoization
    if dp[pos] != -1:
        return dp[pos]
    ans = 0
    for i in range(1, arr[pos]+1):
        ans += jumpWays(arr, n, pos + i, dp)
    dp[pos] = ans
    return ans
 
 
arr = [1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9]
n = len(arr)
dp = [-1]*n
print(jumpWays(arr, n, 0, dp))


Javascript




// Define a function to find the number of ways to reach the end of an array
function jumpWays(arr, n, pos, dp) {
    // Base case
    if(pos == n-1) {
        return 1;
    }
    if(pos >= n) {
        return 0;
    }
    // Memoization
    if(dp[pos] != -1) {
        return dp[pos];
    }
    let ans = 0;
    for(let i = 1; i <= arr[pos]; i++) {
        ans += jumpWays(arr, n, pos + i, dp);
    }
    dp[pos] = ans;
    return ans;
}
 
// Driver Code
let arr = [1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9];
let n = arr.length;
let dp = new Array(n).fill(-1);
console.log(jumpWays(arr, n, 0, dp));


C#




using System;
 
class Gfg {
    static int jumpWays(int[] arr, int n, int pos, int[] dp) {
        // Base case
        if(pos == n-1) {
            return 1;
        }
        if(pos >= n) {
            return 0;
        }
        // Memoization
        if(dp[pos] != -1) {
            return dp[pos];
        }
        int ans = 0;
        for(int i = 1; i <= arr[pos]; i++) {
            ans += jumpWays(arr, n, pos + i, dp);
        }
        dp[pos] = ans;
        return ans;
    }
 
    static void Main() {
        int[] arr = {1, 3, 5, 8, 9, 1, 0, 7, 6, 8, 9};
        int n = arr.Length;
        int[] dp = new int[n];
        Array.Fill(dp, -1);
        Console.WriteLine(jumpWays(arr, n, 0, dp));
    }
}


Output

52

Time Complexity:
The time complexity of the algorithm using recursion and memoization is O(n^2), where n is the number of cells in the given array. This is because we need to compute the number of ways to reach the end from each cell, and there are n cells in the array. Each computation takes O(n) time, as there are at most n valid jumps from each cell.

Space Complexity:
The space complexity of the algorithm using recursion and memoization is O(n), where n is the number of cells in the given array. This is because we need to create a memoization table of size n+1 to store the results of the recursive function calls. Each cell in the memoization table stores an integer value, so the total space required is O(n).



Last Updated : 18 Apr, 2023
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