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Count number of ways to jump to reach end

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  • Difficulty Level : Medium
  • Last Updated : 23 Sep, 2022
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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.


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
?>


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>


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.


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