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 number of ways jumps can be made from that element to reach the end of the array. If an element is 0, then move cannot be made through that element. The element that cannot reach to the end should have a count “-1”.
Examples:
Input : {3, 2, 0, 1} Output : 2 1 -1 0 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 Minimum number of jumps to reach end(Method 3). Here we need to count all ways to reach end from every cell.
The solution is a modified version of the solution to the problem of Minimum number of jumps to reach end(Method 3).
This problem aims to count the different ways to jump from each element so as to reach to the end. For each element the count is being calculated by adding the counts of all those forward elements that can reach to the end and to which the current element could reach + 1(if the element can directly reach to 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]
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 ?> |
Output:
52 52 28 16 8 -1 -1 4 2 1 0
Time Complexity: O(n2) in worst case.
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