Minimum number of Fibonacci jumps to reach end

Given an array of 0s and 1s, If any particular index i has value 1 then it is a safe index and if the value is 0 then it is an unsafe index. A man is standing at index -1(source) can only land on a safe index and he has to reach the Nth index (last position). At each jump, the man can only travel a distance equal to any Fibonacci Number. You have to minimize the number of steps, provided man can jump only in forward direction.
Note: First few Fibonacci numbers are - 0, 1, 1, 2, 3, 5, 8, 13, 21....
Examples: 
 

Input: arr[]= {0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0} 
Output: 3
The person will jump from: 
1) index -1 to index 4 (a distance of jump = 5) 
2) index 4 to index 6 (a distance of jump = 2) 
3) index 6 to the destination (a distance of jump = 5)
Input: arr[]= {0, 0} 
Output: 1
The person will jump from: 
1) index -1 to destination (a distance of jump = 3) 
 

Approach: 
 

• First, we will create an array fib[] for storing fibonacci numbers.
• Then, we will create a DP array and initialize it with INT_MAX and the 0th index DP[0] = 0 and will move in the same way as Coin Change Problem with minimum number of coins.
• The DP definition and recurrence is as followed:

   
   DP[i] = min( DP[i], 1 + DP[i-fib[j]] )
  
  where i is the current index and j denotes
  the jth fibonacci number of which the
  jump is possible

  • Here DP[i] denotes the minimum steps required to reach index i considering all Fibonacci numbers.

  
  Below is the implementation of the approach: 
   

  
  Output: 
  

  3

  
   

  Time Complexity: