Solved Jumping on the Clouds

Challenge:

Emma is playing a new mobile game that starts with consecutively numbered clouds. Some of the clouds are thunderheads and others are cumulus. She can jump on any cumulus cloud having a number that is equal to the number of the current cloud plus  or . She must avoid the thunderheads. Determine the minimum number of jumps it will take Emma to jump from her starting postion to the last cloud. It is always possible to win the game.

For each game, Emma will get an array of clouds numbered  if they are safe or  if they must be avoided. For example,  indexed from . The number on each cloud is its index in the list so she must avoid the clouds at indexes  and . She could follow the following two paths:  or . The first path takes jumps while the second takes .

Function Description

Complete the jumpingOnClouds function in the editor below. It should return the minimum number of jumps required, as an integer.

jumpingOnClouds has the following parameter(s):

• c: an array of binary integers

Input Format

The first line contains an integer , the total number of clouds. The second line contains  space-separated binary integers describing clouds  where .

Constraints

Output Format

Print the minimum number of jumps needed to win the game.

Sample Input 0

7
0 0 1 0 0 1 0

Sample Output 0

4

Explanation 0:
Emma must avoid  and . She can win the game with a minimum of  jumps:

Sample Input 1

6
0 0 0 0 1 0

Sample Output 1

3

Explanation 1:
The only thundercloud to avoid is . Emma can win the game in  jumps:

Approach:

Note: Before start coding resolve the algorithm, The better way is to trace the logic that your mind followed already. Because on looking into the problem, your mind will give the solution. You have to know how the mind got that solution, in which way ?
1. Here we should not go with step by step by  increment of for loop, we should custom increment and should break the flow using if at the end.

Note: We have to remember any time you are not going with Standard increment or decrements, you will loose the control on the flow then you have take it to your hand by using a if condition and break.

2.  Before validating whether next cloud is safe or not, we need to validate whether the next cloud is available or not.

Note: Here we need to talk about logical AND operator. It will not validate the second condition if the first condition fails. We effectively utilized that feature here. Suppose A && B is there. assume expression A is false, then it will not validate for B, it will go with the false result.
Note: I feel we should always be strict to the boundaries, of the problem while developing the solution. 

Solution:

static int jumpingOnClouds(int[] c) {

        int jumps = 0;

        for(int index = 0 ; index < c.length ; ){

           if(c[index] != 1){

               if((index+2)<c.length && c[index+2] != 1){

                   index = index+2;

                   jumps++;

               }else if((index+1)<c.length && c[index+1] != 1){

                   index = index+1;

                   jumps++;

               }else if((index+1)>c.length || (index+2)>c.length){

                   break;

               }

           }

        }

        return jumps;

    }