Jumping On The Clouds HackerRank Solution [One Liner Solution] 

If you find this tutorial helpful, please do not forget to like , comment, share and It would be great if you can leave your feedback about the tutorial, as I have put a lot of hard work to make things easy for you. Thanks ..!! 🙏🙏🙏

Man, I can't thank you enough. I couldn't understand the problem right through the website description but once I heard you explain I got it. Your explanations are great, simple to understand and helpful to improve my skills. Keep going with those tutorials, they are of great help!

Thanks for guiding us through this exercise and breaking down the steps! ^^ And yes, I do indeed want to be a cool dude in front of my friends.

Kanhaiya you are doing really good work, it shows you are selfless and you care about us. This is how our country is gonna develop. Keep uploading such videos.

Thanks for the tutorial.. We also can write a more concise code using java 8 string s = abc; int n=10; while(s.length() (char)c).filter(ch -> ch.equals('a')).count();

Hello sir, instead of doing count = -1 We can use the condition on for loop that is i< array.length-1 So that time take to execute one time for loop will be saved.

OMG I love you man, lol! I was complicating myself a lot and never thought of this solution. Thank you so much

Great explanation Sir.really commendable.....Thank you for your effort and making us learning....

Wow, you do a very good job in cracking all of those HackerRank questions. Very impressed. Thank you.

Where you gone , Please come back and explain us more Hackerrank question :'( .You're the only RU-vidr who does it.Please Come Back.

i easily understand this logic..thanks for helping..

Dear sir you are doing great job the students like me.Please sir complete the hackerrank playlist.

Wow 🤩 .. Exellent teaching .. Thank you very much

I was able to write the code on my own but i was getting array out of bound, but after getting proper explanation i was able to solve my issue no need to wait for coding part in video

Amazing. Im a fan now!

brilliant solution thank you so much

hi what if the array has continuous '1's, i+1 and i+2 both are '1' in this case would you jump and count the jump or skip ? [0,1,0,0,0,1,0,0,1,1,1,1] - what is the output (if you avoid 1's in the last short distance would be 4, if you count on 1's it would be 8

i have found a solution but idk wether its optimistic or not public static void main(String[] args) { int arr[] = {0,0,1,0,1,0,1,0,0,0,1,0}; //6 int step =0; int i =0; while(i

thanks a lot for such a great explanation, and for such helpful solution videos.

Thanku sir your teaching style very easy and understandable thnks sir😍❤️ love for ur teaching

We can also use continue keyword in JAVA, For example i use this in C# Here my solution private static int JumpinClouds(int[] n) { int numOfJump = 0; for (int i = 0; i < n.Length; i++) { int newIndex = i + 1; if (newIndex < n.Length) { if (n[newIndex] == 0) { i++; numOfJump++; continue; } } numOfJump++; } return numOfJump; }

Your excellent brother plz do vedios on interview bit questions also brother 👌👌👌👌 explanation

Myfriend ) thank you! yoa are awesome! if you have any extra time , pleas do video about "cavity map" problem , and "Matrix Layer Rotation" ))

Gr8 explanations...one questions...which editor/tool you have used for the text and drawings?

Hi sir you are doing very great job for us, do you have playlist in which we have topic wise solutions. Like we start with greedy and its problems,game theory and problems, recursion and problems like this.

Thanks a lot for this explanation..💯💯

Thank you sir 😊

Bro plz make a video on repeated string

Can you please solve fradulent activity notification and Sherlock and the valid string?

Please make a video on "Between Two Sets" hackerrank problem

problem in my python code def jumpingOnClouds(c): count = -1 for i in range(len(c)-2): if i+2 < n and c[i+2]==0: i += 2 print() else: i+=1 count+=1 return count if __name__ == '__main__': n = int(input()) c = list(map(int, input().rstrip().split())) result = jumpingOnClouds(c) print(result) test case 0 and 1 only passed rest all failed

Tqsm sir

Thanks

great explanation

Nice Explanation :)

How can I solve any problem

When u are making upcoming videos

if there is a complex program. can I use the inbuilt function as an online test (java)?like Arrays.sort......?

Program starts @13:00

Please solve the angry children 2 problem..

Can you make video on manacher's algorithm

Is it cover all the questions from hacker rank???

can you solve the poisonous plant problem

dude, this throws me error after the first two test cases.

my solution: ´´´function jumpingOnClouds(c) { // Write your code here let jump=0 let length =c.length c.reduce((prev,curr,index) => {let r= prev +1 ; if (r===2 ){ jump+=1 ;curr===1 ? r= 1 : r=0 } else if( r===1 && index===length-1){jump+=1} return r }) return jump }´´´

Sir I dint get y exactly is count=-1

Tell me output for this input [0 0 0 0 1 1 1 0]

Nic3

I not understand why u put count -1