public class Solution { public int solve(int[] A, int B) { int n = A.length; int i = 0; int count = 0; while(i < n) { boolean isBulbLighted = false; int lr = Math.max(i - B + 1, 0); int rr = Math.min(i + B - 1, n - 1); // 0, 0, 1, 0, 1, 0, 0, 1 for(int j = rr; j >= lr; j--) { if(A[j] == 1) { count++; i = j; i += B; isBulbLighted = true; break; } }
Bro, at first I was reluctant to watch a 19 minute video but the way you deliver the solution is absolute delight. The flow of the code and the explanation were in sync.
for java folks: public static int solve(int[] arr, int k) { if (k==0) return -1; int i = 0, count = 0; while (i < arr.length) { int idx = lightItUp(arr, k, i); if (idx == -1) return -1; else { i = idx + k; count++; } } return count; } public static int lightItUp(int arr[], int k, int i) { int j = Math.min(i + k - 1, arr.length - 1); while (j >= Math.max(i - k + 1, 0)) { if (arr[j] == 1) { return j; } j--; } return -1; }
i was just doing the mistake in the second while loop , btw nice explanation , but in the starting your voice was little crumbling , and have you made a git hub repo for interviewbit questions?
Python Solution: class Solution: # @param A : list of integers # @param B : integer # @return an integer def solve(self, A, B): count = 0 i = 0 n = len(A) while(i < n): right = min(n-1, i + B - 1) left = max(i - B + 1, 0) bulbFound = False while(right >= left): if A[right] == 1: bulbFound = True break right -= 1 if bulbFound == False: return -1 else: count += 1 i = right + B return count
Idiots setting the corridor from starting index 1, while the array of lights starts from 0. here is the one pass for starting index 0; int i, end = -1, cnt = 0; for (i = 0; i < A.size();) { if (A[i] == 1) { if (i - B + 1 > end + 1) return -1; end = i + B - 1; i += i + B; ++cnt; } else ++i; } return cnt;
thank you so much ma'am , I was not getting the problem I was afraid when I have read this problem need DP and backtracking concept and I have not yet studied it thank you so much for the confidante by your explanation I have got the problem
![Rearrange Array so that Arr[i] becomes Arr[Arr[i]] with O(1) extra space #InterviewBit Code C++](http://i.ytimg.com/vi/IEr74xZGNoU/mqdefault.jpg)
public class Solution {
public int solve(int[] A, int B) {
int n = A.length;
int i = 0;
int count = 0;
while(i < n) {
boolean isBulbLighted = false;
int lr = Math.max(i - B + 1, 0);
int rr = Math.min(i + B - 1, n - 1);
// 0, 0, 1, 0, 1, 0, 0, 1
for(int j = rr; j >= lr; j--) {
if(A[j] == 1) {
count++;
i = j;
i += B;
isBulbLighted = true;
break;
}
}
if(!isBulbLighted) {
return -1;
}
}
return count;
}
}
Bro, at first I was reluctant to watch a 19 minute video but the way you deliver the solution is absolute delight. The flow of the code and the explanation were in sync.
This is the best explanation for this question. Great job!
Mam plz aap continue rakhiye ye series, views ko dhyan mat dijiye
Aapke channel bhut precious h
Heere ho har koi tarash nhi sakta
though the video was bit long , but it was worth watching it!! so nicely explained every bit of the line of code
Very nicely explained, thank you!
Great explanation, thanks so much!
Good explanation! Thanks for helping out 🤗
Thanks for the video!
quality of Explanation justifies your hard work !!
Nicely explained. Very easy to understand
Thanks for the explanation 😊
int Solution::solve(vector &A, int B) {
int n = A.size();
int count = 0;
int i = 0;
bool bulbfound = false;
while(i=l){
if(A[r]==1){
count++;
i = B+r;
bulbfound = true;
break;
}
r--;
}
if(!bulbfound) return -1;
bulbfound = false;
}
return count;
}
Great Explanation
Thanks :)
Wow !!!!! Really a good explanations.
Great explanation
solving this question looks a piece of cake when explained by her!!!
Thank you ,that was nice application
Most underrated channel
Very nicely explained...Thanks
MOST AWSOME CONTENT MAAM...
Great Explanation❤
everytime I am stuck, you come up like magic. Thanks ma'm.
R u wall clock alwys stuck after putting battary u charged
Fantastic explanation!! KUDOS
Intuitive solution❤
What an awesome explanation!
Wow an awesome comment
Nice explaination 😊
nicely explained. !
such a good video
for java folks:
public static int solve(int[] arr, int k) {
if (k==0) return -1;
int i = 0, count = 0;
while (i < arr.length) {
int idx = lightItUp(arr, k, i);
if (idx == -1) return -1;
else {
i = idx + k;
count++;
}
}
return count;
}
public static int lightItUp(int arr[], int k, int i) {
int j = Math.min(i + k - 1, arr.length - 1);
while (j >= Math.max(i - k + 1, 0)) {
if (arr[j] == 1) {
return j;
}
j--;
}
return -1;
}
very good explanation the tabs show how much u research for one question🙂
super..
is Time complexity is O(n^2) ?
i was just doing the mistake in the second while loop , btw nice explanation , but in the starting your voice was little crumbling , and have you made a git hub repo for interviewbit questions?
best explanation!!
Glad you liked it
Python Solution:
class Solution:
# @param A : list of integers
# @param B : integer
# @return an integer
def solve(self, A, B):
count = 0
i = 0
n = len(A)
while(i < n):
right = min(n-1, i + B - 1)
left = max(i - B + 1, 0)
bulbFound = False
while(right >= left):
if A[right] == 1:
bulbFound = True
break
right -= 1
if bulbFound == False:
return -1
else:
count += 1
i = right + B
return count
❤
Can you please upload your code also? It will help us with revision purposes.
can anyone tell me what will be the time complexity? Is it O(n^2)?
O(N).
The worst case : -
A = [0, 0, 0, 0], B = 1
We would have to iterate over every element..
Wow thanks for the solution. It was an easy question how can't I do that. ;_;
can you please upload solution of Leetcode problem "127. Word Ladder"?
Surèeeeeeeeeeee dhanajoy i m redy for u
How much avg time did you take to solve questions?
How much u r taking less than that
Make a tutorial on how to be focussed to see the main screen while watching your lectures.
Keep a mirror to the side where you can see your own reflection.
Idiots setting the corridor from starting index 1, while the array of lights starts from 0.
here is the one pass for starting index 0;
int i, end = -1, cnt = 0;
for (i = 0; i < A.size();)
{
if (A[i] == 1)
{
if (i - B + 1 > end + 1)
return -1;
end = i + B - 1;
i += i + B;
++cnt;
}
else
++i;
}
return cnt;
Nicely explained
thank you so much ma'am , I was not getting the problem I was afraid when I have read this problem need DP and backtracking concept and I have not yet studied it thank you so much for the confidante by your explanation I have got the problem