To everyone asking *how* to arrive at the intuition of greedily trying to flip any "0" bits, this was basically my thought process that led me there: 1. Since a flipping operation is reversible, that must mean that if I can go from "nums" to all 1s with flips, it must also be possible to go from all 1s to "nums". 2. If we ASSUME the array starts off as being all 1s, that would mean that the first bit that is 0 would HAVE to be the boundry of a flipping location. Since, those all 1's aren't going to sponaneously go from their initial state of "1" to "0" unless it is the start of a flip. 3. This assumption can pretty much apply recursively, where we can assume that any unexpected transition from "1" to "0" represents another flip. 4. If we reach the end and there are still flips not accounted for, that means there were flip boundaries at intervals other than "k" and that we didn't start with all 1s to begin with.
We will keep track of valid count of flips till current index, and check if flip count is odd (num is in flipped condition) and curr num is 1 (means actual flipped num is 0) hence we need to flip this index to make it 1. Also same for, if current count is even (cancelling flips) and num is 0 means we need to flip this index. we will keep track of flip index in deque for efficient adding a new index to end and removing index which is out of current window from front. Here is java solution: public int minKBitFlips(int[] nums, int k) { int n = nums.length, res = 0; Deque dq = new LinkedList(); for(int r = 0; r < n; r++){ if(!dq.isEmpty() && dq.peekFirst() n - k){ return -1; } res++; dq.addLast(r); } } return res; }
the 2nd problem could be solved easily using three variables, and the 3rd problem could be solved using a bool flag; this is a lot harder than them according to me
Solved it on my own Beats(94%). Finally seeing the fruit of almost year of practise. class Solution: def minKBitFlips(self, nums: List[int], k: int) -> int: n = len(nums) flips = [0]*n flipSum = 0 count = 0 # Count the number of flips efficiently for i in range(n - k + 1): if (nums[i] == 0 and (flipSum == 0 or flipSum % 2 == 0)) or (nums[i] == 1 and flipSum % 2 == 1): count += 1 flips[i] = 1 flipSum += 1 if i + 1 >= k: flipSum -= flips[i - k + 1] # Check if last k character is 1 for i in range(n - k + 1, n): if (nums[i] == 0 and (flipSum == 0 or flipSum % 2 == 0)) or (nums[i] == 1 and flipSum % 2 == 1): return -1 flipSum -= flips[i - k + 1] return count
you have to have seen this before or have read some competitive programming books.. these mod tricks, fenwick tree etc are quite common in those books... i swear companies who ask these can go f*ck themselves.. lol
what if the queue had 2 flips that affected the last element ? we have to pop all the flips that no longer affect our current element , thats why we use while
@@aadityatripathi8363 you're checking at each iteration of i though so an if statement would be enough. Once an index is out of range it will immediately be removed from the queue in the next iteration
Fun fact: in the constant space solution, you don't actually need to check if `i-k >=0`. If it ends up being negative, it'll be a lookup from end of the array (i.e. nums[-2]). For all k values, since we haven't changed any of the numbers in the indexes ahead, it'll still do what we want it to do (which is catch any changes in the past only when it sees a 2. It might still be good practice to ensure you are not going out of bounds though...
I understand how the solution works. I just don't understand why the greedy approach is the best solution here. How can a person who sees this problem guarantee that the greedy approach is the best way to go?
Everything is nice nav. but one suggestion, while u provide the code, can u give a side-by-side animation on what is happening in the list or the code that runs..that would be really helpful.thank you
Hi Neetcode, I’ve been focused on learning algorithms and patterns for about a year now, and normally when I see questions I could usually figure out what pattern or algorithm to use. So my question is how do I get good at solving these kind of questions cause they don’t follow the popular algorithms or patterns
Waaoooo great explanation, is this a pattern? i haven't done any bit manipulation questions, though I knew I can use sliding window here but still just didn't know how....
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To everyone asking *how* to arrive at the intuition of greedily trying to flip any "0" bits, this was basically my thought process that led me there:
1. Since a flipping operation is reversible, that must mean that if I can go from "nums" to all 1s with flips, it must also be possible to go from all 1s to "nums".
2. If we ASSUME the array starts off as being all 1s, that would mean that the first bit that is 0 would HAVE to be the boundry of a flipping location. Since, those all 1's aren't going to sponaneously go from their initial state of "1" to "0" unless it is the start of a flip.
3. This assumption can pretty much apply recursively, where we can assume that any unexpected transition from "1" to "0" represents another flip.
4. If we reach the end and there are still flips not accounted for, that means there were flip boundaries at intervals other than "k" and that we didn't start with all 1s to begin with.
fuck, im still confused..
We will keep track of valid count of flips till current index, and check if flip count is odd (num is in flipped condition) and curr num is 1 (means actual flipped num is 0) hence we need to flip this index to make it 1. Also same for, if current count is even (cancelling flips) and num is 0 means we need to flip this index. we will keep track of flip index in deque for efficient adding a new index to end and removing index which is out of current window from front.
Here is java solution:
public int minKBitFlips(int[] nums, int k) {
int n = nums.length, res = 0;
Deque dq = new LinkedList();
for(int r = 0; r < n; r++){
if(!dq.isEmpty() && dq.peekFirst() n - k){
return -1;
}
res++;
dq.addLast(r);
}
}
return res;
}
yeah bro you are not alone
relatable
very similar to biweekly problems this week.
exactly
Hi 😅 what is biweekly problem?
the biweekly passed naive approach, this one doesn't
@@emmanuelfleurine121they meant a problem from the last biweekly contest
the 2nd problem could be solved easily using three variables, and the 3rd problem could be solved using a bool flag; this is a lot harder than them according to me
First try with brute force and get TLE :D
Same bro tle on test case 101/113
110/113 :'D
Solved it on my own Beats(94%). Finally seeing the fruit of almost year of practise.
class Solution:
def minKBitFlips(self, nums: List[int], k: int) -> int:
n = len(nums)
flips = [0]*n
flipSum = 0
count = 0
# Count the number of flips efficiently
for i in range(n - k + 1):
if (nums[i] == 0 and (flipSum == 0 or flipSum % 2 == 0)) or (nums[i] == 1 and flipSum % 2 == 1):
count += 1
flips[i] = 1
flipSum += 1
if i + 1 >= k:
flipSum -= flips[i - k + 1]
# Check if last k character is 1
for i in range(n - k + 1, n):
if (nums[i] == 0 and (flipSum == 0 or flipSum % 2 == 0)) or (nums[i] == 1 and flipSum % 2 == 1):
return -1
flipSum -= flips[i - k + 1]
return count
holy cow as soon as i understood how to use queue i solved it instantly on my own. Thanks man.
Tell me honestly how do you come up with approach?
I can go waste whole day and still can't figure out the solution
you have to have seen this before or have read some competitive programming books.. these mod tricks, fenwick tree etc are quite common in those books...
i swear companies who ask these can go f*ck themselves.. lol
The more problems you do, the more you'll start to pick up patterns, patterns are the key.
look at the biweekly this week #2 and #3
He’s just smarter…… jk….. you got this
Look at solutions & code them hands-on. Then do others without looking.
i know you made it pretty simple, but coming up with this logic of mod is basically made me go into existential crisis and question my reality
Why use while? you can simply use if statement to pop the left element.
what if the queue had 2 flips that affected the last element ? we have to pop all the flips that no longer affect our current element , thats why we use while
@@aadityatripathi8363 you're checking at each iteration of i though so an if statement would be enough. Once an index is out of range it will immediately be removed from the queue in the next iteration
Fun fact: in the constant space solution, you don't actually need to check if `i-k >=0`. If it ends up being negative, it'll be a lookup from end of the array (i.e. nums[-2]). For all k values, since we haven't changed any of the numbers in the indexes ahead, it'll still do what we want it to do (which is catch any changes in the past only when it sees a 2. It might still be good practice to ensure you are not going out of bounds though...
Very Great Explanation. Thank you so much !!!
If you are still confused, I highly recommend to try with very similar problems that have medium level - 3191. and 3192.
I understand how the solution works. I just don't understand why the greedy approach is the best solution here. How can a person who sees this problem guarantee that the greedy approach is the best way to go?
well i thought of greedy, turned out that it was very simple to think but barrier was N*K time, so this video helped after that
Nice explanation!!
awesome, u changed this Ques from Hard to Easy,
this is why LeetCode ques in interview seem unfair sometimes,
if u know u know
Everything is nice nav. but one suggestion, while u provide the code, can u give a side-by-side animation on what is happening in the list or the code that runs..that would be really helpful.thank you
that's kinda your job with every code that you don't understand
Hi Neetcode, I’ve been focused on learning algorithms and patterns for about a year now, and normally when I see questions I could usually figure out what pattern or algorithm to use. So my question is how do I get good at solving these kind of questions cause they don’t follow the popular algorithms or patterns
The best 🌟
Waaoooo great explanation, is this a pattern? i haven't done any bit manipulation questions, though I knew I can use sliding window here but still just didn't know how....
I like the thumbnail have flipflops, lol.
This problem is hell as hard.
best explanation
🎉🎉
remember: if your love makes you weak its not love .
love only makes your partner stronger and as you
Very beautiful
solution in java
class Solution {
public int minKBitFlips(int[] nums,int k){
Queue q = new LinkedList();
int i = 0;
int n = nums.length;
int oprs = 0;
while( i < n){
if( i > ( n - k) ){
if( (nums[i] + q.size()%2)%2 == 0 ) return -1;
}
else {
if( (nums[i] + q.size()%2)%2 == 0 ){
oprs++;
q.add(i);
}
}
if( !q.isEmpty() && q.peek() == (i - k + 1) ){
q.poll();
}
i++;
}
return oprs;
}
}
You are awesome
Solved it but with O(k) space! Sadly, unable to solve it using O(1) space!
❤
PLZ LEETCODE 2090
Why did you create multiple channels? Why are you no longer uploading video in your old channel?
This channel only focus on Leetcode POTD. 1st channel for Interview Preparation related stuff not solving ranom coding question
How do you even comeup with these 🤔
He is a genius
look at the biweekly this week #2 and #3
😢
first
Really nice solution