L4. Max Consecutive Ones III | 2 Pointers and Sliding Window Playlist
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- Опубликовано: 5 окт 2024
- Notes/Codes/Problem links under step 10 of A2Z DSA Course: takeuforward.o...
Entire playlist: • Two Pointer and Slidin...
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Solved by myself before but can't skip your video. Nice one!
Lol... Me also😅
I may not comment on all your video. But I do watch them till last
I've always had a problem with two pointer + sliding window problems. I've solved a few in Leetcode by reading the editorials. I understood them at that point of time but couldn't apply them again in the future as I just couldn't wrap my head around them. But now the intuition kicked in after watching the first few videos of your playlist and I'm able to visualise the algo while solving problems. Thank you so much!!! :)
i think that is the best part of striver everything start to fall in place
3:43 brute
4:45 brute code
7:55 better
13:45 better code
17:00 better T(0)
19:20 best
24:46 - 26:48 best code
Aala11
Unbelievable! I solved it entirely on my own in just 5 minutes using a priority queue. Now, time to watch the video.
the only uncanny thing about that is a priority Queue other than that everything can be accepted
@@iPunishCode My sol with pq:
int findZeroes(int arr[], int n, int m) {
// code here
priority_queue pq;
int maxLength=1;
int left=0;
int right=0;
for(int i=0; im)
{
left = pq.top()+1;
pq.pop();
}
else {
right=i;
maxLength=std::max(maxLength, right-left+1);
}
}
}
return maxLength;
}
see it was not neccesary in this question thats why i said even i can do this same problem in 10 diff ways but that will not be intuitive but hey who am i to judge everyone has their way of thinking
@@iPunishCode I agree
@@tanujaSangwan can we exchange the number to practice together
sliding window and two pointer approach best playlist, thank you so much raj (our striver)
Good video ! Wasn't expecting the last solution, took me some time to think but definitely made my brain work.
The main logic is that once we have found a subarray with 2 zeros of size 5, as discussed in example, and a subarray with 2 zeros of size 6 exists... then once we reach subarray of size 5, we do not shrink our sliding window. And we keep moving it ahead by moving both left and right pointers. Once we reach the subarray of size 6, our sliding window's right pointer is updated while left keeps calm, and sliding window size is updated to 6. I hope it helps.
Easier way to understand this is that the window will eventually reach a condition (with current highest length) when it just crosses K, so what we were doing is move left pointer until we find another zero effectively reducing the highest length we had gotten so far and we start to make a window freshly in hopes that it'll cross the last highest but this is pointless. The answer now will be either higher or the current highest so we do not let the window shrink beyond current highest until zero count comes within K again to eliminate the useless window refresh
I could implement this myself in the first try, thanks for helping me gain confidence raj.
You are the best striver , solving two pointer became very easy after seeing your videos.
OMG i solved it by myself. Idk if it was an easy question but your lectures are super helpful.
class Solution {
public int longestOnes(int[] nums, int k) {
int l=0,r=0,max=0,zero=0,n=nums.length;
while(rk){
if(nums[l]==0) zero--;
l++;
}
if(zero
00:06 Solving the problem of finding the maximum consecutive ones with at most K zeros.
02:57 Using sliding window to find longest subarray with at most K zeros
07:43 Using sliding window to find maximum consecutive ones with K zeros.
10:13 Using sliding window technique to manage consecutive ones and zeros efficiently
15:10 Use a sliding window technique to handle scenarios with more zeros than K.
17:40 Optimizing Max Consecutive Ones III using sliding window technique
22:01 Illustration of updating max consecutive ones with sliding window approach
24:13 Algorithm to find max consecutive ones after K flips.
28:58 Algorithm works with time complexity of O(n) and space complexity of O(1).
these explaining style is good striver please make more videos like that only
Thanks for the optimal approach.
Another Approach:-
class Solution {
public:
int longestOnes(vector& nums, int k) {
int size = nums.size();
int l = 0;
int r = 0;
vectorind;
int i = 0;
int ans = 0;
for(int i = 0;i
How ppl build such logics like my brain stopped braining for a while when i see question.
Literally me
@@psg9278 us bro us
"UNDERSTOOD BHAIYA!!"
Mza aagya
another approach
class Solution {
public:
int longestOnes(vector& nums, int k) {
int n = nums.size(); // Get the size of the input vector
int ans = 0; // Variable to store the maximum length of subarray with at most k zeros
int ct = 0; // Variable to count the number of zeros encountered
vector v1; // Vector to store the cumulative count of zeros
// Traverse the input vector to fill the cumulative count of zeros
for (int i = 0; i < n; i++) {
if (nums[i] == 0) {
ct++; // Increment the count if the current element is zero
}
v1.push_back(ct); // Add the cumulative count to the vector
}
int j = 0; // Left pointer of the sliding window
int g = k - 1; // Right pointer of the sliding window
if (k == 0) {
g = 0; // Handle edge case when k is 0
}
// Traverse the input vector using the sliding window approach
while (g < n && g < v1.size()) { // Ensure g does not go out of bounds
// Calculate the number of zeros in the current window
int temp = v1[g] - (j > 0 ? v1[j - 1] : 0);
if (temp
Thankyou so much Striver for all you efforts throughout in delivering us so much valuable content. Any student / working professional can now be able to transition their career without paying money for courses.
Would also like your insights on the point :
While preparing for interviews most of the aspirants are going through the videos solely and solving the question after completely watching the video. And also are feeling lazy trying to solve the question on our own. What is the best way to complete any topic without being lazy and how should an aspirant approach any topic/playlist?
another way can be use sum to count no of ones and check if len-sum >k, reduce sum if nums[l]=1, update l and find maxlen
Great job... putting out this playlist. And, I don't see notes out there in TuF website?? for 2P and Sliding Windw problems
class Solution {
public:
int longestOnes(vector& nums, int k) {
// Brute force
int length = 0;
int maxLen = 0;
for(int i=0;i
AWOSOME LECTURE. I SOLVED THIS QUESTION BY MYSELF !!!!
My solution with same logic
class Sliding_window {
public static void main(String[] args) {
int[] arr = {1,1,0,1,1,1,0,1,0,1,1,1,0,1};
int k =2;
int l=0,r=0,zeros =0,max=0;
while(r< arr.length){
if(arr[r] == 0){
zeros++;
}
if(zeros>k){
if(arr[l]==0){
zeros--;
}
l++;
}
if(zerosmax){
max = r-l+1;
}
r++;
}
System.out.println(max);
}
}
we can also use a queue instead of nested loop ,
Here the time complexity is O(N), and the space complexity is O(N)
int max =0, l=0,r=0;
Queue index =new LinkedList();
int zero =0;
while(rk){
l=index.poll() +1;
zero--;
}
}
max =Math.max(max, (r-l+1));
r++;
} hope you like my solution🙂
Quality teaching brother... love you
in worst case lets assume all array elements are zero and k=0 it takes still 0(2n) the most optimal one
No, the optimal answer will never run for O(2n), it will always be O(n).
According to your example, the left and right pointer both get updated for each iteration but their updation takes only O(1) i.e. constant time
😊❤2:@@devanshkumar7723
Solved it on my own after seeing the brute force. Buit I used a deque making it more simple
class Solution {
public:
int longestOnes(vector& nums, int k) {
int i = 0, j = 0, zeroes = 0;
int ans = INT_MIN, len = 0;
int n = nums.size();
deque q;
while(j < n){
if(nums[j] == 0){
zeroes++;
q.push_back(j);
}
if(zeroes > k){
zeroes--;
int x = q.front();
i = ++x;
q.pop_front();
}
len = j - i + 1;
ans = max(ans, len);
j++;
}
return ans;
}
};
Thank you so much, it was very helpful
C++ CODE BASED ON THIS LOGIC.
class Solution {
public:
int longestOnes(vector& nums, int k)
{
int n = nums.size();
int left = 0, right = 0, maxlen = 0, count0 = 0;
for(right = 0; right < n; right++)
{
if(nums[right] == 0)
{
count0++;
}
while(count0 > k)
{
if(nums[left] == 0)
{
count0--;
}
left++;
maxlen = max(maxlen, right - left + 1);
}
return maxlen;
}
};
This was asked to me in Arcesium Intern Technical Round 1
broo ur explanation is greattt but can u please reduce the length of the video size cause even though i am able to solve the problems on my own i am still watching your videos to gain better understanding but it would have been great if the length of the video was smaller
16:21 if condition is not required while is doing the same thing
Even in optimal solution. worst case will take O(2n)
understood bhaiya
Thankyou bhai very Good explanation
very thankful to you
Thank you!
dhanyawad guru ji🙏
class Solution {
public:
int longestOnes(vector& nums, int k) {
int l=0, r=0, maxLen =0, zeros=0,len=0;
int n=nums.size();
while(r k){
if(nums[l] ==0){
zeros--;
}
l++;
}
if(zeros
Buddhi khul gayi bhai. Was stuck on this problem for a long time, I had the solution, but still wasn't able to understand it even after dry running it, why did it work. Thank you.
change while to if and we trim TC from O(2n) to O(n) . VOILA 👍👍
int longestOnes(vector& nums, int k) {
int i = 0, j = 0;
int n = nums.size();
int zero = 0;
int maxi = 0;
while (j < n) {
if (nums[j] == 0) {
zero++;
}
while (zero > k) {
if (nums[i] == 0) {
zero--;
}
i++;
}
maxi = max(maxi, j - i + 1);
j++;
}
return maxi;
}
00:06 Solving the Max Consecutive Ones III problem using two-pointer and sliding window techniques.
02:57 Finding longest subarray with at most K zeros.
07:43 Implementing sliding window technique with two pointers
10:13 Sliding window technique helps in efficiently handling zeros in the array.
15:10 Maintain sliding window to count consecutive ones with up to K flips
17:40 Using sliding window to find maximum consecutive ones with K flips
22:01 Using two pointers and sliding window to find maximum consecutive ones with allowed updates.
24:13 Using 2 pointers and sliding window to find max consecutive ones with k allowed flips
28:58 Algorithm works by avoiding moving L extremely to the right
i didn't understand one thing
in most optimum sol by the time "R" reaches end "L' has traversed N-k (k is some constant)
so shouldn't time complexity be O(N+N-k) which is same as O(2N)🤔🤔
SOLVED BY MYSELF BUT SKIPPING VIDEO MAY COST ME LOSE OF MOST OPTIMAL SOLUTION ❤🔥❤🔥
Can we do left = right-1 instead of while loop for validating condition!!
Do your previous website also exist? I have notes for questions attached there, I am not able to find it
Great 🔥😃
Last wala bahut jada intuitive hai.
Ones you get window size for cntZeros
Thanks Brother💌
Understood. thanks
can somebody explain me what happens when there is brigde of lets say less than k zeroes between two groups of ones
Understood
You are amazing....wowwwwwwwwwwwwwwwwwwwwwwwwwwwww
Hey, I have 1 question .
What if number of 0's are less than K so we have to flip all zeros and then k-no. of zeros times we have to flip 1 as well, right? How will we able to solve that question?
understood
Thanks
UnderStud❤❤❤
sir I can't understand how to take length like j-i+1 and some times other length
can you give me any idea
int main() {
int size_arr , flips ;
cin >> size_arr >> flips ;
vector arr(size_arr,0) ;
vector arr0(size_arr + 2 ,0);
int count = 0 ;
arr0[count] = -1 ;
for(int i = 0 ; i < size_arr ; i++){
cin >> arr[i] ;
if(arr[i] == 0){count++;
arr0[count] = i ;
}
}
count++;
arr0[count] = size_arr ;
int l = 0 ;
int r = l + flips + 1 ;
int max_len = arr0[r] - arr0[l] - 1 ;
while(r < count + 1){
r++ ; l++ ;
max_len= max(max_len,arr0[r]-arr0[l] - 1) ;
}
cout
Thanks❤
Please upload codes for all...
In GFG the most optimized approach is giving out TLE with some 50 Testcases left out of 500, and the better one which uses two while loops is passing out all the test cases without TLE! WHY IS THAT SO?
@@ananyamishra382 the idea is to get the maximum window size possible or Better i should say RETAIN the maximum window size , that's why we are not shrinking L .
@@ananyamishra382 we are retaining the size only for the window which is good but small in size , that means considering that condition won't play any vital role in it , therefore we are not shrinking the size for the window which is smaller than the max size window which we have attained earlier.... simply we are just looking for the window with greater length than maxlen. Hope it makes sense
Optimal Approach Not Working
code have not been updated in striver link
great
TC - O(N)
class Solution {
public int longestOnes(int[] arr, int k) {
int r=0;
int l=0;
int maxlen=0;
int zeroes=0;
while(rk){
if(arr[l]==0){
zeroes--;
}
l++;
}
if(zeroes
How is the optimal code running on an example like:
arr = [1,0,1,0,1,0,1,0], k=1.
Isn't the left pointer traversing near about n times as well?
You are asking
If we access value 3 times in one loop it will be 3n
I hope you got why it would be n
can anyone tell me what's wrong in this code ?
int longestOnes(vector& nums, int k) {
int n = nums.size();
int l = 0, r= 0, maxlen = 0,zeros = 0;
while(r < n){
if(nums[r] == 0)zeros++;
if(zeros k){
if(nums[l] == 0){
zeros--;
}
l++;
}
}
}
return maxlen;
}
18:50 There is a while loop inside another while loop. Then how come Time complexity in worst case is O(n)+O(n) and not O(n^2)?
Because for every element it is not running for n times. Even in worst case , it runs for n times for last element only.
@@RK-Sonide4vr gotcha
@@RK-Sonide4vr still it runs N time inside a N time running while loop so why not n^2?
@@amansaini4969 It does not run N times inside the N.
You need to imagine what an actual n^2 loop is like - It means that for every value till N of outer loop, the inner loop is running N times ALWAYS. It's not always here. The outer loop runs for N times surely, but are you always updating the ith pointer in every value of the outer loop? Let's say there are 8 numbers, 5 1's and then 3 0's and the K = 1. That means your outer loop will move one by one to the first 0 after crossing 5 1's, but did you actually run the inner loop while crossing each and every 1? You only run that inner loop just when the condition is violated which does not happen ALWAYS as it should be in n^2.
Take the worst case, where at the end you have 3 zeroes and K = 2. In that case, the inner loop has to cover till n-2 index which is near about N, but it did in ONE instance of outer loop, not for every instance of outer loop for it to become multiplicative. As it happened for one instance, it got added up and became 2N.
Java Code :
class Solution {
public int longestOnes(int[] nums, int k) {
int n = nums.length;
int maxLen = 0;
int zeros = 0;
int l = 0, r = 0;
while(l
How about this solution
public int longestOnes(int[] nums, int k) {
int i=0;
int l=0;
for(i=0;i
🙌🏻
19:20
Interviewer is never happy 😞
god
C++ solution
TC- O(N)
SC -O(1)
int longestOnes(vector& nums, int k) {
int i =0 , j =0 ;
while(j < nums.size()){
if(nums[j] == 0 ) --k;
if(k < 0){
if(nums[i] == 0) ++k;
++i;
}
++j;
}
return j - i;
}
public int longestOnes(int[]nums,int k){
int l=0,r;
for(r=0;r
us
I don't know why but whenever i am watching these problem statements of two pointer, the first idea striking on my mind is a dp state🥲...then soon understanding dp is having at least 2 states so gotta optimise it
Understood
understood
Understood
understood
understood
understood
understood
understood
Understood