Maximum Ones after Modification InterviewBit Code + Examples C++

Given a binary array A and a number B, we need to find length of the longest subsegment of ‘1’s possible by changing at most B ‘0’s.
A[i]=0 or A[i]=1
Input Format
First argument is an binary array A.
Second argument is an integer B.
Output Format
Return a single integer denoting the length of the longest subsegment of ‘1’s possible by changing at most B ‘0’s.
Example Input
Input 1:
A = [1, 0, 0, 1, 1, 0, 1]
B = 1
Input 2:
A = [1, 0, 0, 1, 0, 1, 0, 1, 0, 1]
B = 2

[@ayushpathak2503]

I don't why she is not that famous....her explanation is way better than every famous youtuber...

[@utkarshasingh1836]

couldn't stop myself from commenting, you explained it with such simplicity and clarity.

[@11iqra]

Best explanation of the problem on the internet!

[@rankitgujjar80]

Crystal clear explanation as always 🙇‍♂🙇‍♂

[@AjayKushwaha-vi5vg]

superb clear explanation.

[@manojkr7554]

Wow what a great explanation I love u dear ❤️💗🥰

[@feelit5386]

truly amazing, aab nahi bhulunga yeh wala concept🥰

[@ayushigupta685]

great explanation .. keep posting .I just love it

[@kirankumbhar2884]

Thanks for the solution! I solved this with a little bit different approach.
1. keep left and right at the start of the string
2. initialize ones_freq = 0 and result = 0
3. increment right by 1 and check if current character is 1, if yes increment ones_freq by 1
4. while right-left+1 - ones_freq > B (current window has more zeros than we can flip)
4.1 increment left by 1 (shrink window) if A[left] == 1 decrement ones_freq by 1
5. Here we have window satisfying the criteria so take max(result, right-left+ 1)
This approach is basically keeping count of 1 in window and based on difference == allowed flips it is calculating the max size

[@ravitalaugh9017]

your explanation is really good

[@LearnWithAnmolll]

great explanation !!!!

[@abhinandanraj535]

just do it by using Sliding window concept
int Solution::solve(vector &A, int B) {
int count=B;
int left=0;
int right=0;
int n= A.size();
int maxLen=0;
while(right=0)
{
count--;

}
if(count>=0)
{
maxLen= max(maxLen,right-left+1);
}
while(count

[@luckydewangan5985]

best explation I got for this question here thankyou so much for this content

[@anmolchauhan3822]

Very Nice Explanation Mam😄😄

[@AdityaYadav-kf4uc]

never commented on any youtube video but this explanation madee me to, thanks

[@kunwarprashant1698]

Nice explanation

[@lostcyrus8578]

Good explanation keep it up

[@lostcyrus8578]

Please make more videos on popular questions.. you can refer to striver SDE sheet medium and hard ones

[@ankushladani496]

Explained so easily....

[@dhruvsanghvi5562]

mam could we also do it by recursion by thinking to either include one zero or not and then calculating maximum calculating for that choice

[@saikrishna872]

Good Explanation What is Time and Space complexity

[@harishkulkarni7244]

Nice explanation, also in order to develop the intuition to solve problems is there any way? Or is this just practice, like solve n number of problems so that if n+1 problem is similar to any problem in those n just think similarly?

[@notyet5009]

Special thanks to alisha's brother who is unable to sleep bcz of these lectures.

[@SaTyamRaWaTvloGs]

💯

[@darkexodus6404]

After watching first 10mins of your video, this is how I tried to implemented it. My thought process was like to maintain a window.
int findZeroes(int arr[], int n, int m) {
vector zeros;
for(int i=0; i

[@collegematerial5348]

Why you minus for calculate length

[@rahulgheware5752]

By using queue it is possible to store previous zero

[@lankeshhj5725]

avoid extra space

[@abhishekkarn8918]

whats the time complexity

[@rankitgujjar80]

it will be o(2n) at worst ig

[@abhishekanand9430]

complixity will be greater than n

[@nostalgiccringeallhailchel3881]

weird explaination. Would have been much easier to understand if you just said you were gonna use sliding point approach

[@rishabhsingh-gw3gf]

woh bnda peeche so raha hai mooh par takiya laga kar🤣🤣