LeetCode Max Consecutive Ones III Solution Explained - Java

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Опубликовано:

5 окт 2024

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Комментарии : 43

@artur4427 8 месяцев назад

what helped me understand this is the realization that i and j are not the actual edges of the longest consecutive segment. the window never shrinks, so the max length is kept track of implicitly by the window's size.

@danielpark4204 4 года назад

took a while but i get it now, however I think its more intuitive to just cache the maxLen in a variable and shrink until K is positive again rather than caching the width throughout.

@aj9706 3 года назад

Yes absolutely

@tapanbasak1453 3 года назад

Can anyone please explain why i-j always gives the correct result without keeping track of the max subarray at all times.

@HoangLe-qr9si 3 года назад

i think that was handled by if k is positive then j will not increase

@user-le6ts6ci7h 3 года назад

It is because , as long as you are able to get contiguous 1 , with maximum k flips of zero, the i value goes on increasing and j remains at the position so far. And once the k becomes negative( that is, k+1 zeros are flipped) then both j and i are incremented , meaning that j-i is still fixed in length and equal to the size of contiguous 1 found so far . So actually this maximum fixed size windows is moving throughout the array which is always expanding rather than contracting

@luz8011 2 года назад

@@user-le6ts6ci7h thank u for explaining it :)

@amansaxena4446 Год назад

not making sense at all

@subhamshaw1726 3 года назад

that was really simple, great explanation

@mehdihassan93 2 месяца назад

for people who are looking for python3 solution: l = 0 for r in range(len(nums)): if nums[r] == 0: k -= 1 if k < 0: if nums[l] == 0 : k += 1 l += 1 r += 1 return r - l

@vijayakumareyunni6010 9 месяцев назад

Key thing to be explained is how the pointers are to be updated which was not explained clearly. He also got confused when coding and somehow it worked.

@SnehaBharathiReddy 5 месяцев назад

Straight forward Explanation. Thanks Nick

@akshittyagi1784 3 года назад

2 mins and 30 seconds into the video, and Nick's like enough talk, let's just get straight to business😂😂 respect man👍

@miaohong601 4 года назад

This is much easier to understand than the most upvoted in discussion, Thanks Nick!!

@MrThepratik 4 года назад

Getting my head around it

@emerylin6793 2 года назад

while the code is simple, you dont get the longest number of ones, which is not the right answer.

@gitanjalikumari9262 3 года назад

So nice solution 👍😊

@mohammedghabyen721 2 года назад

public class Solution { public int LongestOnes(int[] nums, int k) { int left=0; int right=0; int ck = k; int maxNu = 0; while(right0) { ck--; right++; }else{ if(nums[left]==0) { ck++; left++; }else left++; } } maxNu = Math.Max(maxNu,right-left); } return maxNu; } }

@kakwancheng9493 2 года назад

This logic is really clear, thanks a lot!!!

@edwardnewgate2198 5 лет назад

What about the test case "0011111000"- and with K as 2? In that case- j wouldn't be able to reach a 0. How would we get the answer in that case??

@chaoschao9432 5 лет назад

j will reach 0, you may run the test case step by step.

@MrSaiyah007 4 года назад

@@chaoschao9432 it will not

@ArtInMotion6912 4 года назад

Yup but the size is same both j and i increments so no effect on it

@ashutoshchauhan4928 3 года назад

I had the same doubt, thanks for giviing the rationale behind it @brock lesner

@sarscov9854 3 года назад

it works. I put it in the compiler, gave me 7.

@aayush5474 4 года назад

whats the timecompleity?

@mikea3264 3 года назад

O(n) because there is just one loop going through the whole array.

@rajatsrivastava555 3 года назад

clean and not fancy!!

@sujayshanbhag2055 Год назад

What happens if i reaches end before k becomes positive?

@sujayshanbhag2055 Год назад

I think this is more Intuitive : - int longestOnes(vector& nums, int k) { int last=0; int first=0; int n=nums.size(); int maxLen=0; while(last< n) { if(nums[last]==0) k--; while(k

@sujayshanbhag2055 Год назад

Intuition behind the code in the video: This is notes I made for me, but if it helps you, great!. Above code: Whenever the number of zeros taken exceeds k, move the starting point of window until we recover the 0. In video code: We don't run a loop to recover the 0. Whenever we take in a bad zero, we increment the start. By doing this whenever, we increment last, first gets incremented and we maintain the length. [We do not keep track of the actual sub array]. If we recover the 0 in process, only last will start to increment, increasing the length of the subarray.

@comedycentral4333 3 года назад

What would be this test case "1110" if k = 1 the output will be 3 right, but the above code returning 4, can anyone please suggest me?

@user-le6ts6ci7h 3 года назад

No , the answer would be 4. At the end of the loop i will be having a value 4, the reason of which it exits the loop. And j will be at 0 .So returns 4-0