Double-double lang...

gajab..

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The concept is - n-(x+1) => n - x - 1 = 8 - 4 =>4

wtf I thought it was really hard until I saw this video. Thanks for the explanation

Not understood middle for loop. 😢

nice solutions you are posting, will bookmark your channel. 😀

Good procedure

code gives and error

wtf! it is so much simple

Awesome solution bro. Love from America

Thanks for your effort for clear explaination.

THANK U SOO MUCH!!!

How to did we get this formulae? idealSpots=No.of Chunks - cooling time?

Thank you

worst video of all time . get lost from yt.

Crisp and to the point explanation 🙏🙏, Thank you

We all understand the recursion tree. We just don't understand this specific line: (!winnerSquareGame()). Which is the main part of the problem and you didn't explain that.

i/2 i didn't understand any one explain it

great , can you explain , binary search method also. If you are still active, please make a video to explain binary search method for this problem, I am subscribing you for notifications.

you took wrong numbers from the list

3 years after , this problem still exists on leetcode

sir your solution expalnation is best thanks uu...

Below average explanation , Thanks

very nice solution keep it going bro

Thanks

but what if we change 0 to11 and 1 to 12 and 6 to 9 then the minimum value we be 9 and maximum wiil as it is 15 then difference will create of 6 . 15-9=6

if the String is long then we have to use this code : - import java.math.BigInteger; class Solution { public int compareVersion(String version1, String version2) { String v1[] = version1.split("\\."); String v2 []= version2.split("\\."); for (int i=0; i < Math.max(v1.length, v2.length); i++) { BigInteger num1 = i < v1.length ? new BigInteger(v1[i]) : BigInteger.ZERO; BigInteger num2 = i < v2.length ? new BigInteger(v2[i]) : BigInteger.ZERO; int comp = num1.compareTo(num2); if(comp == 0) { continue; } return comp; } return 0; } }

Kadane's Algorithm it is, I no where see DP. Please don't mention DP in heading

naresh bhai !!

Thank you sir! Your video was very helpful to me!

Thank you! Great quick video!

u just know the solution but u cant explain it

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I am also in uae and searching for java developer position having 1.9 yrs of experience in IBM india, if you can help in placement in dubai it will b helpful. Thank you i understand this question approch from your video.

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Very well explained 👏

I didn't even understand the problem description. After your initial beautiful explanation, I was able to code it myself! Thanks :)

wonderful solution sir I was watching 2-3 videos before and none of the approach was understandable and after watching yours I coded myself.

Damn bro! the explanation was so neat!! keep it up

for the algorithm.Thanks

i"ll be using this algorithm in my assignment, i'll edit my comment if i got A+

Thank you very much.

Thank you! Great reasources

class Solution { public: int length (ListNode *&l){ int cnt = 0; ListNode *t = l; while(t){ t = t->next; cnt++; } return cnt; } void addZero(ListNode *&l,int cnt){ ListNode *add = new ListNode(0); ListNode *b = add; cnt--; while(cnt > 0){ cnt--; ListNode *n = new ListNode(0); add -> next = n; add = n; } add -> next = l; l = b; } void addll(ListNode *&l1,ListNode *&l2,int& cnt){ if(!l1 && !l2)return ; addll(l1->next,l2->next,cnt); int sum = l1->val+l2->val+cnt; cnt = sum / 10; l1 -> val = sum % 10; } ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { int c1 = length(l1); int c2 = length(l2); int l = abs(c1-c2); if(c1 > c2) addZero(l2,l); else if(c1 < c2) addZero(l1,l); int c = 0; addll(l1,l2,c); if(c){ ListNode *n = new ListNode(1); n -> next = l1; l1 = n; } return l1; } };

great thank you

While very simple it's pretty elegant to encode unbalanced as -1. The boolean statement could even be height(root) >= 0, which reads really well.

count = (k-1)*m ?? not understood this code

Time complexity here is O(n) is because you do not sort the result. Actually your final answer is not sorted that is not correct they expect final array should be sorted so you need to sort the array at the end and that is why time Complexity is Onlogn Please correct me If I'm wrong