Video

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

NICE SUPER EXCELLENT MOTIVATED

NICE SUPER EXCELLENT MOTIVATED

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.

NICE SUPER EXCELLENT MOTIVATED

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

hey How did you come on the formula for chunk ? whats the logic

what a champ, happy I did not have to do this problem on my own ❤‍🔥

working java code: class Solution { public TreeNode construct(int[] pre, int preStart, int preEnd, int[] post, int postStart, int postEnd, HashMap<Integer, Integer> map) { if(preStart > preEnd) return null; TreeNode root = new TreeNode(pre[preStart]); if(preStart == preEnd) return root; // when there is only one node in left or right subtree int postIdx = postStart; postIdx = map.get(pre[preStart+1]); // preStart+1 bcz we can determine left and right subtree nodes by checking the pos of next node in preOrder in postorder int len = postIdx-postStart +1; root.left = construct(pre, preStart+1, preStart+len, post, postStart, postIdx, map); root.right = construct(pre, preStart+len+1, preEnd, post, postIdx+1, postEnd-1, map); return root; } public TreeNode constructFromPrePost(int[] preorder, int[] postorder) { HashMap<Integer, Integer> map = new HashMap<>(); int n = postorder.length; for(int i = 0; i < n; i++) { map.put(postorder[i], i); // indx, value } return construct(preorder, 0, n-1, postorder, 0, n-1, map); } }

Thanks

ty

The idea is brilliant, but kind of slow. Here are two approaches - approach 1 - def increasingTriplet(self, nums: List[int]) -> bool: num1 = num2 = float('inf') for n in nums: if n <= num1: num1 = n elif n <= num2: num2 = n else: return True return False approach 2 - the one explained in this video [Note that, I am using only 1 pass to find both leftMin array and rightMax array] def increasingTriplet(self, nums: List[int]) -> bool: leftMin, rightMax = [], [] n = len(nums) j = n-1 minVal, maxVal = float('inf'), float('-inf') for i in range(0, n): minVal = min(minVal, nums[i]) maxVal = max(maxVal, nums[j]) leftMin.append(minVal) rightMax.append(maxVal) j-=1 for i in range(n): if nums[i]>leftMin[i] and nums[i]<rightMax[n-i-1]: return True return False

I cant thank you enough sir!!! The explanation was absolutely amazing!!!! I didnt have a doubt even for a second!! Thank you so much sir!!!

Its fine but the explanation part can be better

Insted of break if we return score there than no need of maxScore

I saw 4...5 videos before watching your video sir.....great explanation....tnx👏

This is not dfs but a normal recursive function to iterate the hashmap Dfs deals with a map inside the node and go recursive with the next next nodes.

Thank You So Much for this wonderful answer.............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

class Solution { public: int bitwiseComplement(int n) { if(n==0) return 1; int i = 0; int ans = 0; while(n>0){ int bit = n&1; if(bit==0){ ans = ans+pow(2,i); } n = n>>1; i++; } return ans; } };

Simple Explanation ❤

Can someone explain (p - 1) /k +1 ? I solved it using time = p /k ; if p % k !=0 then time++ My approach of calculating time is worse than this simple formula, which i dont yet understand. I know that if you divide some natural number by number N, that the maximum remainder can be N - 1.

wow. Never throught of that.

You are the king of this problem explanation. Thank you very much.

beauty

Explanation is not good, it could have been better. Please try to improve.

Please come with stack ,queue, linkedlist problems more in java

u r awesome, much better than neetcode, strive solution

naresh you also started vlogging

Great video, thank you!