Looks like he forget about this account password 😂

i think 2nd question is somewhat like mergeIntervals , for every ith pair we will see if the jth pair can be merged with it

Very good explanation

Superb explanation

Great content what ever you said 💯% true m here in Chennai from last 7 months people of chennai is very nice and educated there is no hate for North Indian here even they respect hindi speaking people more what ever we see in media is totally false and fake Tamil people are hard working and honest they are very down-to-earth very helpful no showoff life in Chennai is peaceful you can settle here with family also

in second question, what will be the approach if we can rearrange the elements ?

😪

thank you for making this channel

Bhaiya what about job losses due to AI and being replaced by AI?

where is code

bhaiya konsa pentab and software h , mujhe lena h dsa k notes and question k liye

Is it in c or c++ language?

first question is leetcode 1004 max consecutive ones III. already solved using sliding window.

Very nice explaination

how is kla tencor Chennai?

result[0]=1; why you use this??

i did this in question in o(n^2 + m) because of that i am getting time exceed in leetcode algorithm optimisation is very hard

I think First question is of DP and second question is of dfs with memorization. O(N) complexity for both

can u please provide me pdf of this book

This question has been asked to me in EPAM Company's interview. Thanks Shreyansh for DSA series

But your speaking style is funny 😂

from collections import Counter def minimum_window(str1,str2): window="a"*(len(str1)+1) left=0 count=0 hashmap=Counter(str2) for i in range(len(str1)): if str1[i] in hashmap: hashmap [str1[i]]-=1 if hashmap [str1[i]]>=0: count+=1 while count==len(str2): if len(str1[left:i+1]) < len(window): window=str1[left:i+1] if str1[left] in hashmap: hashmap [str1[left]]+=1 if hashmap [str1[left]]>0: count-=1 left+=1 return window str1='ADOBECODEBANC' str2='ABC' print(minimum_window(str1,str2)) str1 = "PRWSOERIUSFK" str2 = "OSU" print(minimum_window(str1,str2))

Just one for loop in the efficient approach.......use one more index say nonzero index.....and make nums(nonzeroindex) = num and nums(i) = 0 .....???

Ground to earth

if someone needs the code . class Solution { TrieSearch root; public List<List<String>> suggestedProducts(String[] products, String searchWord) { List<List<String>> result = new ArrayList<List<String>>(); root = callFillTrie(products); String prefix=""; for(char c : searchWord.toCharArray()) { List<String> outputforprefix = new ArrayList<String>(); prefix = prefix + c; outputforprefix = searchString(prefix); result.add(outputforprefix); } return result; } public List<String> searchString(String prefix) { TrieSearch searchStartNode = root; List<String> result = new ArrayList<String>(); for (char c : prefix.toCharArray()) { if (searchStartNode.nodes[c - 'a'] == null) { return result; } searchStartNode = searchStartNode.nodes[c - 'a']; } dfs(searchStartNode, result); return result; } public void dfs(TrieSearch rootNode, List<String> outputforprefix) { if (outputforprefix.size() == 3) { return; } if (rootNode.word != null) { outputforprefix.add(rootNode.word); } for (int i = 0; i < 26; i++) { if (rootNode.nodes[i] != null) { dfs(rootNode.nodes[i], outputforprefix); } } } public TrieSearch callFillTrie(String[] products) { TrieSearch result = new TrieSearch(); for (String product : products) { insertIntoTrie(product, result); } return result; } public void insertIntoTrie(String product, TrieSearch trieSearchNode) { for (int i = 0; i < product.length(); i++) { char ch = product.charAt(i); int index = ch - 'a'; TrieSearch searchNode; if (trieSearchNode.nodes[index] == null) { searchNode = new TrieSearch(); trieSearchNode.nodes[index] = searchNode; } searchNode = trieSearchNode.nodes[index]; if (i == product.length() - 1) { searchNode.word = product; } trieSearchNode = searchNode; } } class TrieSearch { String word; TrieSearch nodes[] = new TrieSearch[26]; } }

for 2nd solution eg. " )(* " expected ans - False as per soln - True i think second solution is not valid

What is this man, you dont even know how to explain the algorithm, you are just showing the code. Where is intuition, approach and nothing is explained clearly. Please improve yourself, you are wasting others time.

Hi Shrayansh, I miss this series. I have decided to follow along with you for 100days. But unfortunately, There were no videos for the past 2 weeks now. What happened?

Problem for food in Chennai is a joke.

try to learn south Indian language when you come to south. don’t impose Hindi here.

Millenia Business Park Phase - II Campus 3A 143, Dr. M.G.R Road, N Veeranam Salai, Kandhanchavadi, Perungudi, Chennai, Tamil Nadu 600096 Idhar office ho to kaha rehna sahi rehega aur kitna kharcha ho sakta hai for single living? For pg and 1bhk or 1RK flat? Btw, I have been searching for a good comprehensive vdo regarding this and your video is the proper one I need. So, thank you❣️

Hi Shrayansh, Please post the question by Morning so that we can give it a try before coming to the solution video. Kindly consider.

Promo`SM

2 nd question do check it out and suggest any optimizations #include <iostream> #include <bits/stdc++.h> using namespace std; int ans = INT_MIN; void f(int idx,vector<int> &arr,int cnt,vector<int> &dp); int main() { // Write C++ code here vector<int> arr = {34,23,1,2,42,12,2,4,1,3,4,7,7,8,8,9,9,12}; //output : 3 pairs to be deleted = 34,23 | 42,12 | 1,3 vector<int> dp(arr.size(),-1); for(int i=0;i<arr.size();i+=2){ f(i,arr,1,dp); } cout<<arr.size()/2 - ans; return 0; } void f(int idx,vector<int> &arr,int cnt,vector<int> &dp){ // cout<<"Running"<<endl; // int ans = INT_MIN; if(dp[idx]!=-1){ cnt = dp[idx]; return ; } if(idx>=arr.size()-2){ // cout<<"Running"<<endl; ans = max(ans,cnt); dp[idx] = ans; return ; } for(int k=idx+2;k<arr.size();k+=2){ // cout<<arr[k]<<endl; if(arr[k] == arr[idx+1]){ // cout<<"Runing"<<endl; f(k,arr,cnt+1,dp); } if(k>=arr.size()-2){ ans = max(ans,cnt); dp[idx] = ans; return ; } } }

thanks for the question, dnf algo i think

Amazing questions, thanks for sharing, eagerly waiting for solutions from your side For the 2nd question, as we cannot re-order, as well as rotate, can we do it in O(n*n) complexity ?.. where for every given pair, we check for a matching pair in the index +1 of the given pair, and find the longest / max number of pairs doing the same in short, running two loops would do the work?.. please suggest

this was my solution with dynamic programming. I did not use any extra space as well: // Online C++ compiler to run C++ program online #include <iostream> # include<string> # include <vector> using namespace std; int returnAnswer(vector<int> arr) { int n = arr.size(); int value1 = 0; // dp[i-1] int value2 = 0; // dp[i-2] vector<int> dp(n,0); for(int i=0;i<n;i++){ int value3 =arr[i]; if(i-1>=0) { value3=max(value3,arr[i]+value1); value3=max(value3,stoi(to_string(arr[i-1])+to_string(arr[i]))+((i-2)>=0?value2:0)); } value2=value1; value1=value3; } return n==0? 0 : value1; } int main() { // Write C++ code here std::cout << "Hello world!"; int ans = returnAnswer({3,19,191,91,3}); cout<<ans; return 0; } I hope this helps :)

Good explanation with deep details.. Thank you!!

For the first question, i can add memorization to existing solution and it can reduce the time complexity to O(n)

I wish they asked all this for the interview XD. I guarantee, with my luck, the day I sit for the interview they are going to ask me something along the lines of Graphs DP and all the super tough topics.

can you please share the link where I can purchase the latest addition?

I was also solving using 2 pointers approach . Thanks for the 2nd approach class Solution { public void moveZeroes(int[] nums) { if(nums.length>1){ int firstPointer=0; int secondPointer=1; while(secondPointer<nums.length && firstPointer<nums.length){ if(nums[firstPointer]==0){ if(nums[secondPointer]!=0){ nums[firstPointer]=nums[secondPointer]; nums[secondPointer]=0; firstPointer++; } secondPointer++; }else{ if(nums[secondPointer]!=0){secondPointer++;} firstPointer++; } } } } }

class Solution { public int[] sortByBits(int[] arr) { HashMap<Integer, ArrayList<Integer>> mapList = new HashMap<>(); for (int i : arr) { if(mapList.containsKey(getbitCount(i))) { ArrayList<Integer> list = mapList.get(getbitCount(i)); list.add(i); Collections.sort(list); mapList.put(getbitCount(i), list); }else { ArrayList<Integer> list = new ArrayList<>(); list.add(i); mapList.put(getbitCount(i), list); } } ArrayList<Integer> result =new ArrayList<>(); for(int i=0;i<=32;i++) { if(mapList.containsKey(i)) { result.addAll(mapList.get(i)); } } return result.stream().mapToInt(Integer::intValue).toArray(); } public String getBinary(int n) { StringBuilder s = new StringBuilder(); while(n>0) { s.append(n%2); n=n/2; } return s.reverse().toString(); } public int getbitCount(int n) { String s = getBinary(n); int count=0; for(int i=0;i<s.length();i++) { if(s.charAt(i)=='1') { count++; } } return count; } } brute force

you forgot to share the LC link in description today

Hello sir, can you please give me a tip, how to implement this code using class Scanner, so that I can input expression, and the console gives me an answer, thank you sir

I am on notice period and I have 2 months in my hand till 3rd Jan. I have seriously followed your java series and I have learned a lot and built a lot of confidence in myself , Thank You So much for such amazing contents. I have one offer in my hand with 35% increment but that is with one service based company. I am trying my best and need to start your system design videos. Just want your input 2 months would be enough to crack a good prouct based company if i spend 6-7 hrs daily in preparation and practice?

Hi Shrayansh. It would be good if you add the LC link of the problem in the description. Thank You

one doubt, please guide, i am not getting why we arent doing the following : - while(I>=0){ nums1[k] = nums1[I] ; i--; k-- ;} whereas we are doing it for j / nums2 elements ??

starting from last instead of starting index is an simple, yet amazing and eye-opener approach, kudos