Pick Toys | An Interesting Sliding Window Problem 

Like this comment if your were able to reduce it into the previous problem by your own. (I want to get a count of class toppers :P)

Now I am placed in good company with good package after some rejections.For me you are more than a teacher and I know shukriya to kam hi padega bro but still thanks uuuuuuuuuuuuuuuh soooooooo much 😁😁😁😁😁😊and kabhi waqat mile to Nit Jalandhar me hiring ke liye jarur aana😅😅..Thanks again!!

mummy ji ki conditions: 1. line s uthao --> mtlb substring 2. max k types --> max k unique characters in the substring

This question is just longest substring with at most k unique characters

Samilar Leetcode Question :Fruit and Basket.

@Aditya Verma - 904. Fruit Into Baskets - LeetCode Same kind of problem, with trees and fruit.

Happiest moment of my day when I see the notification that aditya Verma has uploaded video you are legend sir

SIR GRAPH PADHA DO PLEASE PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE JULY AUG ME PLACEMENT HAI SIR JI PLEASE GRAPHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Leetcode Problem: 904. Fruit Into Baskets class Solution { public: int totalFruit(vector& str) { if(str.size()==1) return 1; //edge case int n=str.size(); int i=0, j=0, res =INT_MIN; unordered_map mp; while(j < n) { mp[str[j]]++; //calculation if(mp.size() 2) //condition 3 { while(mp.size() > 2) { mp[str[i]]--; if(mp[str[i]] == 0){ mp.erase(str[i]); } i++; // if(mp.size() == 2) // res = max(res, j-i+1); } j++; } } if(mp.size()==1){ auto it = mp.begin(); return it->second; } return res; } };

A slight improvement, it will not work for (a,a,a,a,a,a), so we need to update it to longest substring with atmost k unique characters.

Fruit into baskets leetcode 👍

Want the Solutions for the playlist, Here you go: github.com/venkateshjamge/Sliding-Window-Algorithm-Aditya-Verma

Happiness is listening to the question at the beginning and then immediately thinking that "Ye to Longest Substring with atmost 2 unique characters hain!"

What if input only contains only one type of toys.... String:aaaaa We need to change code little bit. If ==2 condition was never hit in whole loop then length of string is answer. Edit: thanks for videos

Bhai.. Already one week but not any new video... Please upload a new video

Aditya bhaiya it has been 4 weeks since you uploaded your last video.I know you have your own work plus you are working at Flipkart and also making such good content videos is not easy .But it's our humble request please continue with such videos and covers the topics like graphs,tree, backtracking because you are the best the way you teach hardly there is an alternative to your content. We student really need your support kindly if you can make the videos in the above topics it would be of a great help🥺🙏🙏

Just after the problem statement, click the solution ( Longest Substring with K Unique Characters) Thanks @Aditya bhaiya ❤✨🤙

But here according to the prob statement, k is less than or equal to 2 i.e k

Happy new year Aditya... I was so happy to see your video after such a long gap...Bhai please complete backtracking and other playlist soon :)

What if the array has all one type of toy ,this is a edge case ig if(mp.size()

jab mujhe toys nhi mile, phir mai John ko maximum kyun dilau?

Bro do make one playlist on greedy algorithm too.. U made DP soo easy for us please make greedy also... Plz bro

You are a blessing for the students preparing for the interview man! Can't thank you enough

Sir, please please make video playlist on how to identify the pattern and apply specific DS and algorithm.

Legend is back 💥

bhai backtracking par bnado plz playlist

This problem is similar to longest substring with atmost 2 unique characters

Problem Description: leetcode.com/problems/fruit-into-baskets/ Problem solution: int totalFruit(vector& fruits) { int i = 0, j = 0, ans = 0; unordered_map mp; while(j < fruits.size()){ mp[fruits[j]]++; if(mp.size() 2){ mp[fruits[i]]--; if(mp[fruits[i]] == 0){ mp.erase(fruits[i]); } i++; } j++; } } return ans; }

Sir, please bring in new lectures of the remaining topics .It's really a boon to all of us .

according to this problem unique should be maximum 2 not equal to 2 So, all toys of same type will also work, the correct code would be: #include using namespace std; int main(){ string s="abaccab"; int i=0; int j=0; int k=2; map m; int mx=INT_MIN; while(j

Thank you, Sir, Solve the problem in just the first 4 mins of your video explanation. Longest Subarray with 2 unique characters.

Bhaiya apki Doctor wali writing hai. But your explanation is really cool

So in this question, I will have to make two function calls, one with k=1 and k=2 and find out the maximum of the two, right?

English me nahi Hindi me samjh liye hai 🤪🤪🤪😂😂 Thanks

This is somewhat similar to longest substring with k unique characters.

Longest subarray having atmost 2 unique elements.

I need 4x speed in this video

Wonderful ❤️. Brother btw can you tell is the DP series complete? Cuz I couldn't find videos for LIS, Fibonacci, kadanes and DP on grid. So can you please add those? Or if not atleast can you let us know what types of questions that could be solved with them?

The code for given input- k=2 string s=abaccab //Here's the code if needed #include using namespace std; int maxToys(string s, int k){ unordered_mapm; int i=0, j=0, ct=0, mx=0; while(jk){ m[s[i]]--; ct--; if(m[s[i]]==0){ m.erase(s[i]); } i++; } } if(m.size()==k){ mx=max(mx, ct); } j++; } } return mx; } int main(){ int k; cin>>k; string s; cin>>s; cout

#java import java.util.HashMap; public class Solution { public static int getLongestSubString(String s, int k) { int max = -1; HashMap map = new HashMap(); int i = 0; int j = 0; while(j < s.length()) { char ch = s.charAt(j); if(map.containsKey(ch)) { map.put(ch, map.get(ch) + 1); }else { map.put(ch, 1); } if(map.size() == k) { max = Math.max(max, j - k + 1); j++; } if(map.size() > k) { char ch1 = s.charAt(i); while(map.size() > k) { if(map.containsKey(ch1)) { map.put(ch1, map.get(ch1) - 1); if(map.get(ch1) == 0) { map.remove(ch1); } } i++; } } j++; } return max; } public static void main(String[] args) { String s = "aaaa"; int k = 2; System.out.println(Solution.getLongestSubString(s, k)); } }

i wish, john must watch this SLIDING WINDOW! from aditya bhai..,

Hey, please make more videos if you have time....yours video are like osm

Hi bro, Love your videos. Just a suggestion that you should make a Telegram channel if possible. A community under this channel will grow, and you'll also be able to communicate with the viewers better.

/* John is at a toy store help him pick maximum number of toys. He can only select in a continuous manner and he can select only two types of toys. Example: Input: 1 abaccab */ /* LONGEST SUBSTRING WITH TWO UNIQUE CHARACTERS */ #include using namespace std; int pickToys(string str) { unordered_map mp; int i = 0, j = 0, maxi = 0; for (; j < str.size(); j++) { mp[str[j]]++; if (mp.size() == 2) maxi = max(maxi, j - i + 1); while (mp.size() > 2) { mp[str[i]]--; if (mp[str[i]] == 0) mp.erase(str[i]); i++; } } return maxi; } int main(int argc, char const *argv[]) { string s; cin >> s; cout

Sir are you planning to upload Greedy algorithm playlist??

I didn't watched the complete video , but I think this is similar to the problem of finding longest substring with 2 unique characters. Am I right?

Bhaiyya ab nayi series bna bhi dijiye. Kitna rulayenge bhakto ko?😢

tussi great ho paaji

Thank you so much bhaiyaaa for great explanation.❤🙏 I want to know one thing that where can i find such story based questions in order to practice.

#include using namespace std; int pickToys(string s){ int i=0, j=0; int maxi = INT_MIN; unordered_map mp; while (j < s.size()){ //calculation mp[s[j]]++; //cond < 2 if (mp.size() < 2){ j++; } //cond == 2 else if (mp.size() == 2){ maxi = max(maxi, j-i+1); j++; } //cond > 2 else if (mp.size() > 2){ while (mp.size() > 2){ mp[s[i]]--; if (mp[s[i]] == 0){ mp.erase(s[i]); } i++; } j++; } } return maxi; } int main(){ string s = "abaccab"; cout

bro, giveaway result?

c++ code //my version int atMostKuniqueelements(vector& A, int K) { //we can pass string in place of vector A int i = 0, res = 0; unordered_map count; //use incase of string for (int j = 0; j < A.size(); ++j) { if (!count[A[j]]++) K--; while (K < 0) { if (!--count[A[i]]) K++; i++; } res =max(res,j - i + 1); } return res; }

| C++ solution | #include using namespace std; //This function returns max toys that can be bought as a substring of given string having atmost k types of toys //Time complexity - O(n) and space complexity - O(k) int maxToys(string &toys, int k) { unordered_map mp; //to store count of different types of toys selected int n=toys.length(), ans=0; int i=0, j=0; //sliding window //traversing string while(j

Please start a graph theory series You have helped me so much. I need one more help and that is graph. Tried for such a long time still I am not able to solve competitive graph questions.

if we were count the number of substring with atmost k characters, how would we do it?

Solved before watching the video :) Sep'27, 2023 06:27 pm int totalFruit(vector& fruits) { int n=fruits.size(); unordered_mapmp; int i=0, j=0, ans=0; while(j2){ ans=max(ans, j-i); mp[fruits[i]]--; if(mp[fruits[i]]==0){ mp.erase(fruits[i]); } i++; } j++; } ans=max(ans, j-i); return ans; }

vo to sahi hai.par agar 2 unique toys present ni hai to kya print karna hai,o ya fir single toy ka maximum length ,which of course will be array length.

Thank you very much. You are a genius.

class Solution { public int totalFruit(int[] fruits) { int i=0; int j=0; int n= fruits.length; HashMap map = new HashMap(); int ans =0; while(j2){ while(map.size() > 2){ map.put(fruits[i], map.get(fruits[i])-1); if(map.get(fruits[i])==0){ map.remove(fruits[i]);} i++; }j++; } }return ans; } }

Mummy ji made this question interesting

class Solution { public: int totalFruit(vector &v) { int ans = 0, i = 0, j = 0, sum = 0; unordered_map u; while (i < v.size()) { u[v[i]]++; if (u.size() > 2) { u[v[j]]--; if (u[v[j]] == 0) u.erase(v[j]); j++; } int s = 0; for (auto k : u) s += k.second; // o(1) => loop is running for only 2 elements ans = max(ans, s); i++; } return ans; } };

C++ class Solution { public: int totalFruit(vector& fruits) { int n=fruits.size(); int i=0,j=0; int maxi=INT_MIN; map mpp; while(j2) { while(mpp.size()!=2) { --mpp[fruits[i]]; if(mpp[fruits[i]]==0) { mpp.erase(fruits[i]); } ++i; } } if(mpp.size()

// leetcode fruits into basket class Solution { public: int totalFruit(vector& fruits) { int i=0,j=0; unordered_mapm; int maxi = 0; while(j 2){ m[fruits[i]]--; if(m[fruits[i]] == 0) m.erase(fruits[i]); i++; } } j++; } return maxi; } };

FRUIT IN BASKET LEETCODE JAVA class Solution { public int totalFruit(int[] fruits) { int i=0; int j=0; int n= fruits.length; int ans=0; Map map= new HashMap(); if(fruits.length==1) return 1; while(j2){ int val= map.get(fruits[i]); val=val-1; if(val==0){ map.remove(fruits[i]); } else{ map.put(fruits[i],val); } i++; } if(map.size()==2){ ans= Math.max(ans,j-i+1); j++; } } } return ans; } }

class Solution { public: int totalFruit(vector& fruits) { int n=fruits.size(); int i=0,j=0; int maxi=INT_MIN; map mpp; while(j2) { while(mpp.size()!=2) { --mpp[fruits[i]]; if(mpp[fruits[i]]==0) { mpp.erase(fruits[i]); } ++i; } } if(mpp.size()

class Solution { public: int totalFruit(vector& fruits) { int n=fruits.size(); int i=0,j=0; int maxi=INT_MIN; map mpp; while(j2) { while(mpp.size()!=2) { --mpp[fruits[i]]; if(mpp[fruits[i]]==0) { mpp.erase(fruits[i]); } ++i; } } if(mpp.size()

Chithhi na koi Sandesh .. Jaane vo kaun desh jaha tum chalegaye 😰

bhaiya next kisi topic pr bnado tree, graph, backtracking plz placements aane wali hai

always waiting for your videos. Thanks

Beautifully Taught !

length of longest subarray with two different toys.

solved the question without seeing the video coz i watched the previous videos ...... this guy is legend

Longest increasing subsequence

Happy New year Aditya boss

kaash bhaiya college me hote toh abhi bhi videos bana paate!

fruits into basket is same as this

This question is same as Longest_K_unique_characters_substring but just K=2

Question waighrah to thik ye shraddha ka kya karna hai

Working Code: public static void findMaxToysCount(String str) { int i=0,j=0; HashMap map = new HashMap(); int maxCount = 0; while(j

if anyone is looking for the JAVA solution public class PickToys { public static void main(String[] args) { // Variable window // conditions -> toy ko line se uthana hai aur maximum two type ka toy hi hona chahiye // no of unique toy should be two and we should give john max toys String str="abaccacb"; // each character represents a type of toy // a-> car // b->helicopter // c->elephant int maxToys= Integer.MIN_VALUE; int k=2; HashMap map= new HashMap(); int i=0,j=0; while(jk) { int val= map.get(str.charAt(i)); val--; if(val==0) { map.remove(str.charAt(i)); i++; } else { map.put(str.charAt(i), val); } } j++; } } System.out.println(maxToys); } }

Working Java Solution: public static int findMaxToysToPick(String s) { Map frequencyMap = new HashMap(); int windowStart = 0, windowEnd = 0, maxLength = Integer.MIN_VALUE; while (windowEnd < s.length()) { char rightChar = s.charAt(windowEnd); frequencyMap.put(rightChar, frequencyMap.getOrDefault(rightChar, 0) + 1); if (frequencyMap.size() < 2) { windowEnd++; } else if (frequencyMap.size() == 2) { maxLength = Math.max(maxLength, windowEnd - windowStart + 1); windowEnd++; } else { while (frequencyMap.size() > 2) { char leftChar = s.charAt(windowStart); frequencyMap.put(leftChar, frequencyMap.get(leftChar) - 1); if (frequencyMap.get(leftChar) == 0) { frequencyMap.remove(leftChar); } windowStart++; } } } return maxLength; }

Same question as Longest Substring With K Unique Characters | Variable Size Sliding Window with K=2

Not everyone is blessing with an ADITYA VERMA during an OA

Longest substring with atmost k unique characters - just plotted a story of toy on that 😪

Eagerly waiting for next video.

coded it myself, blessed by prev videos.. Kudos to @TheAdityaVerma :))

Fruit into Baskets

Fruites in basket

1:03 ? shradda sabke saath hai?

There is a problem or maybe a possible problem in last 3 ques...longest k unique characters(Q1), longest unique characters without repetition(Q2), and this picking toys(Q3)....in all the 3 ques when we are exceeding the condition i.e. when(m.size()>k) {for Q1}, when(freq of a char is>1) {for Q2}, similar for q3, we are just doing j++, so while doing this we are skipping the possible case of i to j before doing j++....for ex....take aabacbe an ex of Q1 i=0,j=0 {a} ; m.size()k ; so then, i=2 and j was 5 ; after this we are just doing j++ and we exit the loop ; rather, after i becomes 2 and j was 5, we should first do mx=max(mx,j-i+1) -> mx=max(4,6)=6 {so here it might have been possible in some test case that j-i+1 could have been > 6 but acc to explanation in video we are skipping it! } ; then j++; Please correct me if I am missing something here...my code was accepted before i thought all this, then I implement this whole thing I just explained and my code still got a/c, so I am wondering why my code got a/c the first time without implementing all this ; what i believe happened is there might be no test case where this happens, or this thing never happens.

@Aditya sir, please improve your way of notation, you even denoted continuous as 'contd'......

Bro why are you not uploading videos...bhaiu are the hope please upload yaar...Please consider backtracking

To find whether a subarray is present or not whose sum is equal to zero.. Sir, please include this problem in the playlist..

Longest substring with k unique characters hi hai ye , as far as I have listened the problem statement

This is same as max subarray with k unique characters edit : saying this before seeing the explanation part

In a nutshell we need to find the longest substring with 2 different characters. !!

maximum substring length with two unique characters

thanks bhai, after explanation solution click kar gya mujhe

Hey Aditya would be very helpful if you can add videos of bit manipulation

I think I got the logic , before watching the complete video