Відео

Scs is not their its hidden

Plz kisi ko mat btaya karo is playlist ke bare me competition bhut bad jayega...... 😂😂😂😂

Whenever I need prepare Interview I just come here... And now I am seeing you r back... Great

me to coachings of DSA - DP bhut tough topic h achhe se padha doge..?? They replied- ha ha sabse achha lekin aditya karke koi h usse achha ni padha payenge to be honest 😂😂😂😂 pta pada bad me fee refund mangne lago...

Hello, this approach will not work if we include negative numbers in the array, right ?

I find the Tabular approach less intuitive, and prefer Brute Force Recursive approach. As s2 = (Range - 2*s1) and s1 lies between 0 < s1< range/2. So in order to have a min Range - 2*s1, we need to maximize s1, so can we start searching possible sum in subset of the input array from range sum to 0 in decreasing order, and once we find the first sum is possible to make, that will be our answer? Thoughts?

Nice explanation bhaiya, here is how i did it in Java: import java.util.*; class connectRopes{ public static void main(String[] args) { int[] arr={1,2,3,4,5}; PriorityQueue<Integer> min_heap= new PriorityQueue<>(); for (int val : arr) { min_heap.add(val); } System.out.println(min_heap); int cost=0; while(min_heap.size()>=2){ int f=min_heap.poll(); int s=min_heap.poll(); // System.out.println("//f+s "+(f+s)); cost+=(f+s); // System.out.println("//cost "+cost+" "); min_heap.add(f+s); } System.out.println(cost); } }

AMAZING

I am still confuse on why at last we are returning Temp.

The best explanation in the youtube on DP where I found the explanation even in Matrix wise ,where no one does..

Pls explain me if mid ==k then why we can't return mid +1 element as array is sorted so next element would be in mid +1 Can someone please explain

C++ code: class Solution { public: void solve(string &str, int k, int start, string &ans){ if(k==0 || start==str.size()-1){ return; } char maxs= *max_element(str.begin()+start,str.end()); for(int i=start+1;i<str.size();i++){ if(str[start]<str[i] && str[i]==maxs){ swap(str[start], str[i]); if(str.compare(ans)>0) ans=str; solve(str,k-1,start+1,ans); swap(str[start], str[i]); } } solve(str,k,start+1,ans); } //Function to find the largest number after k swaps. string findMaximumNum(string str, int k) { // code here string ans=str; solve(str,k,0,ans); return ans; } };

a simpler way to understand this problem is as follows: let the sum of the array is S. two subsets having sum S1 and S2. Therefore S1 + S2 = S; we have to minimize the difference of the two subsets sum S1 and S2. we have to minimize S1 - S2 , minimum diff will begin at 0 and then 1, then n go up to S as one full set and one empty set diff. now we know S1 + S2 = S S1 - S2 will be one of 0,1,2,...,S so let's begin with checking 0, that means S1 - S2 = 0; problem becomes equal subset problem. if not that, then we will check 1, while checking1 equations will be S1 + S2 = S S1 - S2 = 1 which results 2S1 = S + 1 S1 = (S+1)/2 while checking 0 as minimum, it as S1 = (S + 0 ) / 2 general eq will be ( S + i )/2. so we have check in loop if there lies a subset in array whose sum is ( S + i )/2. Inside loop call the helper function to check if subset occurs. which can be done as follows. This solution works for arrays with negative elements too. int minimumDifference(vector<int>& nums) { int sum = 0, ans = 0; for(auto & a : nums) sum += std::abs(a);//sum with abs elements int s = std::accumulate(nums.begin(),nums.end(),0);//sum with negative numbers for(int i = 0; i < sum +1; i++) { int res = helper(nums,nums.size(),(s+i)/2,vec); if(res == 1) { ans = i; break; } } return ans; } int helper(vector<int>& nums, int n, int sum, std::vector<std::vector<bool>>& vec) { if(sum == 0 && n == nums.size()) return 0; if(sum == 0) return 1; if(n==0) return 0; if(nums[n-1]<= sum) { return helper(nums, n-1, sum - nums[n-1], vec) || helper(nums, n-1, sum, vec); } else { return helper(nums, n-1, sum, vec); } } please like the comment if it helps, this will help me knowing if I am able to help someone. this playlist has been quite helpful for me so far. Learning DP after leaving college 11 years back.

Khud kr lia vro

this code is optimized and is same as u have told in the video still it is not passing all the test cases showing TLE Can someone help me with this pls //{ Driver Code Starts // Initial Template for c++ #include <bits/stdc++.h> using namespace std; // } Driver Code Ends // User function Template for C++ class Solution{ public: //RECURSIVE TLE bool ispalindrome(string str,int i,int j) { while(i<j) { if(i>=j) return true; if(str[i]!=str[j]) { return false; } i++; j--; } return true; } int solve(string str,int i,int j,vector<vector<int>> &dp) { if(i>=j) { return 0; } //check if already palindrome //no partition required if(ispalindrome(str,i,j)) return 0; if(dp[i][j]!=-1) return dp[i][j]; int ans=INT_MAX; int left=0; int right=0; for(int k=i;k<=j-1;k++) { if(dp[i][k]!=-1) { left=dp[i][k]; } else { left=solve(str,i,k,dp); dp[i][k]=left; } if(dp[k+1][j]!=-1) { right=dp[k+1][j]; } else { right=solve(str,k+1,j,dp); dp[k+1][j]=right; } int temp=1+left+right; if(temp<ans) ans=temp; } return dp[i][j]=ans; } int palindromicPartition(string str) { int n=str.length(); int p=0; int q=n-1; vector<vector<int>>dp(n,vector<int>(n,-1)); return solve(str,p,q,dp); } //{ Driver Code Starts. int main(){ int t; cin>>t; while(t--){ string str; cin>>str; Solution ob; cout<<ob.palindromicPartition(str)<<" "; } return 0; } // } Driver Code Ends

next level stuff

bhai ka solution: problem link: leetcode.com/problems/permutations/ void helper(int start, vector<int>&nums, vector<vector<int>> & ans){ if(start==nums.size()){ ans.push_back(nums); return; } for(int i=start; i<nums.size();i++){ swap(nums[start], nums[i]); helper(start+1, nums, ans); swap(nums[start], nums[i]); } } vector<vector<int>> permute(vector<int>& nums) { vector<vector<int>> ans; int start=0; helper(start,nums, ans); return ans; }

I understood this totally sir. Your way of teaching is reallyy amazing all my concepts are clear now. but i have a question. In the interviews, will they ask us to optimize/ shorten the length of the code ? because that way i might not be able to

thank you so much.🙏

#finished 👍

Java code : public class SortArray { public static void main(String[] args) { List<Integer> list = new ArrayList<>(Arrays.asList(3,2,1,4,5,6,7)); System.out.println(sort(list)); } public static List<Integer> sort(List<Integer> list) { if (list.size() == 1) return list; int temp = list.remove(list.size() - 1); sort(list); insert(list, temp); return list; } public static List<Integer> insert(List<Integer> list, int temp) { if (list.isEmpty() || list.get(list.size() - 1) <= temp) { list.add(list.size(), temp); return list; } int value = list.remove(list.size() - 1); insert(list, temp); list.add(list.size(), value); return list; } }

leetcode 438 class Solution { public: vector<int> findAnagrams(string s, string p) { unordered_map<char,int>mp; for(int i=0;i<p.size();i++) mp[p[i]]++; int i=0; int j=0; int count=mp.size(); int k=p.size(); vector<int>res; while(j<s.size()) { if(mp.find(s[j])!=mp.end()){ mp[s[j]]--; if(mp[s[j]]==0) count--; } if(j-i+1 < k) j++; else if(j-i+1 == k) { if(count==0) res.push_back(i); if (mp.find(s[i]) != mp.end()) { mp[s[i]]++; if (mp[s[i]] == 1) count++; } i++; j++; } } return res; } };

Java Solution: class Solution { static ArrayList<Integer> list; public static ArrayList<Integer> increasingNumbers(int n) { list = new ArrayList<>(); ArrayList<Integer> temp = new ArrayList<>(); if(n == 1){ for(int i=0; i<10; i++) list.add(i); return list; } solve(list, temp, n, 0); return list; } static void solve(ArrayList<Integer> list, ArrayList<Integer> temp, int n, int start){ if(n == 0){ int ans = 0; for(int i=0; i<temp.size(); i++) ans = ans * 10 + temp.get(i); list.add(ans); } for(int i = start + 1; i<10; i++){ temp.add(i); solve(list, temp, n - 1, i); temp.remove(temp.size() - 1); } } }

class Pair{ int num; int index; Pair(int x,int y){ num=x; index=y; } } class Solution { public int largestRectangleArea(int[] heights) { int[] left=LeftIndex(heights); int[] right=RightIndex(heights); int[] width=new int[heights.length]; for(int i=0;i<heights.length;i++){ width[i]=right[i]-left[i]-1; } int[] area=new int[heights.length];int max=Integer.MIN_VALUE; for(int i=0;i<heights.length;i++){ area[i]= width[i]*heights[i]; if(max<area[i]){ max=area[i]; } } return max; } public int[] LeftIndex(int[] h){ int[] left = new int[h.length]; Stack<Pair> stl = new Stack<>(); int pseudoindex = -1; for (int i = 0; i < h.length; i++) { if (stl.isEmpty()) { left[i] = pseudoindex; } else if (!stl.isEmpty() && stl.peek().num < h[i]) { left[i] = stl.peek().index; } else if (!stl.isEmpty() && stl.peek().num >= h[i]) { while (!stl.isEmpty() && stl.peek().num >= h[i]) { stl.pop(); } if (stl.isEmpty()) { left[i] = pseudoindex; } else { left[i] = stl.peek().index; } } stl.push(new Pair(h[i], i)); } return left; } public int[] RightIndex(int[] h){ int[] right = new int[h.length]; Stack<Pair> str = new Stack<>(); int pseudoindex = h.length; for (int i = h.length - 1; i >= 0; i--) { if (str.isEmpty()) { right[i] = pseudoindex; } else if (!str.isEmpty() && str.peek().num < h[i]) { right[i] = str.peek().index; } else if (!str.isEmpty() && str.peek().num >= h[i]) { while (!str.isEmpty() && str.peek().num >= h[i]) { str.pop(); } if (str.isEmpty()) { right[i] = pseudoindex; } else { right[i] = str.peek().index; } } str.push(new Pair(h[i], i)); } return right; } }

7:13 😆

Not good.

The approach i tried in java int minmumSubSetDiff(int arr[], int n) { // first get subsetsum array response boolean[][] subSetSum = subSetSum(arr, n); boolean[] lastRow = subSetSum[n]; int sum1 = 0; int sum = calculateSum(arr); for (int i = (lastRow.length / 2); i >= 0 + 1; i--) { if (lastRow[i] == true) { sum1 = i; break; } } return sum - 2 * sum1; }

Sir ! In which company do you work .

which language you using?

wow great problem, solved excellents, there's little maths involved but yea great explanation.

Can we optimise this using memoization

I have never seen this kind of detailed and clear explanation . of tough questions like this

class Solution { public: vector <int> v_r; vector <int> v_l; stack <pair<int,int>> s_r; stack <pair<int,int>> s_l; vector<int> width; int area= 0; std::vector<int> nextSmallerRight(const std::vector<int>& heights){ int n = heights.size(); for(int i=n-1; i>=0; i--){ if(s_r.size()==0){ v_r.push_back(n); } else if(s_r.size()>0&& s_r.top().first<heights[i]){ v_r.push_back(s_r.top().second); } else if(s_r.size()>0&& s_r.top().first>=heights[i]){ while(s_r.size()>0&& s_r.top().first>=heights[i]){ s_r.pop(); } if(s_r.size()==0){ v_r.push_back(n); } else{ v_r.push_back(s_r.top().second); } } s_r.push({heights[i],i}); } std:: reverse(v_r.begin(), v_r.end()); return v_r; } std::vector<int> nextSmallerLeft(const std::vector<int>& heights) { int n = heights.size(); for(int i=0; i<n; i++){ if(s_l.size()==0){ v_l.push_back(-1); } else if(s_l.size()>0&& s_l.top().first<heights[i]){ v_l.push_back(s_l.top().second); } else if(s_l.size()>0&& s_l.top().first>=heights[i]){ while(s_l.size()>0&& s_l.top().first>=heights[i]){ s_l.pop(); } if(s_l.size()==0){ v_l.push_back(-1); } else{ v_l.push_back(s_l.top().second); } } s_l.push({heights[i],i}); } return v_l; } int largestRectangleArea(vector<int>& heights) { nextSmallerRight(heights); nextSmallerLeft(heights); for(int i=0;i<heights.size();i++){ width.push_back(v_r[i]-v_l[i]-1); } for(int i=0; i<heights.size(); i++){ area = std::max(area, heights[i]*width[i]); } return area; } }; Here is my code

bhai tumhare concept bahut acche clear hai but bahut kachra or bakwas trike se likhte ho, samj nhi ata kaha code hai kaha algo,, aise kaise bnoge acche edutuber, or adha ghante ka time apne online communit ke nhi nikal skte kya ki ye backtracking ki video playlist complete krle, mai to ppori dekne hi aya tha, or ye code to likhr nhi kamse kam psudoocde to lik dia kro, ki jisko ppori video nhi dekni to bas code dekh ki age badh jae. agar meri ye baten manoge fir tum bahut acche teacher bnoge dsa ke utubepe

Here's how i did it in Java: import java.util.*; class Pair { int key; int value; Pair(int key, int value) { this.key = key; this.value = value; } } class top_k_freq { public static void main(String[] args) { int[] arr = {1, 1, 1, 3, 2, 2, 4}; int k = 2; HashMap<Integer, Integer> hm = new HashMap<>(); for (int i : arr) { hm.put(i, hm.getOrDefault(i, 0) + 1); } // Step 2: Use a priority queue to find the top k frequent elements PriorityQueue<Pair> min_heap = new PriorityQueue<>(new Comparator<Pair>() { public int compare(Pair a, Pair b) { return Integer.compare(a.value, b.value); } }); for (Map.Entry<Integer, Integer> entry : hm.entrySet()) { min_heap.add(new Pair(entry.getKey(), entry.getValue())); if (min_heap.size() > k) { min_heap.poll(); } } // Step 3: Extract the results from the priority queue List<Integer> result = new ArrayList<>(); while (!min_heap.isEmpty()) { result.add(min_heap.poll().key); } // Since the smallest frequency elements were removed first, reverse the result list Collections.reverse(result); // Print the result System.out.println(result); } }

bhai please yr app dsa ki videos continue upload kro apse acha explaination maine pure youtube pr nhi dekha pls brother🙏

Mai likhunga kachara ho jayega, BC(base condition) and drawing rat so funny🤣

why the first solve array we're passing same i and k value, don't you thkink both the mattrix have same value?

leetcode 2461 using hashmap to remove duplicate elements class Solution { public: long long maximumSubarraySum(vector<int>& nums, int k) { long long n=nums.size(); int i=0; int j=0; long long sum=0; long long maxsum=0; unordered_map<int,int>mp; while(j<n) { mp[nums[j]]++; sum+=nums[j]; if(j-i+1 < k) j++; else if(j-i+1 == k) { if(mp.size()==k) maxsum=max(maxsum,sum); sum=sum-nums[i]; mp[nums[i]]--; if(mp[nums[i]]==0) mp.erase(nums[i]); i++; j++; } } return maxsum; } };

Thank you so much, one of the best videos on sliding window. Easy to understand and best explanation.

thanks bhaiya , amazing teaching skills !

Yeh code work nahi kar raha hain

Python solution: def cutRod(self, price, n): dp = [[0 for _ in range(n+1)] for _ in range(n+1)] # dp[i][j] = max value obtained by cutting rod of length j into pieces considering the first i lengths for i in range(1, n+1): for j in range(1, n+1): if i > j: dp[i][j] = dp[i - 1][j] else: includeCut = price[i - 1] + dp[i][j - i] excludeCut = dp[i - 1][j] dp[i][j] = max(includeCut, excludeCut) return dp[n][n] However, this runs out of time for some test cases since it can be optimized with 1D array for dp. Here is optimized approach: class Solution: def cutRod(self, price, n): dp = [0] * (n + 1) for i in range(1, n + 1): for j in range(i, n + 1): include = price[i - 1] + dp[j - i] exclude = dp[j] dp[j] = max(include, exclude) return dp[n]

Bhaiya please weak mai atleast ek video dal diya karo DSA/CP related

no one better than u

graphhhhh please

just in case u people needed a code in java.... import java.util.*; public class sw2 { public static void minneg(int arr[], int k) { ArrayList<Integer> list = new ArrayList<>(); int i = 0, j = 0; while (j < arr.length) { if (arr[j] < 0) { list.add(arr[j]); } if (j - i + 1 == k) { if (!list.isEmpty()) { System.out.println(list.get(0)); } else { System.out.println(0); } if (arr[i] < 0 && !list.isEmpty() && arr[i] == list.get(0)) { list.remove(0); } i++; } j++; } } public static void main(String[] args) { int arr[] = {12, -1, -7, 8, -15, 30, 16, 28}; int k = 3; minneg(arr, k); } }

this is the code just in case.... import java.util.*; public class slidingwindow1 { public static int maxsum(int arr[],int k) { int i=0; int j=0; int sum=0; int max = Integer.MIN_VALUE; while(j<arr.length) { sum=sum+arr[j]; if(j-i<k){ j++; } else if(j-i==k) { max=Math.max(sum,max); sum=sum-arr[i]; i++; j++; } } return sum; } public static void main(String[] args) { int arr[]={1,2,3,4,5}; int k=3; int result = maxsum(arr, k); System.out.println("Maximum sum of subarray of size " + k + " is: " + result); } }

Hey Google :- How to watch all the videos in 1 minutes, 1 by 1 without skipping and high speed. I want to become master of dp in 1 mintues 😅😅

we need series for graph T.T