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Code GitHub Link: github.com/UdaySharmaGitHub/GFG-160-Challenge.git

class Solution { public: // Function to find the sum of contiguous subarray with maximum sum. int maxSubarraySum(vector<int> &arr) { // code here... // Time Complexity O(n) | Space Complexity: O(1); int res = INT_MIN , max_so_far = 0; for(int i=0;i<arr.size();i++){ max_so_far += arr[i]; if(res < max_so_far) res = max_so_far; if(max_so_far < 0) max_so_far = 0; // kadane's } return res; } };

bhai maina aj se 24 nov se start kiya ky ma starting question kr skta or sarre question 1 din me kr skta 24 tak ke question ya muza per day one question krna padega ?

or woman k liya compulsory bag ha but iska kya matlab agar humne complte kiya humko bag nehi milaga kya

vi maine aj start kiya ha ,to mereko sirf day 1 ka Second Largest wala problem hi sollve karna padega kya?or day 2 ka kal ,aise karke 160 din tak?

Hey, I just had a confusion. Please help me out, Actually jab me 160 days ke actual panel se question solve karta hu jaisa aapne video me bataya hai to wo POTD Streak me count nahi ho raha, udhar streak 0 hi show ho rahi hai. So mujhe POTD Streak ko hi 80 tak lana hoga means Bag claim karne keliye right? PS: actually I enrolled in this course just 3 days before, that's why this problem is happening. I think mujhe directly wahi karne hai jo POTD me specific day keliye listed hai?

Code Link (GitHub): github.com/UdaySharmaGitHub/GFG-160-Challenge/blob/main/Arrays/Day8_Stock_Buy_and_Sell_Max_one_Transaction_Allowed.cpp

Let's crack it.

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class Solution { public: int maximumProfit(vector<int> &prices) { // code here // Accumate Profit // Time Complexity O(n) | Space Complexity O(1); int maxProfit = 0 ; for(int i=1;i<prices.size();i++){ if(prices[i] > prices[i-1]) maxProfit += prices[i] - prices[i-1]; } return maxProfit; } };

GitHub Link: github.com/UdaySharmaGitHub/GFG-160-Challenge.git

there is no such Repository Named GFG-160-Challenge in your Github.

class Solution { public: // Function to find the majority elements in the array vector<int> findMajority(vector<int>& arr) { // Your code goes here. unordered_map<int,int> map; int n = arr.size() /3; for(int i:arr) map[i]++; arr.clear(); for(auto i:map) if(i.second > n ) arr.push_back(i.first); if(arr.size() == 2 && arr[0] > arr[1]) swap(arr[0],arr[1]); return arr; } };

✨✨Apply Link🔗: www.geeksforgeeks.org/courses/gfg-160-series

We have to solve just one problem a day or we have to complete the whole 5 given questions in 1 day(like i registered today itself and its showing 5 problems in array)? And what if the given questions is already solved before the start of challenge, will we still eligible?

Impressive... Bro.. Thanks.. Fr sharing

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class Solution { public: // Function to rotate an array by d elements in counter-clockwise direction. void rotateArr(vector<int>& arr, int d) { // code here // reverse(arr_starting_postion , arrr_ending_postion); d%=arr.size(); // faltu rotate operation unhe avoid kar sake reverse(arr.begin(),arr.begin()+d); reverse(arr.begin()+d,arr.end()); reverse(arr.begin(),arr.end()); } };

class Solution { public: void reverseArray(vector<int> &arr) { // code here int i = 0 , j = arr.size() -1; while(i<j) swap(arr[i++],arr[j--]); } };

class Solution { public: // Function to find intersection point in Y shaped Linked Lists. // length function int getLen(Node* head){ int len = 0 ; while(head){ len++; head= head->next;} return len; } // head1 -> longest , head2 =smallest; int getIntersectionPoint(int diff,Node* head1,Node* head2){ for(int i = 0 ;i<diff;i++){ head1 = head1->next; } while(head1 && head2){ if(head1 == head2) return head1->data; head1= head1->next; head2= head2->next; } return -1; } int intersectPoint(Node* head1, Node* head2) { // O(m+n) | O(1) int n1 = getLen(head1) ; int n2 = getLen(head2); int diff =0; if(n1 > n2){ diff = n1 - n2; return getIntersectionPoint(diff,head1,head2); } diff = n2 - n1; return getIntersectionPoint(diff,head2,head1); } };

class Solution { public: // Function to find intersection point in Y shaped Linked Lists. int intersectPoint(Node* head1, Node* head2) { // Your Code Here // 2nd Approach // O(m+n) | O(m+n) unordered_map<Node*,bool> map; while(head1){ map[head1] = 1; head1 = head1->next; } while(head2){ if(map[head2]) return head2->data; head2 = head2->next; } return -1; } };

Nice 👍 explanation

Good going Uday!!

class Solution { public: bool canAttend(vector<vector<int>> &arr) { // Your Code Here int count = 1; int st = arr[0][0] , en = arr[0][1]; for(int i=1;i<arr.size();i++){ if(st <= arr[i][0] && en <= arr[i][0]) count++; st = arr[i][0] ,en = arr[i][1]; } return count == arr.size(); } };

bhaiya time complexity O(nlog(n)) nahi hogi? kyon ki insertion in set is O(log(N)) he to??

class Solution { public: // a,b : the arrays // Function to return a list containing the union of the two arrays. vector<int> findUnion(vector<int> &a, vector<int> &b) { // Your code here // return vector with correct order of elements set<int> set; // storing for(int i:a) set.insert(i); // O(m) for(int i:b) set.insert(i); // o(n) a.clear(); for(auto i:set) a.push_back(i); // O(n+m) return a; } };

// User function template for C++ class Solution { public: string addString(string s1,string s2){ int i = s1.size()-1 ,j= s2.size()-1; string ans = ""; int carry = 0; while(i>=0 || j>=0 || carry){ int a = 0 , b = 0 ; if(i>=0) a = s1[i]-'0'; if(j>=0) b = s2[j]-'0'; int sum = a+b+carry; int digit = sum%10; carry =sum/10; if(sum) ans.push_back('0'+digit); i--;j--; } reverse(ans.begin(),ans.end()); return ans; } string minSum(vector<int> &arr) { // code here sort(arr.begin(),arr.end()); string s1 = "" ,s2 =""; for(int i =0;i<arr.size();i++){ if(i&1) s2.push_back('0'+arr[i]); else s1.push_back('0'+arr[i]); } return addString(s1,s2); } };

Nice... Explaination

Nice explanation

void rotate(vector<vector<int> >& mat) { // Your code goes here // Reverse -> Transpose reverse(mat.begin(),mat.end()); // reverse // Transpose for(int i=0;i<mat.size()-1;i++){ for(int j = i+1;j<mat.size();j++) swap(mat[i][j],mat[j][i]); } }

Nice explanation bro, keep it up. Will you see again tomorrow with next gfg daily challenge problem.

// User function Template for C++ class Solution { public: vector<vector<int>> findTriplets(vector<int> &arr) { // Code here unordered_map<int,vector<int>> map; vector<vector<int>> res ; for(int i=0;i<arr.size();i++) map[arr[i]].push_back(i); for(int j=1;j<arr.size()-1;j++){ for(int k=j+1;k<arr.size();k++){ int sum = -1*(arr[j] +arr[k]); for(int i:map[sum]){ if(i<j) res.push_back({i,j,k}); } } } return res; } };

class Solution { public: bool isLengthEven(struct Node **head) { // Code here int count = 0 ; while(*head){ count++; (*head) = (*head)->next; } return !(count&1); // return count%2==0; } };

class Solution { public: bool checkDuplicatesWithinK(vector<int>& arr, int k) { // your code unordered_map<int,int> map; map[0] = -1; // best practice for(int i=0;i<arr.size();i++){ // main condition if(map.find(arr[i]) != map.end()){ // duplicate value within the k distance if(i - map[arr[i]] <= k) return 1; } // updating the old index value with new Value map[arr[i]] = i; // insert value } return 0; } };

class Solution { public: long long maxSum(vector<int>& arr) { // code here sort(arr.begin(),arr.end()); int n = arr.size(); long long int diff = 0; for(int i =0;i<n/2;i++){ // Main Iteration // Array Circular hai tabhi multiple 2 // L -> R || R -> L diff += (2*(arr[n-1-i] - arr[i])); } return diff; } };

Best concepts 🎉❤❤❤❤

class Solution { public: Node* sortedInsert(Node* head, int x) { // Code here Node* curr = head,* nxt = head->next; Node* node = getNode(x); if(!head->next){ if(head->data <x){ head->next = node; node->prev = head; return head; } node->next= head; head ->prev = node; return node; } if(head->data> x){ node->next = head; head -> prev = node; head = node; return head; } while(x>nxt->data && nxt){ curr = nxt; nxt = nxt->next; if(!nxt) break; } if(!nxt){ curr->next = node; node->prev = curr; return head; } curr->next = node; node->prev = curr; node->next= nxt; nxt -> prev = node; return head; } };

Age 20 years and engineering ke 7th sem me hu to kya apply kar sakta hu?

Bsc cs walo ke liye kuch hai kya? Cs Mai?

// 3rd Approach class Solution { public: vector<int> removeDuplicate(vector<int>& arr) { // code here unordered_set<int> set; vector<int> ans; for(int i:arr){ if(set.find(i) == set.end()){ ans.push_back(i); } set.insert(i); } return ans; } };

// 2nd Approach class Solution { public: vector<int> removeDuplicate(vector<int>& arr) { // code here unordered_map<int,bool> map; vector<int> ans; for(int i:arr){ if(map[i] ==0){ ans.push_back(i); map[i] = 1; } } return ans; } };

// 1 Approach class Solution { public: vector<int> removeDuplicate(vector<int>& arr) { // code here // Freq stored vector<int> freq(101,0); vector<int> ans; for(int i=0;i<arr.size();i++){ if( freq[arr[i]] == 0){ ans.push_back(arr[i]); freq[arr[i]] = 1; } } return ans; } };

// Triplet Family Similar to 3 Sum or Triplets Sum in the Array class Solution { public: bool findTriplet(vector<int>& arr) { // Your code sort(arr.begin(),arr.end()); // O(n*log(n)) // O(n^2) for(int i = arr.size()-1;i>=0 ;i--){ int j = 0, k = i-1; while(j<k){ int sum = arr[j] + arr[k]; if(sum==arr[i]) return 1; else if(sum > arr[i]) k--; else if(sum <arr[i]) j++; } } return 0; } };

Very good bhai Keep going 🎉👏🏼👏🏼👏🏼

// Using Set class Solution { public: int count(struct Node* head, int key) { // add your code here set<int> set; int count = 0 ; while(head){ if(set.find(key) != set.end()){ count++; set.erase(key); } set.insert(head->data); head = head->next; } if(set.find(key) != set.end()){ count++; set.erase(key); } return count ; } };

// Using HaspMap or HashTable class Solution { public: int count(struct Node* head, int key) { // add your code here map<int,int> map; while(head){ map[head->data]++; head = head->next; } return map[key]; } };

// Using Counting class Solution { public: int count(struct Node* head, int key) { // add your code here int count = 0; while(head){ if(head->data == key )count++; head = head->next; } return count; } };

class Solution { public: vector<int> alternateSort(vector<int>& arr) { // Your code goes here sort(arr.begin(),arr.end()); int n = arr.size(),k=0; vector<int> ans(n,0); for(int i =0 ;i<n;i++){ if(i&1) ans[i] = arr[k++]; else ans[i] = arr[n-k-1]; } return ans; } };

// Every Possible modification for this type of question class Solution { public: vector<int> modifyAndRearrangeArray(vector<int> &arr) { // Complete the function vector<int> ans(arr.size(),0); for(int i = 0 ; i<arr.size()-1;i++){ if(arr[i] != 0 && arr[i+1]!=0){ if(arr[i] == arr[i+1]){ ans[i] = arr[i]*2; arr[i+1] = 0;} else{ ans[i] = arr[i] ; ans[i+1] = arr[i+1]; } } else if(arr[i] == 0 && arr[i+1] != 0){ ans[i+1] = arr[i+1]; } else if(arr[i] != 0 && arr[i+1] == 0){ ans[i] = arr[i] ; ans[i+1] = arr[i+1]; } arr[i] = 0; } arr[arr.size()-1]= 0; int j = 0 ; for(int i:ans) if(i>0) arr[j++] = i; return arr; } };

class Solution { public: vector<int> modifyAndRearrangeArray(vector<int> &arr) { // Complete the function vector<int> ans; for(int i=0;i<arr.size();i++){ if(arr[i] == arr[i+1]){ arr[i]*=2; arr[i+1] = 0; i++; } } for(int i =0;i<arr.size();i++){ if(arr[i]) ans.push_back(arr[i]); } for(int i =0;i<arr.size();i++){ if(arr[i] == 0) ans.push_back(arr[i]); } return ans; } };