Keshav Arora @_keshav_arora_

🔥Day 74/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find all the unique permutations of str s; 📷[Approach] : Use recursion & next_permutation function to achieve this. #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 72/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Bellman Floyd Alogrithm 📷[Approach] : do (V-1) relaxations and then you will get the shortest path. If there is loop, the distance array will gets reduced on Nth relaxation #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 73/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Remove loop in LL 📷[Approach] : Use floyd detection to detect cycle, and set last node of the cycle (last node == node before where ptrs first meet) to Null. The cycle gets removed #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 72/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find the first node in LL 📷[Approach] : Use floyd warshel to check cycle. #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 71/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Detect Cycle in a Linked List. 📷[Approach] : Floyd's cycle detection algo. Use slow pointer & fast pointer , where it matches ==> cycle else if fastpointer reaches to end, no cycle. #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 70/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Clone the LL 📷[Approach] : get to editorial for detailed approach #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 69/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Add numbers in 2 given LL. 📷[Approach] : Reverse list, add list and then remove the trailing zeroes. #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 68/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Reverse every kth node of the LL. 📷[Approach] : Just iterate till the end of the LL, and reverse every kth node (By normal reversal Technique). Simple a O(n) solution !!. Solved !! #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 67/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Merge 2 sorted LL. 📷[Approach] : Just simple as merging two arrays. maintain head/tail ptr, and append smallers values out of 2 lists and move forward. Simple as that !! #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 66/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Rotate a LL. 📷[Approach] : Just reach to the end, connect end (tail) to the head (list made circular, connected). now get to (k-1) pos, and set head = (k-1)->next & disconnect tail #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 65/160 - GeeksforGeeks Streak Challenge 📷[Problem]: How to reverse a linked list 📷[Approach] : Declare ptrs (curr, nxt, prev). Prev=null, curr = head while(curr != null) nxt = curr->next curr->next = nxt nxt = curr curr = nxt #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 64/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Return arr, where arr[i] = product of all number in array except arr[i]; 📷[Approach] : Just basic Implementation technique; #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 63/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Largest Subarray with sum = 0 📷[Approach] : Set 0s = -1 & 1s = 1; Then take prefix sum. Check for prefix[i] = 0 => len = i + 1. & check if sum previously exists, len = i - mp[prev]. #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 62/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find longest subarray length with sum k 📷[Approach] : Make prefix array. Then check if(arr[i] == k) len = i + 1; then check if (arr[i] - arr[j] == k) then len = i - j; return len; #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 61/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find idx i : such that sum of first i elements == sum of elements from i to n. 📷[Approach] : Easy problem of prefix, suffix. Just initialize and that idx == (suffix[i] == prefix[i]) #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 60/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find container with most water 📷[Approach] : Use 2 ptr. water = max(water, min(arr[left], arr[right]) * (right - left)); water = minimum of left / right * distance b/w left/right. #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 59/160 - GeeksforGeeks Streak Challenge 📷[Problem]: How much water is stored b/w walls; 📷[Approach] : l = 0, r = n -1 ; while(l < r). We know a[l] < a[r] : implies for all walls atleast L height of water is stored - arr[i]. same fr r #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 58/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find the longest Substring with diff char 📷[Approach] : Use 2 ptr, l = 0, r = (0, s.len) and iterate. If you find s[r] in map, reset l = mp[r] + 1 & len = max(r- l + 1). Solved !!! #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 57/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Count distinct element in every window 📷[Approach] : Just iterate in the window of size k & add mp.size() to the vector. Simple, Easy problem of Hashing & Sliding Window. Solved !! #geekstreak2025 @geeksforgeeks #gfg160

🔥Day 56/160 - GeeksforGeeks Streak Challenge 📷[Problem]: Find Index of Subarr with sum=t. return first subArr 📷[Approach] : Use 3 ptr, left=0, right=0, sum=0; while(r < n) w(s > t & r >= l) s -= a[l] l++, s = t ; return l + 1, r + 1 r++; #geekstreak2025 @geeksforgeeks #gfg160