Notes/Codes/Problem links under step 10 of A2Z DSA Course: takeuforward.org/strivers-a2z... Entire playlist: • Two Pointer and Slidin... Follow us on our other social media handles: linktr.ee/takeuforward
I've always had a problem with two pointer + sliding window problems. I've solved a few in Leetcode by reading the editorials. I understood them at that point of time but couldn't apply them again in the future as I just couldn't wrap my head around them. But now the intuition kicked in after watching the first few videos of your playlist and I'm able to visualise the algo while solving problems. Thank you so much!!! :)
Good video ! Wasn't expecting the last solution, took me some time to think but definitely made my brain work. The main logic is that once we have found a subarray with 2 zeros of size 5, as discussed in example, and a subarray with 2 zeros of size 6 exists... then once we reach subarray of size 5, we do not shrink our sliding window. And we keep moving it ahead by moving both left and right pointers. Once we reach the subarray of size 6, our sliding window's right pointer is updated while left keeps calm, and sliding window size is updated to 6. I hope it helps.
Another Approach:- class Solution { public: int longestOnes(vector& nums, int k) { int size = nums.size(); int l = 0; int r = 0; vectorind; int i = 0; int ans = 0; for(int i = 0;i
Thankyou so much Striver for all you efforts throughout in delivering us so much valuable content. Any student / working professional can now be able to transition their career without paying money for courses. Would also like your insights on the point : While preparing for interviews most of the aspirants are going through the videos solely and solving the question after completely watching the video. And also are feeling lazy trying to solve the question on our own. What is the best way to complete any topic without being lazy and how should an aspirant approach any topic/playlist?
another approach class Solution { public: int longestOnes(vector& nums, int k) { int n = nums.size(); // Get the size of the input vector int ans = 0; // Variable to store the maximum length of subarray with at most k zeros int ct = 0; // Variable to count the number of zeros encountered vector v1; // Vector to store the cumulative count of zeros // Traverse the input vector to fill the cumulative count of zeros for (int i = 0; i < n; i++) { if (nums[i] == 0) { ct++; // Increment the count if the current element is zero } v1.push_back(ct); // Add the cumulative count to the vector } int j = 0; // Left pointer of the sliding window int g = k - 1; // Right pointer of the sliding window if (k == 0) { g = 0; // Handle edge case when k is 0 } // Traverse the input vector using the sliding window approach while (g < n && g < v1.size()) { // Ensure g does not go out of bounds // Calculate the number of zeros in the current window int temp = v1[g] - (j > 0 ? v1[j - 1] : 0); if (temp
Solved it on my own after seeing the brute force. Buit I used a deque making it more simple class Solution { public: int longestOnes(vector& nums, int k) { int i = 0, j = 0, zeroes = 0; int ans = INT_MIN, len = 0; int n = nums.size(); deque q; while(j < n){ if(nums[j] == 0){ zeroes++; q.push_back(j); } if(zeroes > k){ zeroes--; int x = q.front(); i = ++x; q.pop_front(); } len = j - i + 1; ans = max(ans, len); j++; } return ans; } };
00:06 Solving the problem of finding the maximum consecutive ones with at most K zeros. 02:57 Using sliding window to find longest subarray with at most K zeros 07:43 Using sliding window to find maximum consecutive ones with K zeros. 10:13 Using sliding window technique to manage consecutive ones and zeros efficiently 15:10 Use a sliding window technique to handle scenarios with more zeros than K. 17:40 Optimizing Max Consecutive Ones III using sliding window technique 22:01 Illustration of updating max consecutive ones with sliding window approach 24:13 Algorithm to find max consecutive ones after K flips. 28:58 Algorithm works with time complexity of O(n) and space complexity of O(1).
i didn't understand one thing in most optimum sol by the time "R" reaches end "L' has traversed N-k (k is some constant) so shouldn't time complexity be O(N+N-k) which is same as O(2N)🤔🤔
Hey, I have 1 question . What if number of 0's are less than K so we have to flip all zeros and then k-no. of zeros times we have to flip 1 as well, right? How will we able to solve that question?
int longestOnes(vector& nums, int k) { int i = 0, j = 0; int n = nums.size(); int zero = 0; int maxi = 0; while (j < n) { if (nums[j] == 0) { zero++; } while (zero > k) { if (nums[i] == 0) { zero--; } i++; } maxi = max(maxi, j - i + 1); j++; } return maxi; }
In GFG the most optimized approach is giving out TLE with some 50 Testcases left out of 500, and the better one which uses two while loops is passing out all the test cases without TLE! WHY IS THAT SO?
TC - O(N) class Solution { public int longestOnes(int[] arr, int k) { int r=0; int l=0; int maxlen=0; int zeroes=0; while(rk){ if(arr[l]==0){ zeroes--; } l++; } if(zeroes
I don't know why but whenever i am watching these problem statements of two pointer, the first idea striking on my mind is a dp state🥲...then soon understanding dp is having at least 2 states so gotta optimise it
