L10. Count number of Nice subarrays | 2 Pointers and Sliding Window Playlist 

i have been watching your video since 2 week, now I solved this without watching your video, whole credit goes to you

The way u change this problem to the previous question.. amazing. Thanks a lot😀💯

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?

add this to your title - 1248. leetcode question number , such that it will come on top search .

this is the best channel

Do add Leetcode question number as well in video title for better search results.

very good striver

There is a type the inner while loop should check while(map.size() > k ) then only we need to reduce the frequency

Understood ❤

Mindblown

nice explaination

Striver on a different level

but it wont work for the test case: [2,2,2,1,2,2,1,2,2,2] k=2 Because at l=0 r=0, r will reach upto last index because sum

here we are using O(4N) TC at worst case but there's a better approach to solve it with [TC = O(N) & SC = O(1)]

understood

striver we want strings and stack playlist

there is no need for modifying array to 0 and 1. We can just use sliding window with concept of finding subarrays with odd numbers less the k - subarrays with odd numbers less then k-1

Finally mil gya 🥲

3:15 AM at night 🌃

❤

1248. Leetcode problem

solutions: class Solution { public: int solve(vector& nums, int k){ if(k

public int numberOfSubarrays(int[]nums,int k){ int n=nums.length; int[]cnt=new int[n+1]; cnt[0]=1; int sum=0,ans=0; for(int num:nums){ sum+=num%2; if(sum>=k) ans+=cnt[sum-k]; cnt[sum]++; } return ans; } 🎉❤

I think it will fail for a=[2,4,6] and k=2

class Solution { public int helpMe(int[] nums,int goal){ int l=0; int r=0; int sum=0; int cnt=0; if(goal