Awesome Explanation Brother.... I was waiting for your video for this problem solution. This was my 2nd biweekly contest and I got stuck at this problem. It is a bit tough to digest for me but I really appreciate your time and efforts for complete in depth solution for these problems. Stay Blessed Bhaiya and thank you ..........
for a subarray [i ... j] the sum is prefSum[j] - prefSum[i - 1]. Now to maximise the sum of all subarrays we need to minimise the prefSum[i - 1] part as prefSum[j] will be fixed. That's why we are storing the minimised prefSum[i - 1] for each nums[i] in the map so that we dont need to traverse for all the duplicates.
Thanks for providing the intuition for the duplicate values. Appreciate your work
Awesome Explanation Brother.... I was waiting for your video for this problem solution. This was my 2nd biweekly contest and I got stuck at this problem. It is a bit tough to digest for me but I really appreciate your time and efforts for complete in depth solution for these problems. Stay Blessed Bhaiya and thank you ..........
Were you able to come up with the 2nd?
@@ritishrai581 I couldn't submit it as I was getting stuck on one or two corner cases. That's why I was super Curious to know the solution
THNX bro, well explained.
i used vector of indices but still passed all test cases 🙃🙃
Now it will not
ye minimize kya kr rahe hai samjh nahi aaya
for a subarray [i ... j] the sum is prefSum[j] - prefSum[i - 1]. Now to maximise the sum of all subarrays we need to minimise the prefSum[i - 1] part as prefSum[j] will be fixed. That's why we are storing the minimised prefSum[i - 1] for each nums[i] in the map so that we dont need to traverse for all the duplicates.
Jo part easy hai wo 10 baar samjata hai ye
wahi uski toh repeat karega baar 😁😁😁😁 ,tough wale part ka direct code karta hai
bhai thoda dheere bol le