XOR Queries of a Subarray - Leetcode 1310 - Python

[@anirudhbhat3099]

Hey Neet, I just got placed in a company through the on-campus drive and want to thank you for your videos and the roadmap you created. There are quite a few teachers who use Python for LeetCoding but I found out that your explanations really helped me understand the concepts deeply and was able to give the interviews with confidence🙏

[@NeetCodeIO]

LFG!!! congratulations you earned it my man

[@anirudhbhat3099]

@@NeetCodeIO Thanks a lot man😄

[@M.V.CHOWDARI]

Can please share your resume or suggest some projects ​@@NeetCodeIO

[@anonymousgreen5080]

Congratulations man. Could u give us some suggestions about interview ? Like are they same or a bit different from the old videos ?

[@rohangupta1266]

What college ?

[@freecourseplatformenglish2829]

Solved it on my own. Huge credit goes to your teaching style. Thanks man.

[@adilansari-hq9ge]

Bro, I Liked so much the style of having multiple solutions . It is useful to learn mutiple techniques.

[@MehmetDemir-xi3yy]

Hey man, I don't know if you did this but could you upload a video that you solve a segment tree problem. That way, you could teach us basics and usage of segment tree like you did for union find in the past. Thanks 🎉

[@petercheng1725]

For the second solution, can you add a 0 to the end of arr so if we go out of bounds we would index the last element which is 0. Very specific to python but similar thought to adding a 0 at the front.

[@MP-ny3ep]

Great explanation as always. Thank you for doing the daily problems

[@abhinavnair4577]

a simpler way to optimize this code is
for left, right in queries:
if left == 0:
ans.append(prefix[right])
else:
ans.append(prefix[right]^prefix[left-1])
# in ur case maybe do prefix[right+1]^prefix[left] , if u have len(prefix) =len(arr)+1
-this also gives runtime 300 ms
Complete solution -
class Solution:
def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
l = len(arr)
ans =[]
prefix = [0]*l
prefix[0] = arr[0]
for i in range(1,l):
prefix[i] = prefix[i-1] ^ arr[i]
for left, right in queries:
if left == 0:
ans.append(prefix[right])
else:
ans.append(prefix[right]^prefix[left-1])
return ans

[@sirk3v]

Thank you for providing. What math topics does one need be familiar with for DSA?

[@albin_joby]

for i in range(1,len(arr)):
arr[i] = arr[i-1]^arr[i]
res = []
for query in queries:
left,right = query[0],query[1]
if left == 0:
res.append(arr[right])
else:
res.append(arr[right]^arr[left-1])
return res

[@ShinAkuma]

You don't need extra space. Just store the prefix xor value in the original array.

[@NeetCodeIO]

Isn't that what I did?

[@ShinAkuma]

@@NeetCodeIO Nevermind, I commented before watching the second approach.

[@business_central]

I thought the time complexity would be O(n * m) since we are going through each query and also the input arr for the calculations.
Can anyone please clarify?

[@eshukla15]

thank you neetcode

[@M.V.CHOWDARI]

Hey, neet Code please suggest some advance projects

[@VijayKrishnaKumar-o4o]

Hii bro please make video on open Ai o1

[@MWKING]

second