Neighboring Bitwise XOR | Leetcode 2683

[@viralAdda213]

good morning sir,amazing just potd is live and you have uploaded the video,hatsoff to your dedication & consistency🙏First viewer

[@techdose4u]

nice :)

[@AlbinSabu-y2v]

Early morning.your efforts are really appreciated ❤🎉

[@techdose4u]

Thank you so much 😀

[@logeshv2125]

Super explanation brother, keep going✨

[@techdose4u]

thanks

[@sailendrachettri8521]

Thank you sir :)

[@techdose4u]

welcome 🤗

[@wierdo9514]

I counted the number of 1’s in the derived if their odd means false else zero 🙂

[@techdose4u]

nice work :)

[@godFather-123-b8g]

question dekha nhi solution phele aa gya lol

[@rajavignesh7216]

It also works by checking freq of 1 is even or not , idk y but it passed all cases😅

[@techdose4u]

great

[@opkrchauhan_1]

class Solution {
public boolean doesValidArrayExist(int[] derived) {
int n = derived.length;
int ans =0;
for(int i:derived){
ans ^=i;
}
if(ans==0){
return true;
}
return false;
}
}

[@techdose4u]

👍🏼

[@eshandhok2591]

why is this wrong?
class Solution {
public:
bool doesValidArrayExist(vector& d) {
if (d.size() == 0) {
return true;
}
if (d.size() == 1) {
return !d[0];
}
int n = size(d);
vector o1(n, 0), o2 = o1;
if (d[n - 1] == 0) {
o1[0] = 1, o2[0] = 0;
o1[n - 1] = 1, o2[n - 1] = 0;
} else {
o1[0] = 0, o2[0] = 0;
o1[n - 1] = 1, o2[n - 1] = 1;
}
for (int i = n - 2; i >= 1; i--) {
if (d[i] == 1) {
o1[i] = 0;
o2[i] = 1;
} else {
o1[i] = 1;
o2[i] = 0;
}
}
return ((o1[0] ^ o1[1]) == d[0]) or ((o2[0] ^ o2[1]) == d[0]);
}
};