Find Missing and Repeated Values

"""
Intuition:
XOR all grid values with all expected numbers from 1..n^2. Every correct number cancels out,
so we get xor = missing ^ repeated. Now pick a set bit from xor (mask), which guarantees that
missing and repeated differ at this bit. Using this mask, split both the grid values and the
numbers 1..n^2 into two XOR groups; all matching numbers cancel inside their group, leaving
two candidates (group1, group2) = {missing, repeated}. Finally, check which candidate actually
exists in the grid: the one present is the repeated value, and the other is the missing value.
"""

class Solution:
    def findMissingAndRepeatedValues(self, grid: List[List[int]]) -> List[int]:
        xor=0
        k=1
        n=len(grid)
        for i in grid:
            for j in i:
                xor=xor^j^k
                k+=1
        mask=1
        k=1 #xor=10

        while not xor & mask:  #for this you can also use: mask = xor & -xor to get the first right most set bit, for eg 100-> 4, it doesn't gives the index but it gives the value
            mask=mask<<1
        group1,group2=0,0
        k=1
        for i in grid:
            for j in i:
                if mask & j:
                    group1=group1^j
                else:
                    group2=group2^j
                if mask &k:
                    group1=group1^k
                else:
                    group2=group2^k
                k+=1

        #check which is present from group1 and group 2 in the array for their position in the return array
        check1=False
        for i in grid:
            if group1 in i:
                check1=True

        if check1:
            return [group1,group2]
        else:
            return [group2,group1]

        return [group1,group2]