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A very easy solution in O(n)
- arka.bhowmik95 June 11, 2022since l2 already has space to accomodate l1, lets push all l2 elements to its end, keeping l2's front empty.
shiftL2ToEnd() in O(n) time
Now use 2 pointer approach to fill up l1 in l2's blank space . for cases where the pointer shifts to l2, we can empty that space in l2 and move it to front.