If n % 4 == 0, no matter how you act, your rival can make sure the next time you choose you still have n % 4 == 0.

1. Use hash table to mark element that shows before.
2. Use math, 2 * sum of all unique number - sum of all.
3. Use a XOR 0 = a, a XOR a =0.

s1. n<<1 + n should be all 1s.

s2. similar to s1 but more pythonic

s3. check by bit

Root height = max(left height, right height) +1

DFS will do.

If they are not equal, the answer will be the longer string; otherwise, answer is not-existed(-1).

s1. Depth first traverse, record every node’s (value, level)

s2. Width first traverse, keep two queues as parent level and kid level.

Use python list slice and mod.

Skip the first row and first column and traverse the matrix, check if m[i][j] = m[i-1][j-1].

s1: traverse all “1”, add the number of adjacent “0”s to the output. O(n) time.

s2: check every adjacent pairs horizontally and vertically, add 1 to the ouput if the pair is (“0”, “1”). In other words, we are checking every edge to check if it is a border. O(n) time but neater.

useful skill: map(list, zip(*grid)) = transpose(grid).