deleteorder.py

def deleteorder(x):
    n = len(x)
    if n%2 != 0:
        return
    if any(x.count(i) > n//2 for i in set(x)):
        return
    x = sorted(x)
    assert all(x[i] < x[i + n//2] for i in range(n//2))
    # since x[0] <= x[1] <= x[2] <= ... <= x[n-1]
    # there must be x[i] <= x[i + n//2]
    # suppose if x[i] == x[i + n//2]
    #     then x[i] == x[i+1] == x[i+2] == ... == x[i + n//2]
    #     then x.count(x[i]) >= n//2 + 1 > n//2
    #     which is impossible via L5
    return list(zip(x[:n//2], x[n//2:]))