Another approach to the Missing Number problem

Idea: get the sorted version of L, then find the gap: find an i such that L[i+1]-L[i] is 2 rather than 1.Then the answer is L[i]+1, the number between L[i] and L[i+1].

For example, if the sorted version of L is

[1, 2, 3, 4, 6, 7, 8, 9], then L[4]-L[3] == 2, and we can conclude that the missing number is L[3]+1, i.e., 4+1=5.

def missing_k_C(L):
    
    sorted_L = sorted(L)   #L stays the same. sorted_L stores the sorted 
                           #version of the list L
    
    n = len(sorted_L) + 1  #Since the original list had n elmenents, and L
                           #is one element shorter than the original list,
                           #to get n, we must add 1 to len(sorted_L)
    
    
    #Take care of the special cases: the missing element is in the beginning
    #or the missing element is at the end
    if sorted_L[0] != 1:
        return 1
    
    #sorted_L[len(sorted_L)-1] is the same as sorted_L[-1]: it's the last
    #element of sorted_L
    if sorted_L[-1] != n:
        return n
    
    #The main loop: find the i such that there is a gap between
    #L[i] and L[i+1]
    for i in range(len(sorted_L)-1):
        if sorted_L[i+1]-sorted_L[i] != 1:
            return sorted_L[i] + 1

One more thing

Here's another idea: sort the list, and then check of sorted_L[0] is 1, if sorted_L[1] is 2, .... and stop when you find a mismatch.

def missing_k_D(L):
    sorted_L = sorted(L)            
    
    for i in range(len(sorted_L)):
        #i = 0, ...., n-2
        if sorted_L[i] != i+1:
            return i+1
    
    return len(sorted_L) + 1