Henry's Algorithm

시간 복잡도: O(N**2) --> 54%의 정답률

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(X, A):
    # write your code in Python 3.6
    
    # step 1 - input x,a
    # already succeed
    
    # step 2 - if 1,2,3,.. in arr then, change the index into 0
    # else: continue
    
    arr = list(range(1,X+1))
    
    for i in range(len(A)):
        if A[i] in arr:
            arr[A[i]-1] = 0
            if sum(arr) == 0:
                return i
           
    pass

시간 복잡도: O(N) --> 100%의 정답률

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(X, A):
    # write your code in Python 3.6
    
    # step 1 - input x,a
    # already succeed
    
    # step 2 - # step 2 - if 1,2,3,.. in arr then, change the index into 0
    # else: continue
    
    check = [0] * X
    check_sum = 0
    
    for i in range(len(A)):
        if check[A[i]-1] == 0:
            check[A[i]-1] = 1
            check_sum += 1
            
            if check_sum == X:
                return i
                
    return -1
        
    pass

처음 코드(53% 정답률) 

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(A):
    # write your code in Python 3.6

    # step 1 - if length of A ==  2
    if len(A) == 2:
        return abs(A[0]-A[1])
        
    # step 2 - create a array
    arr_ = []
    
    # step 3 - get a difference between the sum of two part using iteration
    for i in range(1,len(A)):
        tmp = abs(sum(A[:i])-sum(A[i:]))
        arr_.append(tmp)
        
    return min(arr_)
        
    pass

정답 코드

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(A):
    # write your code in Python 3.6

    # step 1 - if length of A ==  2
    if len(A) == 2:
        return abs(A[0]-A[1])
        
    # step 2 - create a array
    arr_ = []
    
    tmp_1 = 0
    tmp_2 = sum(A)
    
    # step 3 - get a difference between the sum of two part using iteration
    for i in range(len(A)-1):
        tmp_1 = tmp_1 + A[i]
        tmp_2 = tmp_2 - A[i]
        arr_.append(abs(tmp_1 - tmp_2))
        
        
    return min(arr_)
        
        
    pass

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(A):
    # write your code in Python 3.6
    
    # step 1 - sort the array by acsend way
    A = sorted(A)
    
    # step 2 - find the missing one in range 1 ~ len(A)+1
    for i in range(0,len(A)):
        if i+1 != A[i]:
            return i+1
            
    return len(A)+1
            
    pass

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(X, Y, D):
    # write your code in Python 3.6
    
    # step 1 - 방정식 문제로 풀 수 있겠다
    
    ########################
    # frog X >= frog Y
    # a를 구하면 되는 문제
    
    # X+a*D >= Y
    # a*D >= Y - X
    # a >= (Y - X)/D
    # ex) (85 - 10)/30
    # ex) (75)/30 = 2.5
    ########################
    
    
    a = (Y - X) / D
    
    # 딱 맞아떨어지면, int() 내장 함수를 써서, 2.0 -> 2로 변환 후 반환해줌
    if a % 1 == 0:
        return int(a)
    # 2.4 이렇게 소수로 떨어지면 int()를 통해 정수로 변환하는 '버림'을 하고 2로 만든 후 +1 해서 반환 
    else:
        return int(tmp) + 1
    
    
    pass

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(A):
    # write your code in Python 3.6
    
    # step 1 - make a dictionary
    dict_= {}
    
    # step 2 - fill the dictionary
    for i in A:
        try: 
            dict_[i] += 1
        except: 
            dict_[i] =1
    
    # step 3 - find the odd value
    for i in dict_:
        if dict_[i] % 2 == 1:
            return i
    
    pass

mac os 에서 이맥스 사용방법 모음

1. 상하좌우 -> 기본 화살표 or control + p(상) / n(하) / b(좌) / f(우)

2. 커서가 있는 줄의 맨 앞,끝 -> control + a(맨 앞) / e(맨 끝) 

기존 4200에 포트가 제대로 닫히지 않았을때,

MAX OS 환경에서는 

sudo lsof -t -i tcp:4200 | xargs kill -9

위의 코드를 입력하면 4200 포트를 강제로 닫을 수 있음.

# you can write to stdout for debugging purposes, e.g.  
# print("this is a debug message")  
  
def solution(A, K):  
    # write your code in Python 3.6  
      
    # step 1 - take input A,K  
    # already succeed  
      
    # step 2 - create new array  
    arr_ = list(range(len(A)))  
      
    # step 3 - take iteration to solve the prob
    for i in range(len(A)):  
        arr_[(i+K)%len(A)] = A[i]  
      
    return arr_
    pass