def solution(arr):
i=0
while True:
a = 2 ** i
if a == len(arr):
break
elif a < len(arr):
i += 1
elif a> len(arr):
for i in range(a-len(arr)):
arr.append(0)
break
return arr
def solution(arr1, arr2):
if len(arr1)> len(arr2):
return 1
elif len(arr1) < len(arr2):
return -1
elif len(arr1) == len(arr2):
if sum(arr1) > sum(arr2):
return 1
elif sum(arr1) < sum(arr2):
return -1
elif sum(arr1) == sum(arr2):
return 0
set()
: 중복값 없애기
count
: set에 있는 원소들이 각각 몇개있는지 세는거
#예시
a = [1,2,3,2,1,1,2]
for i in set(a):
print(a.count(i))
#결과
#3 3 1
def solution(strArr):
arr= []
answer = []
for i in strArr:
arr.append(len(i))
for i in set(arr):
answer.append(arr.count(i))
return max(answer)
def solution(arr, n):
if len(arr) % 2 == 0:
for i in range(1,len(arr),2):
arr[i]+= n
else:
for i in range(0,len(arr),2):
arr[i]+= n
return arr
def solution(num_list):
num_list.sort()
return num_list[0:5]
def solution(num_list):
return sorted(num_list)[5::]