Tasks Details
easy
1.
Dominator
Find an index of an array such that its value occurs at more than half of indices in the array.
Task Score
100%
Correctness
100%
Performance
100%
对于一个给定的整数数组, "支配者"是在这个数组中出现的频率超过一半的整数.
例如:
A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3数值"3"出现过5次, 5/8 > 0.5, 所以数值"3"是一个"支配者";
而在这个数组中, 这个"支配者"出现在数组下标:
- 0, 2, 4, 6 , 7.
请写一个函数
def solution(A)
对给定数组返回其任意一个支配者的数组下标。
例如,对上述数组,函数可以返回0,2,4,6,7中的任意一个。 如果没有支配者,函数应该返回 −1。
假定:
- N 是 [0..100,000] 内的 整数;
- 数组 A 每个元素是取值范围 [−2,147,483,648..2,147,483,647] 内的 整数 .
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Solution
Programming language used Python
Time spent on task 4 minutes
Notes
not defined yet
Code: 08:00:20 UTC,
java,
autosave
Code: 08:01:50 UTC,
py,
autosave
Code: 08:02:04 UTC,
py,
autosave
Code: 08:02:34 UTC,
py,
autosave
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2]
Code: 08:02:48 UTC,
py,
autosave
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
return dominator[0]
Code: 08:02:49 UTC,
py,
verify,
result: Passed
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
return dominator[0]
Analysis
Code: 08:03:19 UTC,
py,
autosave
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
if A.count(dominator[1]) <
return dominator[0]
Code: 08:03:33 UTC,
py,
autosave
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
if A.count(dominator[1]) <= len(A) / 2:
return -1
return dominator[0]
Code: 08:03:40 UTC,
py,
verify,
result: Passed
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
if A.count(dominator[1]) <= len(A) / 2:
return -1
return dominator[0]
Analysis
User test case 1:
[3, 1, 2, 3, 4]
Code: 08:03:49 UTC,
py,
verify,
result: Passed
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
if A.count(dominator[1]) <= len(A) / 2:
return -1
return dominator[0]
Analysis
User test case 1:
[3, 1, 2, 3, 4]
Code: 08:03:51 UTC,
py,
final,
score: 
100
# 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
if A == []:
return -1
dominator = [0, A[0], 1]
for i in range(1, len(A)):
if A[i] == dominator[1]:
dominator[2] += 1
elif dominator[2] == 0:
dominator[1] = A[i]
dominator[2] = 1
dominator[0] = i
else:
dominator[2] -= 1
if A.count(dominator[1]) <= len(A) / 2:
return -1
return dominator[0]
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N*log(N)) or O(N)
expand all
Correctness tests
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK
1.
0.036 s
OK
2.
0.036 s
OK