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%
An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.
For example, consider array A such that
A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.
Write a function
def solution(A)
that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.
For example, given array A such that
A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3the function may return 0, 2, 4, 6 or 7, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Python
Total time used 2 minutes
Effective time used 2 minutes
Notes
not defined yet
Task timeline
Code: 12:24:33 UTC,
java,
autosave
Code: 12:24:52 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(A):
stack = [0]*len(A)
size = 0
index = -1
count = 0
for i in range(len(A)):
if size is 0:
stack[size] = A[i]
size += 1
else:
if stack[size-1] != A[i]:
size -= 1
else:
stack[size] = A[i]
size += 1
# now dominents are remained in the stack
for i in range(len(A)):
if A[i] is stack[size-1]:
index = i
count += 1
if count > len(A)/2:
return index
else:
return -1
Code: 12:25:06 UTC,
py,
autosave
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(A):
stack = [0]*len(A)
size = 0
index = -1
count = 0
for i in range(len(A)):
if size == 0:
stack[size] = A[i]
size += 1
else:
if stack[size-1] != A[i]:
size -= 1
else:
stack[size] = A[i]
size += 1
# now dominents are remained in the stack
for i in range(len(A)):
if A[i] == stack[size-1]:
index = i
count += 1
if count > len(A)/2:
return index
else:
return -1
Code: 12:25:09 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(A):
stack = [0]*len(A)
size = 0
index = -1
count = 0
for i in range(len(A)):
if size == 0:
stack[size] = A[i]
size += 1
else:
if stack[size-1] != A[i]:
size -= 1
else:
stack[size] = A[i]
size += 1
# now dominents are remained in the stack
for i in range(len(A)):
if A[i] == stack[size-1]:
index = i
count += 1
if count > len(A)/2:
return index
else:
return -1
Analysis
Code: 12:25:53 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(A):
stack = [0]*len(A)
size = 0
index = -1
count = 0
for i in range(len(A)):
if size == 0:
stack[size] = A[i]
size += 1
else:
if stack[size-1] != A[i]:
size -= 1
else:
stack[size] = A[i]
size += 1
# now dominents are remained in the stack
for i in range(len(A)):
if A[i] == stack[size-1]:
index = i
count += 1
if count > len(A)/2:
return index
else:
return -1
Analysis
Code: 12:25:55 UTC,
py,
final,
score: 
100
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
def solution(A):
stack = [0]*len(A)
size = 0
index = -1
count = 0
for i in range(len(A)):
if size == 0:
stack[size] = A[i]
size += 1
else:
if stack[size-1] != A[i]:
size -= 1
else:
stack[size] = A[i]
size += 1
# now dominents are remained in the stack
for i in range(len(A)):
if A[i] == stack[size-1]:
index = i
count += 1
if count > len(A)/2:
return index
else:
return -1
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