Tasks Details
easy
1.
Triangle
Determine whether a triangle can be built from a given set of edges.
Task Score
100%
Correctness
100%
Performance
100%
An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if 0 ≤ P < Q < R < N and:
- A[P] + A[Q] > A[R],
- A[Q] + A[R] > A[P],
- A[R] + A[P] > A[Q].
For example, consider array A such that:
A[0] = 10 A[1] = 2 A[2] = 5 A[3] = 1 A[4] = 8 A[5] = 20Triplet (0, 2, 4) is triangular.
Write a function:
def solution(A)
that, given an array A consisting of N integers, returns 1 if there exists a triangular triplet for this array and returns 0 otherwise.
For example, given array A such that:
A[0] = 10 A[1] = 2 A[2] = 5 A[3] = 1 A[4] = 8 A[5] = 20the function should return 1, as explained above. Given array A such that:
A[0] = 10 A[1] = 50 A[2] = 5 A[3] = 1the function should return 0.
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 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Python
Time spent on task 1 minutes
Notes
not defined yet
Code: 13:45:44 UTC,
java,
autosave
Code: 13:45:53 UTC,
py,
verify,
result: Passed
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
# -*- coding:utf-8 -*-
# &Author AnFany
# Lesson 6:Sorting
# P 6.3 Triangle
def solution(A):
"""
判断数组中的任何三元组对应的三个数是否是三角形的,时间复杂度为O(N*log(N))
:param A: 数组
:return: 返回结果
"""
if len(A) < 3:
return 0
else:
A.sort(reverse=True)
for index, value in enumerate(A[:-2]):
if A[index+2] <= 0:
return 0
else:
if value < A[index+1] + A[index+2]:
return 1
return 0
Analysis
Code: 13:45:56 UTC,
py,
final,
score: 
100
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
# -*- coding:utf-8 -*-
# &Author AnFany
# Lesson 6:Sorting
# P 6.3 Triangle
def solution(A):
"""
判断数组中的任何三元组对应的三个数是否是三角形的,时间复杂度为O(N*log(N))
:param A: 数组
:return: 返回结果
"""
if len(A) < 3:
return 0
else:
A.sort(reverse=True)
for index, value in enumerate(A[:-2]):
if A[index+2] <= 0:
return 0
else:
if value < A[index+1] + A[index+2]:
return 1
return 0
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N*log(N))
expand all
Correctness tests
1.
0.036 s
OK
2.
0.036 s
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3.
0.036 s
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4.
0.036 s
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0.036 s
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6.
0.036 s
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1.
0.036 s
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0.036 s
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3.
0.036 s
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0.036 s
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5.
0.036 s
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6.
0.036 s
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1.
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6.
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6.
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expand all
Performance tests
1.
0.048 s
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2.
0.036 s
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3.
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4.
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5.
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6.
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1.
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2.
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4.
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5.
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6.
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1.
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2.
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6.
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1.
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2.
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5.
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6.
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1.
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2.
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5.
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6.
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1.
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2.
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5.
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6.
0.036 s
OK