Tasks Details
easy
1.
MaxSliceSum
Find a maximum sum of a compact subsequence of array elements.
Task Score
100%
Correctness
100%
Performance
100%
A non-empty array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P+1] + ... + A[Q].
Write a function:
int solution(vector<int> &A);
that, given an array A consisting of N integers, returns the maximum sum of any slice of A.
For example, given array A such that:
A[0] = 3 A[1] = 2 A[2] = -6 A[3] = 4 A[4] = 0the function should return 5 because:
- (3, 4) is a slice of A that has sum 4,
- (2, 2) is a slice of A that has sum −6,
- (0, 1) is a slice of A that has sum 5,
- no other slice of A has sum greater than (0, 1).
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [1..1,000,000];
- each element of array A is an integer within the range [−1,000,000..1,000,000];
- the result will be 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 C++
Time spent on task 1 minutes
Notes
not defined yet
Task timeline
Code: 14:10:05 UTC,
java,
autosave
Code: 14:10:07 UTC,
cpp,
autosave
Code: 14:10:11 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
vector<int> max;
int result = 0;
max.push_back(A[0]);
for(int i = 1; i < A.size(); i++)
max.push_back((max[i - 1] + A[i] > A[i]) ? max[i - 1] + A[i] : A[i]);
for(int i = 0; i < max.size(); i++)
result = (result < max[i]) ? max[i] : result;
return result;
}
Code: 14:10:40 UTC,
cpp,
autosave
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
if(A.size() < 2)
return A[0];
vector<int> max;
int result = 0;
max.push_back(A[0]);
for(int i = 1; i < A.size(); i++)
max.push_back((max[i - 1] + A[i] > A[i]) ? max[i - 1] + A[i] : A[i]);
for(int i = 0; i < max.size(); i++)
result = (result < max[i]) ? max[i] : result;
return result;
}
Code: 14:10:58 UTC,
cpp,
verify,
result: Passed
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
if(A.size() < 2)
return A[0];
vector<int> max;
max.push_back(A[0]);
for(int i = 1; i < A.size(); i++)
max.push_back((max[i - 1] + A[i] > A[i]) ? max[i - 1] + A[i] : A[i]);
int result = max[0];
for(int i = 1; i < max.size(); i++)
result = (result < max[i]) ? max[i] : result;
return result;
}
Analysis
Code: 14:11:01 UTC,
cpp,
verify,
result: Passed
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
if(A.size() < 2)
return A[0];
vector<int> max;
max.push_back(A[0]);
for(int i = 1; i < A.size(); i++)
max.push_back((max[i - 1] + A[i] > A[i]) ? max[i - 1] + A[i] : A[i]);
int result = max[0];
for(int i = 1; i < max.size(); i++)
result = (result < max[i]) ? max[i] : result;
return result;
}
Analysis
Code: 14:11:04 UTC,
cpp,
final,
score:
100
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
if(A.size() < 2)
return A[0];
vector<int> max;
max.push_back(A[0]);
for(int i = 1; i < A.size(); i++)
max.push_back((max[i - 1] + A[i] > A[i]) ? max[i - 1] + A[i] : A[i]);
int result = max[0];
for(int i = 1; i < max.size(); i++)
result = (result < max[i]) ? max[i] : result;
return result;
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
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