Test results - Codility

Tasks Details

easy

1. TapeEquilibrium

Minimize the value |(A[0] + ... + A[P-1]) - (A[P] + ... + A[N-1])|.

Task Score

100%

Correctness

100%

Performance

100%

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

We can split this tape in four places:

P = 1, difference = |3 − 10| = 7

P = 2, difference = |4 − 9| = 5

P = 3, difference = |6 − 7| = 1

P = 4, difference = |10 − 3| = 7

Write a function:

int solution(vector<int> &A);

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:

A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [2..100,000];

each element of array A is an integer within the range [−1,000..1,000].

Solution

Programming language used C++

Time spent on task 13 minutes

Notes

not defined yet

Code: 17:46:07 UTC, java, autosave

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// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
    }
}

Code: 17:46:07 UTC, java, autosave

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// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
    }
}

Code: 17:46:34 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    
}

Code: 17:46:52 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //
}

Code: 17:47:23 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 
}

Code: 17:47:47 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    
}

Code: 17:48:18 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i]
    }
}

Code: 17:48:28 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    
}

Code: 17:49:06 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

//
    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }


}

Code: 17:49:27 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    //min = 100000000

}

Code: 17:49:48 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int min = 100000001;

    for()

}

Code: 17:50:01 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int min = 100000001;

    for(int i = 1; i<A.size(); i++){
        
    }

}

Code: 17:50:31 UTC, cpp, autosave

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#include <iostream>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int min = 100000001;
    

    for(int i = 1; i<A.size(); i++){
        
    }

}

Code: 17:50:42 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int min = 100000001;


    for(int i = 1; i<A.size(); i++){
        
    }

}

Code: 17:51:12 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int min = 100000001;


    for(int i = 1; i<A.size(); i++){
        
    }

}

Code: 17:51:35 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int min = 100000001;
    min = min(abs(left - right), min);

    for(int i = 1; i<A.size(); i++){
        
    }

}

Code: 17:51:48 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        
    }

}

Code: 17:52:04 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        left += A[i];
        right -= A[i];
        
    }

}

Code: 17:52:16 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Code: 17:52:28 UTC, cpp, verify, result: Passed

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis

expand all Example tests

▶

example
example test

✔

0.001 s

Code: 17:52:53 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Code: 17:53:20 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Code: 17:53:31 UTC, cpp, verify, result: Passed

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size(); i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis

expand all Example tests

▶

example
example test

✔

0.001 s

expand all User tests

▶

test_data1

✔

0.001 s

OK
function result: 0

▶

test_data2

✔

0.001 s

OK
function result: 0

▶

test_data3

✔

0.001 s

OK
function result: 0

▶

test_data4

✔

0.001 s

OK
function result: 5

User test case 1:

[-1000, 1000]

User test case 2:

[1000, 1000]

User test case 3:

[1, 1, 1, 1]

User test case 4:

[1001, 1002, 1004, -1000]

Code: 17:55:37 UTC, cpp, autosave

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Code: 17:55:39 UTC, cpp, verify, result: Passed

show code in pop-up

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis

expand all Example tests

▶

example
example test

✔

0.001 s

expand all User tests

▶

test_data1

✔

0.001 s

OK
function result: 2000

▶

test_data2

✔

0.001 s

OK
function result: 0

▶

test_data3

✔

0.001 s

OK
function result: 0

▶

test_data4

✔

0.001 s

OK
function result: 5

User test case 1:

[-1000, 1000]

User test case 2:

[1000, 1000]

User test case 3:

[1, 1, 1, 1]

User test case 4:

[1001, 1002, 1004, -1000]

Code: 17:56:16 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);
    

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Code: 17:56:21 UTC, cpp, verify, result: Passed

show code in pop-up

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    cout << answer << endl;

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis

expand all Example tests

▶

example
example test

✔

0.001 s

stdout:

expand all User tests

▶

test_data1

✔

0.001 s

OK
function result: 2000

stdout:

▶

test_data2

✔

0.001 s

OK
function result: 0

stdout:

▶

test_data3

✔

0.001 s

OK
function result: 0

stdout:

▶

test_data4

✔

0.001 s

OK
function result: 5

stdout:

User test case 1:

[-1000, 1000]

User test case 2:

[1000, 1000]

User test case 3:

[1, 1, 1, 1]

User test case 4:

[1001, 1002, 1004, -1000]

Code: 17:56:41 UTC, cpp, autosave

show code in pop-up

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Code: 17:56:55 UTC, cpp, verify, result: Passed

show code in pop-up

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#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis

expand all Example tests

▶

example
example test

✔

0.001 s

expand all User tests

▶

test_data1

✔

0.001 s

OK
function result: 2000

▶

test_data2

✔

0.001 s

OK
function result: 0

▶

test_data3

✔

0.001 s

OK
function result: 0

▶

test_data4

✔

0.001 s

OK
function result: 5

User test case 1:

[-1000, 1000]

User test case 2:

[1000, 1000]

User test case 3:

[1, 1, 1, 1]

User test case 4:

[1001, 1002, 1004, -1000]

Code: 17:58:32 UTC, cpp, verify, result: Passed

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis

expand all Example tests

▶

example
example test

✔

0.001 s

expand all User tests

▶

test_data1

✔

0.001 s

OK
function result: 2000

▶

test_data2

✔

0.001 s

OK
function result: 0

▶

test_data3

✔

0.001 s

OK
function result: 0

▶

test_data4

✔

0.001 s

OK
function result: 5

User test case 1:

[-1000, 1000]

User test case 2:

[1000, 1000]

User test case 3:

[1, 1, 1, 1]

User test case 4:

[1001, 1002, 1004, -1000]

Code: 17:58:35 UTC, cpp, final, score: 100

show code in pop-up

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)

    //숫자가 정해져 있지 않아서 우선 처음으로 나뉘는 부분은 계산하자.
    int left = A[0];
    int right = 0;
    for(int i = 1; i<A.size(); i++){
        right += A[i];
    }

    int answer = 100000001;
    answer = min(abs(left - right), answer);

    for(int i = 1; i<A.size()-1; i++){
        left += A[i];
        right -= A[i];
        answer = min(abs(left - right), answer);
    }

    return answer;

}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

expand all Example tests

▶

example
example test

✔

0.001 s

expand all Correctness tests

▶

double
two elements

✔

0.001 s

▶

simple_positive
simple test with positive numbers, length = 5

✔

0.001 s

▶

simple_negative
simple test with negative numbers, length = 5

✔

0.001 s

▶

simple_boundary
only one element on one of the sides

✔

0.001 s

▶

small_random
random small, length = 100

✔

0.001 s

▶

small_range
range sequence, length = ~1,000

✔

0.001 s

▶

small
small elements

✔

0.001 s

expand all Performance tests

▶

medium_random1
random medium, numbers from 0 to 100, length = ~10,000

✔

0.001 s

▶

medium_random2
random medium, numbers from -1,000 to 50, length = ~10,000

✔

0.001 s

▶

large_ones
large sequence, numbers from -1 to 1, length = ~100,000

✔

0.004 s

▶

large_random
random large, length = ~100,000

✔

0.004 s

▶

large_sequence
large sequence, length = ~100,000

✔

0.004 s

▶

large_extreme
large test with maximal and minimal values, length = ~100,000

✔

0.004 s