Test results - Codility

Tasks Details

easy

1. CyclicRotation

Rotate an array to the right by a given number of steps.

Task Score

100%

Correctness

100%

Performance

Not assessed

An array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the first place. For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7] (elements are shifted right by one index and 6 is moved to the first place).

The goal is to rotate array A K times; that is, each element of A will be shifted to the right K times.

Assume that the following declarations are given:

struct Results { int * A; int N; // Length of the array };

Write a function:

struct Results solution(int A[], int N, int K);

that, given an array A consisting of N integers and an integer K, returns the array A rotated K times.

For example, given

A = [3, 8, 9, 7, 6] K = 3

the function should return [9, 7, 6, 3, 8]. Three rotations were made:

[3, 8, 9, 7, 6] -> [6, 3, 8, 9, 7] [6, 3, 8, 9, 7] -> [7, 6, 3, 8, 9] [7, 6, 3, 8, 9] -> [9, 7, 6, 3, 8]

For another example, given

A = [0, 0, 0] K = 1

the function should return [0, 0, 0]

Given

A = [1, 2, 3, 4] K = 4

the function should return [1, 2, 3, 4]

Assume that:

N and K are integers within the range [0..100];

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

In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.

Solution

Programming language used C

Time spent on task 17 minutes

Notes

not defined yet

Task timeline

Code: 00:30:14 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, int K) {
        // write your code in Java SE 8
    }
}

Code: 00:30:16 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    // write your code in C99 (gcc 6.2.0)
    result.A = ...
    result.N = ...
    return result;
}

Code: 00:30:47 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int * arr = (int *)malloc(N * 4);
    
    // write your code in C99 (gcc 6.2.0)
    result.A = ...
    result.N = ...
    return result;
}

Code: 00:31:17 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    K %= N;
    
    for()
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    result.N = N;
    return result;
}

Code: 00:31:28 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    K %= N;
    
    for(i = 0; i < N; i++){
        
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    result.N = N;
    return result;
}

Code: 00:36:40 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    
    K %= N;
    
    for(i = 0; i < N; i++){
        
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    result.N = N;
    return result;
}

Code: 00:36:50 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    
    
    K %= N;
    
    for(i = 0; i < N; i++){
        
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    result.N = N;
    return result;
}

Code: 00:37:05 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    if(N == 0){
        result.A = A;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    result.N = N;
    return result;
}

Code: 00:37:20 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:38:08 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        ar
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:38:18 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        arr[i] = A[]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:38:28 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        arr[i] = A[K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:40:26 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        arr[i] = A[N]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:40:36 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    for(i = 0; i < N; i++){
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:41:25 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    
    
    for(i = 0; i < N; i++){
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:41:35 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
    int * arr = (int *)malloc(N * seizeof(int));
    
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    for(i = 0; i < N; i++){
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:42:01 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:43:25 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:43:38 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if(N-K )
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:43:48 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if(N-K)
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:44:01 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K)
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:44:11 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N)
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:44:22 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            
            
        }
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:44:39 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            
        }
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:44:51 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[N-K]
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:45:11 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++);
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:45:39 UTC, c, autosave

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:45:52 UTC, c, verify, result: Failed

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j++;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Analysis

Compile error

func.c: In function 'solution':
func.c:6:13: error: expected '=', ',', ';', 'asm' or '__attribute__' before '++' token
     int i, j++;
             ^~
func.c:22:35: warning: implicit declaration of function 'seizeof' [-Wimplicit-function-declaration]
     int * arr = (int *)malloc(N * seizeof(int));
                                   ^~~~~~~
func.c:22:43: error: expected expression before 'int'
     int * arr = (int *)malloc(N * seizeof(int));
                                           ^~~
func.c:25:20: error: 'j' undeclared (first use in this function)
         if((N-K) + j == N){\
                    ^
func.c:25:20: note: each undeclared identifier is reported only once for each function it appears in

Code: 00:46:04 UTC, c, verify, result: Failed

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j=0;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Analysis

Compile error

func.c: In function 'solution':
func.c:22:35: warning: implicit declaration of function 'seizeof' [-Wimplicit-function-declaration]
     int * arr = (int *)malloc(N * seizeof(int));
                                   ^~~~~~~
func.c:22:43: error: expected expression before 'int'
     int * arr = (int *)malloc(N * seizeof(int));
                                           ^~~

Code: 00:46:11 UTC, c, autosave

show code in pop-up

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
#
struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j=0;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Code: 00:46:22 UTC, c, verify, result: Failed

show code in pop-up

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
#include<string.h>

struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j=0;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * seizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Analysis

Compile error

func.c: In function 'solution':
func.c:23:35: warning: implicit declaration of function 'seizeof' [-Wimplicit-function-declaration]
     int * arr = (int *)malloc(N * seizeof(int));
                                   ^~~~~~~
func.c:23:43: error: expected expression before 'int'
     int * arr = (int *)malloc(N * seizeof(int));
                                           ^~~

Code: 00:46:32 UTC, c, verify, result: Passed

show code in pop-up

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// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j=0;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * sizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Analysis

expand all Example tests

▶

example
first example test

✔

0.001 s

▶

example2
second example test

✔

0.001 s

▶

example3
third example test

✔

0.001 s

Code: 00:46:37 UTC, c, 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 28 29 30 31 32 33 34

// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j=0;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * sizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Analysis

expand all Example tests

▶

example
first example test

✔

0.001 s

▶

example2
second example test

✔

0.001 s

▶

example3
third example test

✔

0.001 s

Code: 00:46:39 UTC, c, 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 28 29 30 31 32 33 34

// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
struct Results solution(int A[], int N, int K) {
    struct Results result;
    int i, j=0;
   
    result.N = N;
    
    if(N == 0){
        result.A = A;
        return result;
    }
    
    K %= N;
    
    if(K == 0){
        result.A = A;
        return result;
    }
    
    int * arr = (int *)malloc(N * sizeof(int));
    
    for(i = 0; i < N; i++){
        if((N-K) + j == N){\
            K = N;
            j = 0;
        }
        arr[i] = A[(N-K)+(j++)];
    }
    // write your code in C99 (gcc 6.2.0)
    result.A = arr;
    
    return result;
}

Analysis summary

The solution obtained perfect score.

Analysis

expand all Example tests

▶

example
first example test

✔

0.001 s

▶

example2
second example test

✔

0.001 s

▶

example3
third example test

✔

0.001 s

expand all Correctness tests

▶

extreme_empty
empty array

✔

0.001 s

▶

single
one element, 0 <= K <= 5

✔

0.001 s

▶

double
two elements, K <= N

✔

0.001 s

▶

small1
small functional tests, K < N

✔

0.001 s

▶

small2
small functional tests, K >= N

✔

0.001 s

▶

small_random_all_rotations
small random sequence, all rotations, N = 15

✔

0.001 s

10.

0.001 s

11.

0.001 s

12.

0.001 s

13.

0.001 s

14.

0.001 s

15.

0.001 s

▶

medium_random
medium random sequence, N = 100

✔

0.001 s

▶

maximal
maximal N and K

✔

0.001 s