Test results - Codility

Tasks Details

medium

1. MissingInteger

Find the smallest positive integer that does not occur in a given sequence.

100%

This is a demo task.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.

For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.

Given A = [1, 2, 3], the function should return 4.

Given A = [−1, −3], the function should return 1.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000];

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

Solution

Programming language used Java 8

Time spent on task 24 minutes

Notes

not defined yet

Task timeline

Code: 00:33:38 UTC, java, verify, result: Failed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

// 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) {
		boolean[] checker = new boolean[A.length + 1];
		int num;
		for (int i = 0; i < A.length; i++) {
			num = A[i];
			if (num > 0 && num < checker.length) checker[num] = true;   //양수일때만 체크
		}

		for (int i = 1; i < checker.length; i++) {
			if (false == checker[i]) return i;
		}
		return 1;   //위에 해당하지 않는 경우 음수만 이루어진 경우 이므로 양의 최소수 1 리턴
	}
}

Analysis

▶

example1
first example test

✔

▶

example2
second example test

✘

WRONG ANSWER
got 1 expected 4

▶

example3
third example test

✔

Code: 00:37:57 UTC, java, verify, result: Passed

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

// 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) {
		boolean[] checker = new boolean[A.length + 1];
		int checkCount = 0;
		int num;
		for (int i = 0; i < A.length; i++) {
			num = A[i];
			if (num > 0 && num < checker.length) checker[num] = true;   //양수 일 때만 체크
		}

		for (int i = 1; i < checker.length; i++) {
			if(checker[i]) checkCount++;
			else return i;
		}
		return checkCount == (checker.length - 1) ? checker.length : 1;   //모두 만족하면 그 다음수 리턴 아니면 모두 음수이므로 양의 최소값 1 리턴
	}
}

Analysis

▶

example1
first example test

✔

▶

example2
second example test

✔

▶

example3
third example test

✔

Code: 00:38:08 UTC, java, verify, result: Passed

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

// 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) {
		boolean[] checker = new boolean[A.length + 1];
		int checkCount = 0;
		int num;
		for (int i = 0; i < A.length; i++) {
			num = A[i];
			if (num > 0 && num < checker.length) checker[num] = true;   //양수 일 때만 체크
		}

		for (int i = 1; i < checker.length; i++) {
			if(checker[i]) checkCount++;
			else return i;
		}
		return checkCount == (checker.length - 1) ? checker.length : 1;   //모두 만족하면 그 다음수 리턴 아니면 모두 음수이므로 양의 최소값 1 리턴
	}
}

Analysis

▶

example1
first example test

✔

▶

example2
second example test

✔

▶

example3
third example test

✔

Code: 00:38:12 UTC, java, final, score: 100

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

// 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) {
		boolean[] checker = new boolean[A.length + 1];
		int checkCount = 0;
		int num;
		for (int i = 0; i < A.length; i++) {
			num = A[i];
			if (num > 0 && num < checker.length) checker[num] = true;   //양수 일 때만 체크
		}

		for (int i = 1; i < checker.length; i++) {
			if(checker[i]) checkCount++;
			else return i;
		}
		return checkCount == (checker.length - 1) ? checker.length : 1;   //모두 만족하면 그 다음수 리턴 아니면 모두 음수이므로 양의 최소값 1 리턴
	}
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N) or O(N * log(N))

▶

example1
first example test

✔

▶

example2
second example test

✔

▶

example3
third example test

✔

▶

extreme_single
a single element

✔

▶

simple
simple test

✔

▶

extreme_min_max_value
minimal and maximal values

✔

▶

positive_only
shuffled sequence of 0...100 and then 102...200

✔

▶

negative_only
shuffled sequence -100 ... -1

✔

▶

medium
chaotic sequences length=10005 (with minus)

✔

▶

large_1
chaotic + sequence 1, 2, ..., 40000 (without minus)

✔

▶

large_2
shuffled sequence 1, 2, ..., 100000 (without minus)

✔

▶

large_3
chaotic + many -1, 1, 2, 3 (with minus)

✔