Test results - Codility

Tasks Details

easy

1. EquiLeader

Find the index S such that the leaders of the sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N - 1] are the same.

100%

A non-empty array A consisting of N integers is given.

The leader of this array is the value that occurs in more than half of the elements of A.

An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.

For example, given array A such that:

A[0] = 4 A[1] = 3 A[2] = 4 A[3] = 4 A[4] = 4 A[5] = 2

we can find two equi leaders:

0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.

2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.

The goal is to count the number of equi leaders.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the number of equi leaders.

For example, given:

A[0] = 4 A[1] = 3 A[2] = 4 A[3] = 4 A[4] = 4 A[5] = 2

the function should return 2, as explained above.

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 [−1,000,000,000..1,000,000,000].

Solution

Programming language used Java 8

Time spent on task 3 minutes

Notes

not defined yet

Code: 07:22:30 UTC, java, autosave

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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

// 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
        if (A.length == 1) {
            return 0;
        }

        // 배열 전체 갯수 중 반+1 이상이 되어야 equi leader 존재
        int conditionCnt = (A.length/2)+1;
        int leaderValue = 0;    // 기준이 되는 leader 값
        int leaderCnt = 0;    // 기준이 되는 leader 갯수

        // 요소 갯수 세기
        HashMap<Integer, Integer> hMap = new HashMap<Integer, Integer>();
        for (int n : A) {
            if (hMap.containsKey(n)) {
                hMap.put(n, hMap.get(n)+1);
            } else {
                hMap.put(n, 1);
            }
        }

        // 기준이 되는 leader값 추출
        for (Map.Entry<Integer, Integer> entry : hMap.entrySet()) {
            if (conditionCnt == 0 || entry.getValue().compareTo(conditionCnt) >= 0) {
                leaderValue = entry.getKey();
                leaderCnt = entry.getValue();
            }
        }

        int equiLeaderCnt = 0;
        int LeftLeaderCnt = leaderCnt;
        int RightLeaderCnt = 0;  // 리더 숫자의 갯수를 넣어줌

        for (int i=A.length-1; i>=0; i--) {
            if (A[i] == leaderValue) {
                LeftLeaderCnt--;
                RightLeaderCnt++;
            }

            if (LeftLeaderCnt > i/2 && RightLeaderCnt > ((A.length-1)/2)) {
                equiLeaderCnt++;
            }
        }

        return equiLeaderCnt;
    }
}

Code: 07:22:33 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 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

// you can also use imports, for example:
// import java.util.*;
import java.util.HashMap;
import java.util.Map;
// 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
        if (A.length == 1) {
            return 0;
        }

        // 배열 전체 갯수 중 반+1 이상이 되어야 equi leader 존재
        int conditionCnt = (A.length/2)+1;
        int leaderValue = 0;    // 기준이 되는 leader 값
        int leaderCnt = 0;    // 기준이 되는 leader 갯수

        // 요소 갯수 세기
        HashMap<Integer, Integer> hMap = new HashMap<Integer, Integer>();
        for (int n : A) {
            if (hMap.containsKey(n)) {
                hMap.put(n, hMap.get(n)+1);
            } else {
                hMap.put(n, 1);
            }
        }

        // 기준이 되는 leader값 추출
        for (Map.Entry<Integer, Integer> entry : hMap.entrySet()) {
            if (conditionCnt == 0 || entry.getValue().compareTo(conditionCnt) >= 0) {
                leaderValue = entry.getKey();
                leaderCnt = entry.getValue();
            }
        }

        int equiLeaderCnt = 0;
        int LeftLeaderCnt = leaderCnt;
        int RightLeaderCnt = 0;  // 리더 숫자의 갯수를 넣어줌

        for (int i=A.length-1; i>=0; i--) {
            if (A[i] == leaderValue) {
                LeftLeaderCnt--;
                RightLeaderCnt++;
            }

            if (LeftLeaderCnt > i/2 && RightLeaderCnt > ((A.length-1)/2)) {
                equiLeaderCnt++;
            }
        }

        return equiLeaderCnt;
    }
}

Analysis

▶

example
example test

✘

WRONG ANSWER
got 1 expected 2

Code: 07:25:01 UTC, java, autosave

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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

// you can also use imports, for example:
// import java.util.*;
import java.util.HashMap;
import java.util.Map;
// 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
        if (A.length == 1) {
            return 0;
        }

        // 배열 전체 갯수 중 반+1 이상이 되어야 equi leader 존재
        int conditionCnt = (A.length/2)+1;
        int leaderValue = 0;    // 기준이 되는 leader 값
        int leaderCnt = 0;    // 기준이 되는 leader 갯수

        // 요소 갯수 세기
        HashMap<Integer, Integer> hMap = new HashMap<Integer, Integer>();
        for (int n : A) {
            if (hMap.containsKey(n)) {
                hMap.put(n, hMap.get(n)+1);
            } else {
                hMap.put(n, 1);
            }
        }

        // 기준이 되는 leader값 추출
        for (Map.Entry<Integer, Integer> entry : hMap.entrySet()) {
            if (conditionCnt == 0 || entry.getValue().compareTo(conditionCnt) >= 0) {
                leaderValue = entry.getKey();
                leaderCnt = entry.getValue();
            }
        }

        int equiLeaderCnt = 0;
        int LeftLeaderCnt = leaderCnt;
        int RightLeaderCnt = 0;  // 리더 숫자의 갯수를 넣어줌

        for (int i=A.length-1; i>=0; i--) {
            if (A[i] == leaderValue) {
                LeftLeaderCnt--;
                RightLeaderCnt++;
            }

            if (LeftLeaderCnt > i/2 && RightLeaderCnt > ((A.length-ㅑ)/2)) {
                equiLeaderCnt++;
            }
        }

        return equiLeaderCnt;
    }
}

Code: 07:25:11 UTC, java, autosave

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 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

// you can also use imports, for example:
// import java.util.*;
import java.util.HashMap;
import java.util.Map;
// 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
        if (A.length == 1) {
            return 0;
        }

        // 배열 전체 갯수 중 반+1 이상이 되어야 equi leader 존재
        int conditionCnt = (A.length/2)+1;
        int leaderValue = 0;    // 기준이 되는 leader 값
        int leaderCnt = 0;    // 기준이 되는 leader 갯수

        // 요소 갯수 세기
        HashMap<Integer, Integer> hMap = new HashMap<Integer, Integer>();
        for (int n : A) {
            if (hMap.containsKey(n)) {
                hMap.put(n, hMap.get(n)+1);
            } else {
                hMap.put(n, 1);
            }
        }

        // 기준이 되는 leader값 추출
        for (Map.Entry<Integer, Integer> entry : hMap.entrySet()) {
            if (conditionCnt == 0 || entry.getValue().compareTo(conditionCnt) >= 0) {
                leaderValue = entry.getKey();
                leaderCnt = entry.getValue();
            }
        }

        int equiLeaderCnt = 0;
        int LeftLeaderCnt = leaderCnt;
        int RightLeaderCnt = 0;  // 리더 숫자의 갯수를 넣어줌

        for (int i=A.length-1; i>=0; i--) {
            if (A[i] == leaderValue) {
                LeftLeaderCnt--;
                RightLeaderCnt++;
            }

            if (LeftLeaderCnt > i/2 && RightLeaderCnt > ((A.length-i)/2)) {
                equiLeaderCnt++;
            }
        }

        return equiLeaderCnt;
    }
}

Code: 07:25:13 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 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

// you can also use imports, for example:
// import java.util.*;
import java.util.HashMap;
import java.util.Map;
// 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
        if (A.length == 1) {
            return 0;
        }

        // 배열 전체 갯수 중 반+1 이상이 되어야 equi leader 존재
        int conditionCnt = (A.length/2)+1;
        int leaderValue = 0;    // 기준이 되는 leader 값
        int leaderCnt = 0;    // 기준이 되는 leader 갯수

        // 요소 갯수 세기
        HashMap<Integer, Integer> hMap = new HashMap<Integer, Integer>();
        for (int n : A) {
            if (hMap.containsKey(n)) {
                hMap.put(n, hMap.get(n)+1);
            } else {
                hMap.put(n, 1);
            }
        }

        // 기준이 되는 leader값 추출
        for (Map.Entry<Integer, Integer> entry : hMap.entrySet()) {
            if (conditionCnt == 0 || entry.getValue().compareTo(conditionCnt) >= 0) {
                leaderValue = entry.getKey();
                leaderCnt = entry.getValue();
            }
        }

        int equiLeaderCnt = 0;
        int LeftLeaderCnt = leaderCnt;
        int RightLeaderCnt = 0;  // 리더 숫자의 갯수를 넣어줌

        for (int i=A.length-1; i>=0; i--) {
            if (A[i] == leaderValue) {
                LeftLeaderCnt--;
                RightLeaderCnt++;
            }

            if (LeftLeaderCnt > i/2 && RightLeaderCnt > ((A.length-i)/2)) {
                equiLeaderCnt++;
            }
        }

        return equiLeaderCnt;
    }
}

Analysis

▶

example
example test

✔

Code: 07:25:17 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 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55

// you can also use imports, for example:
// import java.util.*;
import java.util.HashMap;
import java.util.Map;
// 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
        if (A.length == 1) {
            return 0;
        }

        // 배열 전체 갯수 중 반+1 이상이 되어야 equi leader 존재
        int conditionCnt = (A.length/2)+1;
        int leaderValue = 0;    // 기준이 되는 leader 값
        int leaderCnt = 0;    // 기준이 되는 leader 갯수

        // 요소 갯수 세기
        HashMap<Integer, Integer> hMap = new HashMap<Integer, Integer>();
        for (int n : A) {
            if (hMap.containsKey(n)) {
                hMap.put(n, hMap.get(n)+1);
            } else {
                hMap.put(n, 1);
            }
        }

        // 기준이 되는 leader값 추출
        for (Map.Entry<Integer, Integer> entry : hMap.entrySet()) {
            if (conditionCnt == 0 || entry.getValue().compareTo(conditionCnt) >= 0) {
                leaderValue = entry.getKey();
                leaderCnt = entry.getValue();
            }
        }

        int equiLeaderCnt = 0;
        int LeftLeaderCnt = leaderCnt;
        int RightLeaderCnt = 0;  // 리더 숫자의 갯수를 넣어줌

        for (int i=A.length-1; i>=0; i--) {
            if (A[i] == leaderValue) {
                LeftLeaderCnt--;
                RightLeaderCnt++;
            }

            if (LeftLeaderCnt > i/2 && RightLeaderCnt > ((A.length-i)/2)) {
                equiLeaderCnt++;
            }
        }

        return equiLeaderCnt;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

▶

single
single element

✔

▶

double
two elements

✔

▶

simple
simple test

✔

▶

small_random
small random test with two values, length = ~100

✔

▶

small
random + 200 * [MIN_INT] + random ,length = ~300

✔

▶

large_random
large random test with two values, length = ~50,000

✔

▶

large
random(0,1) + 50000 * [0] + random(0, 1), length = ~100,000

✔

▶

large_range
1, 2, ..., N, length = ~100,000

✔

▶

extreme_large
all the same values

✔