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 26 minutes

Notes

not defined yet

Task timeline

Code: 02:25:07 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

// 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
        int size=0, leader=-1, count=0, fcnt=0, res=0;
        for(int i=0; i<A.length; i++) {
            if(size == 0) {
                leader = A[i];
            }
            if(leader == A[i]) {
                size++;
            } else {
                size--;
            }
        }
        if(size == 0) return 0;
        for(int i=0; i<A.length; i++) {
            if(leader == A[i]) {
                count++;
            }
        }
        for(int i=0; i<A.length; i++) {
            if(leader == A[i] ) {
                fcnt++;
            }
            if((i+1) < fcnt*2) && (A.length-i-1) < (count-fcnt)*2) {
                res++;
            }
        }
        return res;
    }
}

Analysis

Compile error

Solution.java:31: error: illegal start of expression
            if((i+1) < fcnt*2) && (A.length-i-1) < (count-fcnt)*2) {
                               ^
Solution.java:31: error: ';' expected
            if((i+1) < fcnt*2) && (A.length-i-1) < (count-fcnt)*2) {
                                                                 ^
2 errors

Code: 02:25:57 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 24 25 26 27 28 29 30 31 32 33 34 35 36 37

// 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
        int size=0, leader=-1, count=0, fcnt=0, res=0;
        for(int i=0; i<A.length; i++) {
            if(size == 0) {
                leader = A[i];
            }
            if(leader == A[i]) {
                size++;
            } else {
                size--;
            }
        }
        if(size == 0) return 0;
        for(int i=0; i<A.length; i++) {
            if(leader == A[i]) {
                count++;
            }
        }
        for(int i=0; i<A.length; i++) {
            if(leader == A[i] ) {
                fcnt++;
            }
            if((i+1) < fcnt*2 && (A.length-i-1) < (count-fcnt)*2) {
                res++;
            }
        }
        return res;
    }
}

Analysis

▶

example
example test

✔

Code: 02:26:08 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 24 25 26 27 28 29 30 31 32 33 34 35 36 37

// 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
        int size=0, leader=-1, count=0, fcnt=0, res=0;
        for(int i=0; i<A.length; i++) {
            if(size == 0) {
                leader = A[i];
            }
            if(leader == A[i]) {
                size++;
            } else {
                size--;
            }
        }
        if(size == 0) return 0;
        for(int i=0; i<A.length; i++) {
            if(leader == A[i]) {
                count++;
            }
        }
        for(int i=0; i<A.length; i++) {
            if(leader == A[i] ) {
                fcnt++;
            }
            if((i+1) < fcnt*2 && (A.length-i-1) < (count-fcnt)*2) {
                res++;
            }
        }
        return res;
    }
}

Analysis

▶

example
example test

✔

Code: 02:26:16 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

// 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
        int size=0, leader=-1, count=0, fcnt=0, res=0;
        for(int i=0; i<A.length; i++) {
            if(size == 0) {
                leader = A[i];
            }
            if(leader == A[i]) {
                size++;
            } else {
                size--;
            }
        }
        if(size == 0) return 0;
        for(int i=0; i<A.length; i++) {
            if(leader == A[i]) {
                count++;
            }
        }
        for(int i=0; i<A.length; i++) {
            if(leader == A[i] ) {
                fcnt++;
            }
            if((i+1) < fcnt*2 && (A.length-i-1) < (count-fcnt)*2) {
                res++;
            }
        }
        return res;
    }
}

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

✔