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 52 minutes

Notes

not defined yet

Task timeline

Code: 10:35:24 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.*;

// 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
        Stack<Integer> equiLeaderStack = new Stack<Integer>();
        int dominatorOccur=0;
        for(int i=0; i<A.length; i++) {
            if (equiLeaderStack.isEmpty()) {
                equiLeaderStack.push(A[i]);
                continue;
            }
            if (equiLeaderStack.peek() == A[i]) {
                equiLeaderStack.push(A[i]);
            } else {
                equiLeaderStack.pop();
            }
        }

        if(equiLeaderStack.isEmpty()) {
            return 0;
        }

        for(int i=0; i<A.length; i++) {
            if(equiLeaderStack.peek()==A[i]){
                dominatorOccur++;
            }
        }

        if (dominatorOccur<=A.length/2) {
            return 0;
        }

        int dominator = equiLeaderStack.peek();
        int equiCount = 0;
        int dominatorCnt = 0;
        boolean leftCondition, rightCondition;
        for(int i=0;i<A.length;i++) {
            if (A[i] == dominator) {
                dominatorCnt++;  //i번째 까지의 dominatorCnt 체크
            }
            //둘로 나눴을 때 왼쪽 오른쪽 둘 다 만족하는지 체크
            leftCondition = dominatorCnt>(i+1)/2;
            rightCondition = (dominatorOccur-dominatorCnt)>(A.length-i-1)/2;
            if( leftCondition && rightCondition ) {
                equiCount++;
            }
        }
        return equiCount;
    }
}

Code: 10:35:31 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 56

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

// 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
        Stack<Integer> equiLeaderStack = new Stack<Integer>();
        int dominatorOccur=0;
        for(int i=0; i<A.length; i++) {
            if (equiLeaderStack.isEmpty()) {
                equiLeaderStack.push(A[i]);
                continue;
            }
            if (equiLeaderStack.peek() == A[i]) {
                equiLeaderStack.push(A[i]);
            } else {
                equiLeaderStack.pop();
            }
        }

        if(equiLeaderStack.isEmpty()) {
            return 0;
        }

        for(int i=0; i<A.length; i++) {
            if(equiLeaderStack.peek()==A[i]){
                dominatorOccur++;
            }
        }

        if (dominatorOccur<=A.length/2) {
            return 0;
        }

        int dominator = equiLeaderStack.peek();
        int equiCount = 0;
        int dominatorCnt = 0;
        boolean leftCondition, rightCondition;
        for(int i=0;i<A.length;i++) {
            if (A[i] == dominator) {
                dominatorCnt++;  //i번째 까지의 dominatorCnt 체크
            }
            //둘로 나눴을 때 왼쪽 오른쪽 둘 다 만족하는지 체크
            leftCondition = dominatorCnt>(i+1)/2;
            rightCondition = (dominatorOccur-dominatorCnt)>(A.length-i-1)/2;
            if( leftCondition && rightCondition ) {
                equiCount++;
            }
        }
        return equiCount;
    }
}

Analysis

▶

example
example test

✔

Code: 10:35:39 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 56

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

// 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
        Stack<Integer> equiLeaderStack = new Stack<Integer>();
        int dominatorOccur=0;
        for(int i=0; i<A.length; i++) {
            if (equiLeaderStack.isEmpty()) {
                equiLeaderStack.push(A[i]);
                continue;
            }
            if (equiLeaderStack.peek() == A[i]) {
                equiLeaderStack.push(A[i]);
            } else {
                equiLeaderStack.pop();
            }
        }

        if(equiLeaderStack.isEmpty()) {
            return 0;
        }

        for(int i=0; i<A.length; i++) {
            if(equiLeaderStack.peek()==A[i]){
                dominatorOccur++;
            }
        }

        if (dominatorOccur<=A.length/2) {
            return 0;
        }

        int dominator = equiLeaderStack.peek();
        int equiCount = 0;
        int dominatorCnt = 0;
        boolean leftCondition, rightCondition;
        for(int i=0;i<A.length;i++) {
            if (A[i] == dominator) {
                dominatorCnt++;  //i번째 까지의 dominatorCnt 체크
            }
            //둘로 나눴을 때 왼쪽 오른쪽 둘 다 만족하는지 체크
            leftCondition = dominatorCnt>(i+1)/2;
            rightCondition = (dominatorOccur-dominatorCnt)>(A.length-i-1)/2;
            if( leftCondition && rightCondition ) {
                equiCount++;
            }
        }
        return equiCount;
    }
}

Analysis

▶

example
example test

✔

Code: 10:35:43 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 56

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

// 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
        Stack<Integer> equiLeaderStack = new Stack<Integer>();
        int dominatorOccur=0;
        for(int i=0; i<A.length; i++) {
            if (equiLeaderStack.isEmpty()) {
                equiLeaderStack.push(A[i]);
                continue;
            }
            if (equiLeaderStack.peek() == A[i]) {
                equiLeaderStack.push(A[i]);
            } else {
                equiLeaderStack.pop();
            }
        }

        if(equiLeaderStack.isEmpty()) {
            return 0;
        }

        for(int i=0; i<A.length; i++) {
            if(equiLeaderStack.peek()==A[i]){
                dominatorOccur++;
            }
        }

        if (dominatorOccur<=A.length/2) {
            return 0;
        }

        int dominator = equiLeaderStack.peek();
        int equiCount = 0;
        int dominatorCnt = 0;
        boolean leftCondition, rightCondition;
        for(int i=0;i<A.length;i++) {
            if (A[i] == dominator) {
                dominatorCnt++;  //i번째 까지의 dominatorCnt 체크
            }
            //둘로 나눴을 때 왼쪽 오른쪽 둘 다 만족하는지 체크
            leftCondition = dominatorCnt>(i+1)/2;
            rightCondition = (dominatorOccur-dominatorCnt)>(A.length-i-1)/2;
            if( leftCondition && rightCondition ) {
                equiCount++;
            }
        }
        return equiCount;
    }
}

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

✔