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:

function solution(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 JavaScript

Time spent on task 2 minutes

Notes

not defined yet

Task timeline

Code: 03:38:52 UTC, js, verify, result: Passed

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// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A) {
    if (A.length === 1) return 0
    let maxRepetition = 1;
    let maxIndex = -1
    let occurance = new Object()
    for (let i = 0; i < A.length; i++){
        if (occurance.hasOwnProperty(A[i])) {
            occurance[A[i]][0]++
            if (occurance[A[i]][0] > maxRepetition){ 
                if (occurance[A[i]][0] > A.length / 2) {
                    maxRepetition = occurance[A[i]][0]
                    maxIndex =  occurance[A[i]][1]
                }
            }
        }
        else {
            occurance[A[i]] = new Array()

            occurance[A[i]][0] = 1
            occurance[A[i]][1] = i
        }
    }

    leader = A[maxIndex]
    
    let equiLeader = 0
    let stack = []
    let stackIndex = -1
    for (let i = 0; i < A.length; i++){

        if (stack.length > (Math.floor(i / 2)  ) && ( maxRepetition - stack.length > Math.floor((A.length - i)/2))){
            equiLeader++
        }
        if (A[i] === leader) stack.push(i)

    }

    
    return equiLeader
    
}

Analysis

▶

example
example test

✔

Code: 03:38:57 UTC, js, verify, result: Passed

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// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A) {
    if (A.length === 1) return 0
    let maxRepetition = 1;
    let maxIndex = -1
    let occurance = new Object()
    for (let i = 0; i < A.length; i++){
        if (occurance.hasOwnProperty(A[i])) {
            occurance[A[i]][0]++
            if (occurance[A[i]][0] > maxRepetition){ 
                if (occurance[A[i]][0] > A.length / 2) {
                    maxRepetition = occurance[A[i]][0]
                    maxIndex =  occurance[A[i]][1]
                }
            }
        }
        else {
            occurance[A[i]] = new Array()

            occurance[A[i]][0] = 1
            occurance[A[i]][1] = i
        }
    }

    leader = A[maxIndex]
    
    let equiLeader = 0
    let stack = []
    let stackIndex = -1
    for (let i = 0; i < A.length; i++){

        if (stack.length > (Math.floor(i / 2)  ) && ( maxRepetition - stack.length > Math.floor((A.length - i)/2))){
            equiLeader++
        }
        if (A[i] === leader) stack.push(i)

    }

    
    return equiLeader
    
}

Analysis

▶

example
example test

✔

Code: 03:39:02 UTC, js, final, score: 100

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// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A) {
    if (A.length === 1) return 0
    let maxRepetition = 1;
    let maxIndex = -1
    let occurance = new Object()
    for (let i = 0; i < A.length; i++){
        if (occurance.hasOwnProperty(A[i])) {
            occurance[A[i]][0]++
            if (occurance[A[i]][0] > maxRepetition){ 
                if (occurance[A[i]][0] > A.length / 2) {
                    maxRepetition = occurance[A[i]][0]
                    maxIndex =  occurance[A[i]][1]
                }
            }
        }
        else {
            occurance[A[i]] = new Array()

            occurance[A[i]][0] = 1
            occurance[A[i]][1] = i
        }
    }

    leader = A[maxIndex]
    
    let equiLeader = 0
    let stack = []
    let stackIndex = -1
    for (let i = 0; i < A.length; i++){

        if (stack.length > (Math.floor(i / 2)  ) && ( maxRepetition - stack.length > Math.floor((A.length - i)/2))){
            equiLeader++
        }
        if (A[i] === leader) stack.push(i)

    }

    
    return equiLeader
    
}

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

✔