Test results - Codility

Tasks Details

easy

1. Dominator

Find an index of an array such that its value occurs at more than half of indices in the array.

100%

An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that

A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

function solution(A);

that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

For example, given array A such that

A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3

the function may return 0, 2, 4, 6 or 7, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

Solution

Programming language used JavaScript

Time spent on task 2 minutes

Notes

not defined yet

Task timeline

Code: 03:47:13 UTC, js, 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

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

      occurrence[A[i]][0] = 1;
      occurrence[A[i]][1] = i;
    }
  }

  return maxIndex;
    
}

Analysis

▶

example
example test

✔

Code: 03:47:36 UTC, js, 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

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

      occurrence[A[i]][0] = 1;
      occurrence[A[i]][1] = i;
    }
  }

  return maxIndex;
    
}

Analysis

▶

example
example test

✔

Code: 03:47:41 UTC, js, 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

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

      occurrence[A[i]][0] = 1;
      occurrence[A[i]][1] = i;
    }
  }

  return maxIndex;
    
}

Analysis

▶

example
example test

✔

Code: 03:47:45 UTC, js, 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

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

      occurrence[A[i]][0] = 1;
      occurrence[A[i]][1] = i;
    }
  }

  return maxIndex;
    
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

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

▶

example
example test

✔

▶

small_nondominator
all different and all the same elements

✔

▶

small_half_positions
half elements the same, and half + 1 elements the same

✔

▶

small
small test

✔

▶

small_pyramid
decreasing and plateau, small

✔

▶

extreme_empty_and_single_item
empty and single element arrays

✔

▶

extreme_half1
array with exactly N/2 values 1, N even + [0,0,1,1,1]

✔

▶

extreme_half2
array with exactly floor(N/2) values 1, N odd + [0,0,1,1,1]

✔

▶

extreme_half3
array with exactly ceil(N/2) values 1 + [0,0,1,1,1]

✔

▶

medium_pyramid
decreasing and plateau, medium

✔

▶

large_pyramid
decreasing and plateau, large

✔

▶

medium_random
random test with dominator, N = 10,000

✔

▶

large_random
random test with dominator, N = 100,000

✔