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:

int solution(vector<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 C++

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 14:17:41 UTC, cpp, 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

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {

  size_t size = A.size();
  // get leader of whole thing
  int stackSize = 0;
  int stackVal = -1;
  for (size_t i = 0; i < size; i++) {
    int val = A[i];
    if (stackSize == 0) {
      stackVal = val;
      stackSize = 1;
    } else {
      if (val == stackVal) {
        stackSize++;
      } else {
        stackSize--;
      }
    } //  if (stackSize == 0)
  }   // for

  // if stackSize>0, stackval = possible 'leader'
  if (stackSize > 0) {
    int count = 0;
    vector<int> partial(size);
    for (size_t i = 0; i < size; i++) {
      if (A[i] == stackVal)
        count++;
      partial[i] = count;
    }

    if (count > (size / 2)) {
      // leader found
      // count results
      int res = 0;
      for (size_t i = 1; i < size; i++) {
        if ((partial[i - 1] > (i / 2)) && ((count - partial[i - 1]) > ((size - i) / 2))) {
          res++;
        }
      }
      return res;
    } else {
      return 0;
    }
  } else {
    return 0;
  }
}

Analysis

▶

example
example test

✔

Code: 14:17:48 UTC, cpp, 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

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {

  size_t size = A.size();
  // get leader of whole thing
  int stackSize = 0;
  int stackVal = -1;
  for (size_t i = 0; i < size; i++) {
    int val = A[i];
    if (stackSize == 0) {
      stackVal = val;
      stackSize = 1;
    } else {
      if (val == stackVal) {
        stackSize++;
      } else {
        stackSize--;
      }
    } //  if (stackSize == 0)
  }   // for

  // if stackSize>0, stackval = possible 'leader'
  if (stackSize > 0) {
    int count = 0;
    vector<int> partial(size);
    for (size_t i = 0; i < size; i++) {
      if (A[i] == stackVal)
        count++;
      partial[i] = count;
    }

    if (count > (size / 2)) {
      // leader found
      // count results
      int res = 0;
      for (size_t i = 1; i < size; i++) {
        if ((partial[i - 1] > (i / 2)) && ((count - partial[i - 1]) > ((size - i) / 2))) {
          res++;
        }
      }
      return res;
    } else {
      return 0;
    }
  } else {
    return 0;
  }
}

Analysis

▶

example
example test

✔

Code: 14:17:51 UTC, cpp, 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

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {

  size_t size = A.size();
  // get leader of whole thing
  int stackSize = 0;
  int stackVal = -1;
  for (size_t i = 0; i < size; i++) {
    int val = A[i];
    if (stackSize == 0) {
      stackVal = val;
      stackSize = 1;
    } else {
      if (val == stackVal) {
        stackSize++;
      } else {
        stackSize--;
      }
    } //  if (stackSize == 0)
  }   // for

  // if stackSize>0, stackval = possible 'leader'
  if (stackSize > 0) {
    int count = 0;
    vector<int> partial(size);
    for (size_t i = 0; i < size; i++) {
      if (A[i] == stackVal)
        count++;
      partial[i] = count;
    }

    if (count > (size / 2)) {
      // leader found
      // count results
      int res = 0;
      for (size_t i = 1; i < size; i++) {
        if ((partial[i - 1] > (i / 2)) && ((count - partial[i - 1]) > ((size - i) / 2))) {
          res++;
        }
      }
      return res;
    } else {
      return 0;
    }
  } else {
    return 0;
  }
}

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

✔