7
\$\begingroup\$

A zero-indexed 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

int solution(int A[], int N);

that, given a zero-indexed 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.

Assume that:

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].

 int solution(int A[], int N)  { 
        int size = 0; 
        int value = 0; 
        for(int i = 0 ; i < N ; i++)  { 
                if(size == 0)  { 
                   size++; value = A[i]; 
                } 
                else  { 
                   if(value != A[i]) 
                       size--; 
                } 
         } 
         int candidate = -1; 
         if(size > 0) { 
            candidate = value; 
         } 
         int leader = -1; 
         int count = 0; 
         for(int j = 0 ; j < N ; j++)  { 
            if(A[j] == candidate) {
                count++;
            }
         } 
         if(count > (N/2)) {
            leader = candidate; 
         }
         else {
            return -1;  
         }
         int index[count]; 
         int cnt=0; 
         for(int k = 0 ; k < N ; k++) { 
            if(A[k] == leader) { 
               index[cnt] = k; 
               cnt++; 
            } 
         } 
         return index[0]; 
    }
\$\endgroup\$
5
\$\begingroup\$

I see a number of things that may help you improve your program.

Test thoroughly

There is a flaw in your program which causes it to fail on certain kinds of input. Here's the test code I used:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

// solution() goes here

struct test {
    int expected;
    int N;
    int numbers[20];
};

struct test tests[] = {
    { -1, 6, { 3, 3, 0, 6, 9, 3 }},
    { 3, 8, { 3, 4, 3, 2, 3, -1, 3, 3 }},
    { 3, 8, { 3, 3, 4, 2, 3, 3, -1, 3 }},
    { 3, 7, { 3, 3, 4, 3, 3, -1, 3}},
    { 3, 7, { 3, 3, 4, 3, 3, -1, 0}},
    { 3, 6, { 3, 3, 4, 3, 3, -1 }},
    { 6, 2, { 6, 6}},
    { 8, 1, { 8}},
};

bool doTest(struct test* t) {
    int index = solution(t->numbers, t->N);
    if (t->expected == -1) {
        return index == -1;
    }
    return t->expected == t->numbers[index];
}

int main()
{
    int testcount = sizeof(tests) / sizeof(tests[0]);
    for (int i=0; i < testcount; ++i) {
        printf("Test %d: %s\n", i+1, doTest(&tests[i]) ? "OK " : "BAD");
    }
}

With your code as posted, tests 5 and 6 (of 8) fail.

Fix your formatting

The indentation of the code is not consistent, which makes it a little harder to read and understand than if you format the code consistently.

Read the problem statement carefully

The problem statement says that the return value can be any of the index values that point to the dominator number, so there's no need to store all of them as the code does in the index[] array.

Reduce redundant variables

The value, candidate and leader could all be collapsed to a single local variable which would make the code shorter and more understandable.

Minimize loops through arrays

The code attempts to identify a candidate and then does several more loops through the code to find the count and an index value to return. The code could be simplified to this (although the bug still remains):

int solution(int A[], int N)
{
    int size = 0;
    int candidate = 0;
    for (int i = 0; i < N; i++) {
        if (size == 0) {
            size++;
            candidate = A[i];
        } else {
            if (candidate != A[i])
                size--;
        }
    }
    if (size <= 0) {
        return -1;
    }
    int count = 0;
    int index = -1;
    for (int i = 0; i < N; i++) {
        if (A[i] == candidate) {
            count++;
            index = i;
        }
    }
    if (count > (N / 2)) {
        return index;
    }
    return -1;
}

Consider a different algorithm

There is another approach that passes all of the tests I've tried. It first sorts the array. Because the array is now sorted, we know that if there is a dominator, more than half the sorted array must be the dominator value. In other words, for an array size of N, the index A[N/2] of the sorted array must contain the dominator value if one exists. Then all that remains is to verify that it is indeed a dominator and return an index into this sorted version of the original array. (Alternatively, this can be done with a copy of the original array if you make the assumption that the passed array can't be modified. The problem statement doesn't say and doesn't specify const for the array parameter.) Here's the code:

int comp(const void *a, const void *b)
{
    return *(const int *)a - *(const int *)b;
}

int solution(int A[], int N)
{
    qsort(A, N, sizeof(int), comp);
    int count = 0;
    for (int i = 0; i < N && count <= N / 2; ++i) {
        if (A[i] == A[N / 2]) {
            ++count;
        }
    }
    return (count > N / 2) ? N / 2 : -1;
}

Note that the index of N / 2 is guaranteed to point to the dominator if it exists, and the specification says that one may return any element of A in which the dominator exists.

Put another way, if one has a 10 integer array, there is no way to place 6 contiguous numbers within that array without having one of them be at index 5.

The corrected original algorithm

As @vnp and @janos have pointed out, the algorithm above, while simple and correct, is not optimal in terms of performance, so by popular demand, here's a corrected version of your original algorithm:

int solution(int A[], int N)
{
    int count = 1;
    int candidate = A[0];
    int index = 0;
    for (int i = 1; i < N; i++) {
        if (A[i] == candidate) {
            ++count;
        } else if (count > 0) {
            --count;
        } else {
            candidate = A[i];
            index = i;
            count = 1;
        }
    }
    if (count > 0) { 
        count = 0;
        for (int i = 0; i < N && count <= N / 2; ++i) {
            if (A[i] == candidate) {
                count++;
            }
        }
        if (count > (N / 2)) {
            return index;
        }
    }
    return -1;
}

Thanks to @JS1 who pointed out that an earlier version of this failed on an all zeroes input, reinforcing my first point: Test thoroughly! :) I've added the following line to my test array:

{0, 5, {0, 0, 0, 0, 0}},

Timing tests

I modified the test routines a bit to have them copy each test and then sort rather than using a pointer. Then I had each version run 100000 (1e6) times over the test suite of 8 tests. Timing was then noted on each version and, as expected, the qsort algorithm is slower. Tests were all done on a 64-bit Linux machine with a quad-core i7 running at 3.40GHz. Compiler was gcc version 5.3.1 with -O2 optimization.

qsort version: 1.25 seconds
Moore version: 0.77 seconds

Results are fairly consistently repeatable and independent of which algorithm is tested first. Since qsort is \${\mathcal O}(n \log n)\$, it's likely that the difference would tilt even more toward the Moore algorithm if longer strings had been used.

\$\endgroup\$
  • \$\begingroup\$ This is one of the rare cases where sorting hurts rather then helps. The Moore algorithm (which OP tried to implement) runs in linear time and constant space. \$\endgroup\$ – vnp Mar 17 '16 at 19:51
  • \$\begingroup\$ @vnp Good point. It has the advantage that it's simple enough to easily see that it works. And the other advantage is that it actually does. ;) \$\endgroup\$ – Edward Mar 17 '16 at 20:02
  • \$\begingroup\$ +1 overall, though I really really really disagree with the sorting idea. The return (count > N / 2) ? N / 2 : -1; has an error, and in any case this solution loses the index in the original array of the dominating element, so you're not quite finished there ;-) Which is ironic given your Read the problem statement carefully section ;-) \$\endgroup\$ – janos Mar 17 '16 at 21:50
  • \$\begingroup\$ Actually, I did read carefully. It says that one needs to return the index of the original array (which it does) -- but it doesn't say that the original array can't be modified (which it is). In any case, I'll also post the corrected Moore algorithm so we can compare and contrast. \$\endgroup\$ – Edward Mar 17 '16 at 21:52
  • \$\begingroup\$ Your corrected version has a bug where if the array is all zeroes, it will return an index of -1. Other than that, nice review. \$\endgroup\$ – JS1 Mar 18 '16 at 5:49
4
\$\begingroup\$

Breaking large methods into smaller methods

Some of the variables are only used in the first half of the function, while the others in the second half. This suggests that you should separate the one big function to multiple smaller functions, for example:

int findCandidate(int A[], int N)  { 
    int size = 0; 
    int value = 0; 
    for(int i = 0 ; i < N ; i++)  { 
        if(size == 0)  { 
            size++; value = A[i]; 
        } 
        else  { 
            if(value != A[i]) 
                size--; 
        } 
    } 
    return size > 0 ? value : -1;
} 

Your implementation is really doing 3 different things:

  1. Find a candidate
  2. Count the occurrences
  3. Find all the indexes

These should really be in their own methods. Except the last step. You really don't need to find all the indexes (and definitely not store them), you could just return the first one immediately.

Finding a candidate

On closer look, the way you look for a candidate has a bug. Consider the input 5 5 1, let's walk through the loop:

  • first element: 5
    • size is 0, so increment it, and set value to 5
  • second element: 5
    • size is not 0, value is the same, so do nothing
  • third element: 1
    • size is not 0, value is different, so decrement size to 0

Since size is 0, the candidate is set to -1. Which is wrong, as the candidate in this example should be 5.

The problem is that the algorithm doesn't increment the size when seeing the same value. The fix is simple: if value == A[i], then increment size. However, the size++ statement will be appear twice inside the loop. The logic in the loop body can be a bit better organized, like this:

int findCandidate(int A[], int N) {
    int candidate = -1;
    int count = 0;
    for (int i = 0; i < N; ++i) {
        if (A[i] == candidate) {
            ++count;
        } else if (count > 0) {
            --count;
        } else {
            candidate = A[i];
            count = 1;
        }
    }
    return count > 0 ? candidate : -1;
}

Putting it together

With the above suggestions, the main method should be more like this:

int findLeaderIndex(int A[], int N) {
    int candidate = findCandidate(A);
    int count = countElement(A, N, candidate);
    if (count <= N / 2) {
        return -1;
    }
    return findFirstIndex(A, N, candidate);
}

Note that we don't check the value of candidate returned by findCandidate. That's fine. The problem is that -1 is valid value for elements of the array. It may very well be the leader, so the value of -1 can be a false positive. That's not a problem, because everything will become clear based on the result of countElement.

\$\endgroup\$
1
\$\begingroup\$

Your solution() function is too complex, you should create sub-functions to implement your solution. The problem statement does not state that you can't create sub-functions for the solution() function. Good candidates for your sub-functions are each of the for loops in the code. Using sub-functions will make your code easier to read and maintain. The names of the sub-functions should describe what their are doing:

findCandidates(A[], N)
findDominator(A[], N)

To optimize the code create a copy of the array and use qsort() to sort the values in the copy. Find the dominator using the sorted array, then find the indexes using the dominator value.

I don't know if this is a result of how you entered the code, but the following line of code should be 2 lines of code:

size++; value = A[i];

size++;
value = A[i];

If you use sub-functions your code will be self documenting as it is now it needs comments to explain what it is doing.

\$\endgroup\$
1
\$\begingroup\$
int index[count]; 

Please, don't do it. You are allocating space in stack, which is rather limited. Use instead

int solution(int A[], int N) 
{
    int* index = malloc(count * sizeof(int));
    ...
    int ret = index[0];
    free(index);
    return ret;
}
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.