How can we find second maximum from array efficiently?

Question

Is it possible to find the second maximum number from an array of integers by traversing the array only once?

As an example, I have a array of five integers from whi

Answer 1

Other way to solve this problem, is to use comparisons among the elements. Like for example,

a[10] = {1,2,3,4,5,6,7,8,9,10}

Compare 1,2 and say max = 2 and second max = 1

Now compare 3 and 4 and compare the greatest of them with max.

if element > max
     second max = max
     element = max
else if element > second max
     second max = element

The advantage with this is, you are eliminating two numbers in just two comparisons.

Let me know, if you have any problem understanding this.

Answer 2

Your initialization of max and second_max to -1 is flawed. What if the array has values like {-2,-3,-4}?

What you can do instead is to take the first 2 elements of the array (assuming the array has at least 2 elements), compare them, assign the smaller one to second_max and the larger one to max:

if(arr[0] > arr[1]) {
 second_max = arr[1];
 max = arr[0];
} else {
 second_max = arr[0];
 max = arr[1];
}

Then start comparing from the 3rd element and update max and/or second_max as needed:

for(int i = 2; i < arr_len; i++){
    // use >= n not just > as max and second_max can hav same value. Ex:{1,2,3,3}   
    if(arr[i] >= max){  
        second_max=max;
        max=arr[i];          
    }
    else if(arr[i] > second_max){
        second_max=arr[i];
    }
}

Answer 3

Here you are:

std::pair<int, int> GetTwoBiggestNumbers(const std::vector<int>& array)
{
    std::pair<int, int> biggest;
    biggest.first = std::max(array[0], array[1]);  // Biggest of the first two.
    biggest.second = std::min(array[0], array[1]); // Smallest of the first two.

    // Continue with the third.
    for(std::vector<int>::const_iterator it = array.begin() + 2;
        it != array.end();
        ++it)
    {
        if(*it > biggest.first)
        {
            biggest.second = biggest.first;
            biggest.first = *it;
        }
        else if(*it > biggest.second)
        {
            biggest.second = *it;
        }
    }

    return biggest;
}

Answer 4

How about the following below. make_heap is O(n) so this is efficient and this is 1-pass We find the second max by taking advantage that it must be one of the heap children of the parent, which had the maximum.

#include <algorithm>
#include <iostream>

int main()
{
    int arr[6]={0,1,2,3,4,5};

    std::make_heap(arr, arr+6);
    std::cout << "First Max: " << arr[0] << '\n';
    std::cout << "Second Max: " << std::max(arr[1], arr[2]) << '\n';
    return 0;
}

Answer 5

Check this solution.

max1 = a[0];
max2 = a[1];

for (i = 1; i < n; i++)
{
    if (max1 < a[i])
    {
        max2 = max1;
        max1 = a[i];
    }

    if (max2 == max1) max2 = a[i + 1];

    if (max2 == a[n])
    {
        printf("All numbers are the same no second max.\n");
        return 0;
    }

    if (max2 < a[i] && max1 != a[i]) max2 = a[i];
}

Answer 6

Can't we just sort this in decreasing order and take the 2nd element from the sorted array?