details of finding 2nd smallest element in unsorted array

Question

Let me make things clear: I want to know more about the implementation of the "best" ALGORITHM how to find 2nd smallest element in array, which is described here:

Algorithm: Find index of 2nd smallest element from an unknown array

First, I am completely clear about the ALGORITHM how to find 2nd smallest element in array. The best solution is O(n+lg(n)-2) which can be achieved by first obtaining the smallest element S and then searching the smallest element of S's competitors. Read here: Algorithm: Find index of 2nd smallest element from an unknown array

What I cannot understand is how to implement it.

Especially, after finding the smallest element, how to track back those competitors of the smallest element such that I can find 2nd smallest element in those competitors?

NO NEED TO SORT. Quicksort is O(nlg(n)) which is worse than O(n+lg(n)-2).

There have been too many people talking about the best solution while nobody actually gave an implementation. Besides, I found the best solution is just theoretical best. It is very hard to implement it following that "best" solution.

With respect, then it sounds like you are not completely clear on the algorithm. — Oliver Charlesworth, Jun 07 '11 at 21:54
Which way do you want to sort it? Quicksort? http://en.wikipedia.org/wiki/Sorting_algorithm — Dair, Jun 07 '11 at 21:54
If you sort the array to find the smallest element, you can just go to the appropriate index of the array to get the smallest, either index 1 or (n-1)-1 where n is the number that you put for the size of the array. — Dair, Jun 07 '11 at 21:57
@Chris: O(n) can be guaranteed for the worst case, indeed, not just in average. — Vlad, Jun 07 '11 at 22:04

score 3 · Answer 1 · answered Jun 08 '11 at 02:13

3

int smallest, secondsmallest;
if (array[0] < array[1]) {
    smallest = array[0];
    secondsmallest = array[1];
} else {
    smallest = array[1];
    secondsmallest = array[0];
}
for(int i = 2; i < array_size; i++) {
    if (array[i] < smallest) {
        secondsmallest = smallest;
        smallest = array[i];
    } else if (array[i] < secondsmallest) {
        secondsmallest = array[i];
    }
}

answered Jun 08 '11 at 02:13

Puppy

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Your method is O(n^2). Read this: http://stackoverflow.com/questions/3715935/algorithm-find-index-of-2nd-smallest-element-from-an-unknown-array – deryk Jun 08 '11 at 02:23
1

@deryk: I quite clearly only loop through the array once. That makes it kind of impossible for my algorithm to be O(n^2). – Puppy Jun 08 '11 at 02:28
Sorry, your algorithm is 2N > n+lg(n)-2 – deryk Jun 08 '11 at 04:18
2

@deryk: You should brush up on complexity analysis. O(n) is the same as O(n + lgn -2), because you are only concerned with asymtotic behavior. – Dennis Zickefoose Jun 08 '11 at 05:05

score 1 · Answer 2 · answered Jun 08 '11 at 02:26

Odd that no one has bothered to give this its name. Including the original questioner. For those unfamiliar with it, it is called the i th order statistic, and is discussed in many fine algorithm books. It is indeed O(n) (actually theta of n) for n elements.

sabbibJAVA · Answer 3 · 2013-04-08T10:33:41.620

0

Try this, should work

min1 = a[0]; 
min2 = 0;
int i = 0;
for (i = 0; i < a.length; i++){
   if (min1 < a[i]){
      min2 = min1;
      min1 = a[i];
   }
}

a.length is a Java instance of array. I'm not sure if C++ has the same one.

edited Apr 08 '13 at 10:33

answered Jul 28 '12 at 22:05

sabbibJAVA

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the c++ vector has a **.size()** – mks Nov 10 '16 at 06:16
Wow, how is wrong answer like this get undetected for so long? try [110, 10, 20, 120] – zcahfg2 Mar 21 '19 at 08:55

score 0 · Answer 4 · answered Sep 21 '15 at 15:11

// A simple implementation with O(n) runtime complexity;
int findSecondMin(vector<int>& nums) {
    if (nums.size() < 2) throw runtime_error("no second minimum !");
    int min1 = nums[0], min2 = INT_MAX;
    for(int i=1;i<nums.size();i++) {
      if (nums[i] < min1) {
        min2 = min1;
        min1 = nums[i];
      } else if (nums[i] < min2) {
        min2 = nums[i];
      }
    }
    return min2;
}

score 0 · Answer 5 · answered Jun 07 '11 at 21:57

0

You should keep the running "candidates" (and maybe number of their occurrences, depending on your strict definition of "2nd") for the smallest and second smallest value all along the way.

answered Jun 07 '11 at 21:57

Vlad

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I can't find it in my heart to -1 this but we really shouldn't do this for homework problems... – Andrei Jun 07 '11 at 22:01
@Andrei: I assume my hint is not too strong, why? – Vlad Jun 07 '11 at 22:02
It is not actually a homework. There have been too many people talking about the best solution while nobody actually gave an implementation. Besides, I found the best solution is just theoretical best. It is very hard to implement it following that "best" solution. – deryk Jun 08 '11 at 01:47

score 0 · Answer 6 · answered Jun 07 '11 at 23:45

0

Well, you kind of answered your own question:

How do you keep track of the smallest element? You store an index to it.
How do you find the second smallest element? Search again and ignore the element you found first.

answered Jun 07 '11 at 23:45

MSN

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That is not the best solution. The best one is to find the smallest element first and then find the 2nd smallest one by comparing competitors of the found smallest element. – deryk Jun 08 '11 at 01:45
@deryk, if you are asking for the "optimal" solution, you could probably do worse than looking at the implementation of `nth_element(...)` in the C++ STL. That is O(n), assuming you can mutate the array in place. – MSN Jun 08 '11 at 22:30

details of finding 2nd smallest element in unsorted array

Algorithm: Find index of 2nd smallest element from an unknown array

NO NEED TO SORT. Quicksort is O(nlg(n)) which is worse than O(n+lg(n)-2).

6 Answers6