Second minimum element using minimum comparisons


Given an array of integers, find the minimum (or maximum) element and the element just greater (or smaller) than that in less than 2n comparisons. The given array is not necessarily sorted. Extra space is allowed.


Input: {3, 6, 100, 9, 10, 12, 7, -1, 10}
Output: Minimum: -1, Second minimum: 3

We have already discussed an approach in below post.
Find the smallest and second smallest element in an array

Comparisons of array elements can be costly if array elements are of large size, for example large strings. We can minimize the number of comparisons used in above approach.

The idea is based on tournament tree. Consider each element in the given array as a leaf node.


First, we find the minimum element as explained by building a tournament tee. To build the tree, we compare all adjacent pairs of elements (leaf nodes) with each other. Now we have n/2 elements which are smaller than their counterparts (from each pair, the smaller element forms the level above that of the leaves). Again, find the smaller elements in each of the n/4 pairs. Continue this process until the root of the tree is created. The root is the minimum.
Next, we traverse the tree and while doing so, we discard the sub trees whose root is greater than the smallest element. Before discarding, we update the second smallest element, which will hold our result. The point to understand here is that the tree is height balanced and we have to spend only logn time to traverse all the elements that were ever compared to the minimum in step 1. Thus, the overall time complexity is n + logn. Auxiliary space needed is O(2n), as the number of leaf nodes will be approximately equal to the number of internal nodes.

// C++ program to find minimum and second minimum
// using minimum number of comparisons
#include <bits/stdc++.h>
using namespace std;

// Tournament Tree node
struct Node
    int idx;
    Node *left, *right;

// Utility function to create a tournament tree node
Node *createNode(int idx)
    Node *t = new Node;
    t->left = t->right = NULL;
    t->idx = idx;
    return t;

// This function traverses tree across height to
// find second smallest element in tournament tree.
// Note that root is smallest element of tournament
// tree.
void traverseHeight(Node *root, int arr[], int &res)
    // Base case
    if (root == NULL || (root->left == NULL &&
                         root->right == NULL))

    // If left child is smaller than current result,
    // update result and recur for left subarray.
    if (res > arr[root->left->idx] &&
       root->left->idx != root->idx)
        res = arr[root->left->idx];
        traverseHeight(root->right, arr, res);

    // If right child is smaller than current result,
    // update result and recur for left subarray.
    else if (res > arr[root->right->idx] &&
             root->right->idx != root->idx)
        res = arr[root->right->idx];
        traverseHeight(root->left, arr, res);

// Prints minimum and second minimum in arr[0..n-1]
void findSecondMin(int arr[], int n)
    // Create a list to store nodes of current
    // level
    list<Node *> li;

    Node *root = NULL;
    for (int i = 0; i < n; i += 2)
        Node *t1 = createNode(i);
        Node *t2 = NULL;
        if (i + 1 < n)
            // Make a node for next element
            t2 = createNode(i + 1);

            // Make smaller of two as root
            root = (arr[i] < arr[i + 1])? createNode(i) :
                                       createNode(i + 1);

            // Make two nodes as children of smaller
            root->left = t1;
            root->right = t2;

            // Add root

    int lsize = li.size();

    // Construct the complete tournament tree from above
    // prepared list of winners in first round.
    while (lsize != 1)
        // Find index of last pair
        int last = (lsize & 1)? (lsize - 2) : (lsize - 1);

        // Process current list items in pair
        for (int i = 0; i < last; i += 2)
            // Extract two nodes from list, make a new
            // node for winner of two
            Node *f1 = li.front();

            Node *f2 = li.front();
            root = (arr[f1->idx] < arr[f2->idx])?
                createNode(f1->idx) : createNode(f2->idx);

            // Make winner as parent of two
            root->left = f1;
            root->right = f2;

            // Add winner to list of next level
        if (lsize & 1)
        lsize = li.size();

    // Traverse tree from root to find second minimum
    // Note that minimum is already known and root of
    // tournament tree.
    int res = INT_MAX;
    traverseHeight(root, arr, res);
    cout << "Minimum: " << arr[root->idx]
        << ", Second minimum: " << res << endl;

// Driver code
int main()
    int arr[] = {61, 6, 100, 9, 10, 12, 17};
    int n = sizeof(arr)/sizeof(arr[0]);
    findSecondMin(arr, n);
    return 0;


Minimum: 6, Second minimum: 9

We can optimize above code by avoid creation of leaf nodes, as that information is stored in the array itself (and we know that the elements were compared in pairs).

This article is contributed by Dhruv. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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