Open In App

Minimum element in a max heap

Improve
Improve
Like Article
Like
Save
Share
Report

Given a max heap, find the minimum element present in the heap.

Examples:  

Input :     100
           /    \ 
          75     50 
         /  \    / \
       55   10  2  40
Output : 2

Input :     20
           /   \ 
          4    18
Output : 4

Brute force approach

We can check all the nodes in the max heap to get the minimum element. Note that this approach works on any binary tree and does not makes use of any property of the max heap. It has a time and space complexity of O(n). Since max heap is a complete binary tree, we generally use arrays to store them, so we can check all the nodes by simply traversing the array. If the heap is stored using pointers, then we can use recursion to check all the nodes.

Below is the implementation of the above approach: 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// minimum element in a
// max heap
int findMinimumElement(int heap[], int n)
{
    int minimumElement = heap[0];
 
    for (int i = 1; i < n; ++i)
        minimumElement = min(minimumElement, heap[i]);
 
    return minimumElement;
}
 
// Driver code
int main()
{
    // Number of nodes
    int n = 10;
    // heap represents the following max heap:
    //         20
    //       /    \
    //      18     10
    //   /    \    /  \
    //   12     9  9   3
    //  /  \   /
    // 5    6 8
    int heap[] = { 20, 18, 10, 12, 9, 9, 3, 5, 6, 8 };
 
    cout << findMinimumElement(heap, n);
 
    return 0;
}


Java




// Java implementation of above approach
 
public class Improve {
 
    // Function to find the
    // minimum element in a
    // max heap
    static int findMinimumElement(int heap[], int n)
    {
        int minimumElement = heap[0];
 
        for (int i = 1; i < n; ++i)
            minimumElement
                = Math.min(minimumElement, heap[i]);
 
        return minimumElement;
    }
 
    // Driver code
    public static void main(String args[])
    {
        // Number of nodes
        int n = 10;
        // heap represents the following max heap:
        //         20
        //       /    \
        //      18     10
        //   /    \    /  \
        //   12     9  9   3
        //  /  \   /
        // 5    6 8
        int heap[] = { 20, 18, 10, 12, 9, 9, 3, 5, 6, 8 };
 
        System.out.println(findMinimumElement(heap, n));
    }
    // This Code is contributed by ANKITRAI1
}


Python3




# Python 3 implementation of above approach
 
# Function to find the minimum
# element in a max heap
 
 
def findMiniumumElement(heap, n):
 
    minimumElement = heap[0]
 
    for i in range(1, n):
        minimumElement = min(minimumElement, heap[i])
 
    return minimumElement
 
 
# Driver Code
n = 10  # Numbers Of Node
 
# heap represents the following max heap:
#         20
#         / \
#         18 10
#     / \ / \
#     12 9 9 3
#     / \ /
#     5 6 8
#
 
heap = [20, 18, 10, 12, 8,
        9, 3, 5, 6, 8]
 
print(findMiniumumElement(heap, n))
 
# This code is contributed by SANKAR


C#




// C# implementation of above approach
using System;
class GFG {
 
    // Function to find the
    // minimum element in a
    // max heap
    static int findMinimumElement(int[] heap, int n)
    {
        int minimumElement = heap[0];
 
        for (int i = 1; i < n; ++i)
            minimumElement
                = Math.Min(minimumElement, heap[i]);
 
        return minimumElement;
    }
 
    // Driver code
    public static void Main()
    {
        // Number of nodes
        int n = 10;
 
        // heap represents the following max heap:
        //             20
        //           /     \
    //          18     10
        //       /  \    / \
    //      12   9  9   3
        //     /  \ /
        //    5   6 8
        int[] heap = { 20, 18, 10, 12, 9, 9, 3, 5, 6, 8 };
 
        Console.Write(findMinimumElement(heap, n));
    }
}
 
// This code is contributed by ChitraNayal


PHP




<?php
// PHP implementation of above approach
 
// Function to find the
// minimum element in a
// max heap
function findMinimumElement(&$heap, $n)
{
    $minimumElement = $heap[0];
 
    for ($i = 1; $i < $n; ++$i)
        $minimumElement = min($minimumElement,
                              $heap[$i]);
 
    return $minimumElement;
}
 
// Driver code
 
// Number of nodes
$n = 10;
 
// heap represents the following
// max heap:
//         20
//       /    \
//     18     10
//    /  \   /  \
// 12   9  9   3
// / \ /
// 5 6 8
$heap = array(20, 18, 10, 12, 9,
               9, 3, 5, 6, 8 );
 
echo findMinimumElement($heap, $n);
 
// This code is contributed
// by Shivi_Aggarwal
?>


Javascript




<script>
 
    // Javascript implementation of above approach
     
    // Function to find the
    // minimum element in a
    // max heap
    function findMinimumElement(heap, n)
    {
        let minimumElement = heap[0];
  
        for (let i = 1; i < n; ++i)
            minimumElement = Math.min(minimumElement, heap[i]);
  
        return minimumElement;
    }
     
    // Number of nodes
    let n = 10;
 
    // heap represents the following max heap:
    //             20
    //           /     \
    //          18     10
    //       /  \    / \
    //      12   9  9   3
    //     /  \ /
    //    5   6 8
    let heap = [ 20, 18, 10, 12, 9, 9, 3, 5, 6, 8 ];
 
    document.write(findMinimumElement(heap, n));
     
</script>


Output

3

Efficient approach:

The max heap property requires that the parent node be greater than its child node(s). Due to this, we can conclude that a non-leaf node cannot be the minimum element as its child node has a lower value. So we can narrow down our search space to only leaf nodes. In a max heap having n elements, there is ceil(n/2) leaf nodes. The time and space complexity remains O(n) as a constant factor of 1/2 does not affect the asymptotic complexity.

Below is the implementation of above approach:  

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// minimum element in a
// max heap
int findMinimumElement(int heap[], int n)
{
    int minimumElement = heap[n / 2];
 
    for (int i = 1 + n / 2; i < n; ++i)
        minimumElement = min(minimumElement, heap[i]);
 
    return minimumElement;
}
 
// Driver code
int main()
{
    // Number of nodes
    int n = 10;
    // heap represents the following max heap:
    //        20
    //       /    \
    //     18     10
    //   /    \    /  \
    //   12     9  9   3
    //  /  \   /
    // 5    6 8
    int heap[] = { 20, 18, 10, 12, 9, 9, 3, 5, 6, 8 };
 
    cout << findMinimumElement(heap, n);
 
    return 0;
}


Java




// Java implementation of above approach
import java.util.*;
import java.lang.*;
 
class GFG {
 
    // Function to find the
    // minimum element in a
    // max heap
    static int findMinimumElement(int heap[], int n)
    {
        int minimumElement = heap[n / 2];
 
        for (int i = 1 + n / 2; i < n; ++i)
            minimumElement
                = Math.min(minimumElement, heap[i]);
 
        return minimumElement;
    }
 
    // Driver code
    public static void main(String args[])
    {
        // Number of nodes
        int n = 10;
        // heap represents the following max heap:
        //     20
        //     / \
        //     18     10
        // / \ / \
        // 12     9 9 3
        // / \ /
        // 5 6 8
        int heap[] = new int[] { 20, 18, 10, 12, 9,
                                 93568 };
 
        System.out.println(findMinimumElement(heap, n));
    }
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)


Python3




# Python 3 implementation of
# above approach
 
# Function to find the minimum
# element in a max heap
 
 
def findMiniumumElement(heap, n):
 
    # // -->Floor Division Arithmetic Operator
    minimumElement = heap[n // 2]
 
    for i in range(1 + n // 2, n):
        minimumElement = min(minimumElement,
                             heap[i])
 
    return minimumElement
 
 
# Driver Code
n = 10  # Numbers Of Node
 
# heap represents the
# following max heap:
# 20
# / \
# 18 10
# / \ / \
# 12 9 9 3
# / \ \
# 5 6 8
#
 
heap = [20, 18, 10, 12, 9,
        9, 3, 5, 6, 8]
 
print(findMiniumumElement(heap, n))
 
# This code is contributed by SANKAR


C#




// C# implementation of above approach
using System;
 
class GFG {
 
    // Function to find the minimum element
    // in a max heap
    static int findMinimumElement(int[] heap, int n)
    {
        int minimumElement = heap[n / 2];
 
        for (int i = 1 + n / 2; i < n; ++i)
            minimumElement
                = Math.Min(minimumElement, heap[i]);
 
        return minimumElement;
    }
 
    // Driver code
    static public void Main()
    {
 
        // Number of nodes
        int n = 10;
 
        // heap represents the following
        // max heap:
        // 20
        // / \
    // 18 10
        // / \ / \
    // 12 9 9 3
        // / \ /
        // 5 6 8
        int[] heap = new int[] { 20, 18, 10, 12, 9,
                                 9,  3,  5,  6,  8 };
 
        Console.WriteLine(findMinimumElement(heap, n));
    }
}
 
// This code is contributed
// by ajit


PHP




<?php
// PHP implementation of above approach
 
// Function to find the
// minimum element in a
// max heap
function findMinimumElement($heap, $n)
{
    $minimumElement = $heap[$n/2];
 
    for ($i = 1 + $n / 2; $i < $n; ++$i)
        $minimumElement = min($minimumElement, $heap[$i]);
 
    return $minimumElement;
}
 
    // Driver code
    // Number of nodes
    $n = 10;
    // heap represents the following max heap:
    //     20
    //     / \
    //     18     10
    // / \ / \
    // 12     9 9 3
    // / \ /
    // 5 6 8
    $heap = Array(20, 18, 10, 12, 9, 9, 3, 5, 6, 8 );
 
    echo(findMinimumElement($heap, $n));
 
//  This code is contributed to Rajput-Ji
?>


Javascript




<script>
    // Javascript implementation of above approach
     
    // Function to find the minimum element
    // in a max heap
    function findMinimumElement(heap, n)
    {
        let minimumElement = heap[parseInt(n / 2, 10)];
  
        for (let i = 1 + parseInt(n / 2, 10); i < n; ++i)
            minimumElement
                = Math.min(minimumElement, heap[i]);
  
        return minimumElement;
    }
     
    // Number of nodes
    let n = 10;
 
    // heap represents the following
    // max heap:
    // 20
    // / \
    // 18 10
    // / \ / \
    // 12 9 9 3
    // / \ /
    // 5 6 8
    let heap = [ 20, 18, 10, 12, 9, 9,  3,  5,  6,  8 ];
 
    document.write(findMinimumElement(heap, n));
     
</script>


Output

3


Last Updated : 01 Sep, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads