Maximum element in min heap

Given a min heap, find the maximum element present in the heap.

Examples:

Input :      10 
           /    \ 
          25     23 
         /  \    / \
       45   30  50  40
Output : 50

Input :     20
           /   \ 
          40    28
Output : 40

Brute force approach:
We can check all the nodes in the min heap to get the maximum element. Note that this approach works on any binary tree and does not makes use of any property of the min heap. It has a time and space complexity of O(n). Since min 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 above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the
// maximum element in a
// min heap
int findMaximumElement(int heap[], int n)
{
    int maximumElement = heap[0];
  
    for (int i = 1; i < n; ++i)
        maximumElement = max(maximumElement, heap[i]);
  
    return maximumElement;
}
  
// Driver code
int main()
{
    // Number of nodes
    int n = 10;
  
    // heap represents the following min heap:
    //     10
    //    / \ 
    //  25     23
    //  / \   / \ 
    // 45 50 30 35
    // / \ /
    //63 65 81
    int heap[] = { 10, 25, 23, 45, 50, 30, 35, 63, 65, 81 };
  
    cout << findMaximumElement(heap, n);
  
    return 0;
}

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Java

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// Java implementation of above approach
class GFG {
// Function to find the maximum element 
// in a min heap
  
    static int findMaximumElement(int[] heap, int n) {
        int maximumElement = heap[0];
  
        for (int i = 1; i < n; ++i) {
            maximumElement = Math.max(maximumElement,
                    heap[i]);
        }
  
        return maximumElement;
    }
  
// Driver code
    public static void main(String[] args) {
        // Number of nodes
        int n = 10;
  
        // heap represents the following min heap:
        // 10
        // / \ 
        // 25 23
        // / \ / \ 
        // 45 50 30 35
        // / \ /
        //63 65 81
        int[] heap = {10, 25, 23, 45, 50,
            30, 35, 63, 65, 81};
  
        System.out.print(findMaximumElement(heap, n));
    }
}
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of above approach 
  
# Function to find the maximum element 
# in a min heap 
def findMaximumElement(heap, n):
  
    maximumElement = heap[0]; 
  
    for i in range(1, n):
            maximumElement = max(maximumElement, heap[i]); 
  
    return maximumElement; 
  
# Driver code 
if __name__ == '__main__':
      
    # Number of nodes 
    n = 10
  
    # heap represents the following min heap: 
    # 10 
    # / \ 
    # 25     23 
    # / \ / \ 
    # 45 50 30 35 
    # / \ / 
    #63 65 81 
    heap = [ 10, 25, 23, 45, 50
             30, 35, 63, 65, 81 ]; 
  
    print(findMaximumElement(heap, n)); 
  
# This code is contributed by Princi Singh

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C#

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// C# implementation of above approach
using System;
  
class GFG
{
// Function to find the maximum element 
// in a min heap
static int findMaximumElement(int[] heap, int n)
{
    int maximumElement = heap[0];
  
    for (int i = 1; i < n; ++i)
        maximumElement = Math.Max(maximumElement, 
                                        heap[i]);
  
    return maximumElement;
}
  
// Driver code
public static void Main()
{
    // Number of nodes
    int n = 10;
  
    // heap represents the following min heap:
    // 10
    // / \ 
    // 25 23
    // / \ / \ 
    // 45 50 30 35
    // / \ /
    //63 65 81
    int[] heap = { 10, 25, 23, 45, 50, 
                   30, 35, 63, 65, 81 };
  
    Console.Write(findMaximumElement(heap, n));
}
}
  
// This code is contributed by Akanksha Rai

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Output:

81

Efficient approach:
The min heap property requires that the parent node be lesser than its child node(s). Due to this, we can conclude that a non-leaf node cannot be the maximum element as its child node has a lower value. So we can narrow down our search space to only leaf nodes. In a min heap having n elements, there are 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++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the
// maximumelement in a
// max heap
int findMaximumElement(int heap[], int n)
{
    int maximumElement = heap[n / 2];
  
    for (int i = 1 + n / 2; i < n; ++i)
        maximumElement = max(maximumElement, heap[i]);
  
    return maximumElement;
}
  
// Driver code
int main()
{
    // Number of nodes
    int n = 10;
  
    // heap represents the following min heap:
    //     10
    //    / \ 
    //  25     23
    //  / \   / \ 
    // 45 50 30 35
    // / \ /
    //63 65 81
    int heap[] = { 10, 25, 23, 45, 50, 30, 35, 63, 65, 81 };
  
    cout << findMaximumElement(heap, n);
  
    return 0;
}

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Java

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// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
  
class GFG{
      
  
// Function to find the
// maximumelement in a
// max heap
static int findMaximumElement(int heap[], int n)
{
    int maximumElement = heap[n / 2];
   
    for (int i = 1 + n / 2; i < n; ++i)
        maximumElement = Math.max(maximumElement, heap[i]);
   
    return maximumElement;
}
   
// Driver code
public static void main(String args[])
{
    // Number of nodes
    int n = 10;
   
    // heap represents the following min heap:
    //     10
    //    / \ 
    //  25     23
    //  / \   / \ 
    // 45 50 30 35
    // / \ /
    //63 65 81
    int heap[] = { 10, 25, 23, 45, 50, 30, 35, 63, 65, 81 };
   
    System.out.println(findMaximumElement(heap, n));
   
}
}

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Python 3

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# Python 3 implementation of
# above approach
  
# Function to find the maximum 
# element in a max heap
def findMaximumElement(heap, n):
      
    maximumElement = heap[n // 2]
      
    for i in range(1 + n // 2, n):
        maximumElement = max(maximumElement, 
                             heap[i])
    return maximumElement
  
# Driver Code
n = 10 # Numbers Of Node 
  
# heap represents the following min heap: 
#     10 
# / \ 
# 25     23 
# / \ / \ 
# 45 50 30 35 
# / \ \
#63 65 81 
  
heap = [10, 25, 23, 45, 50
        30, 35, 63, 65, 81]
print(findMaximumElement(heap, n))
  
# This code is contributed by Yogesh Joshi 

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C#

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// C# implementation of above approach
using System;
  
class GFG
{
  
// Function to find the
// maximumelement in a
// max heap
static int findMaximumElement(int[] heap, 
                              int n)
{
    int maximumElement = heap[n / 2];
  
    for (int i = 1 + n / 2; i < n; ++i)
        maximumElement = Math.Max(maximumElement, 
                                        heap[i]);
  
    return maximumElement;
}
  
// Driver code
public static void Main()
{
    // Number of nodes
    int n = 10;
  
    // heap represents the following min heap:
    // 10
    // / \ 
    // 25 23
    // / \ / \ 
    // 45 50 30 35
    // / \ /
    //63 65 81
    int[] heap = { 10, 25, 23, 45, 50,
                   30, 35, 63, 65, 81 };
  
    Console.WriteLine(findMaximumElement(heap, n));
}
}
  
// This code is contributed 
// by Akanksha Rai

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Output:

81


My Personal Notes arrow_drop_up

Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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