Maximum element in min heap
Last Updated :
15 Mar, 2023
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.
Algorithm:
Step 1: Create a function named “findMaximumElement” which takes the heap array and the number of nodes n as input parameter with the int return type.
Step 2: Create a variable named “maximumElement” and initialize it with the first element of the heap array.
Step 3: Start a for loop and traverse over the elements of the heap array starting from the second element.
Step 4: In each iteration, compare the current element with the “maximumElement” and update it if the current element is greater.
Step 5: After the loop completes, return the “maximumElement”.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findMaximumElement( int heap[], int n)
{
int maximumElement = heap[0];
for ( int i = 1; i < n; ++i)
maximumElement = max(maximumElement, heap[i]);
return maximumElement;
}
int main()
{
int n = 10;
int heap[] = { 10, 25, 23, 45, 50, 30, 35, 63, 65, 81 };
cout << findMaximumElement(heap, n);
return 0;
}
|
Java
class GFG {
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;
}
public static void main(String[] args) {
int n = 10 ;
int [] heap = { 10 , 25 , 23 , 45 , 50 ,
30 , 35 , 63 , 65 , 81 };
System.out.print(findMaximumElement(heap, n));
}
}
|
Python3
def findMaximumElement(heap, n):
maximumElement = heap[ 0 ];
for i in range ( 1 , n):
maximumElement = max (maximumElement, heap[i]);
return maximumElement;
if __name__ = = '__main__' :
n = 10 ;
heap = [ 10 , 25 , 23 , 45 , 50 ,
30 , 35 , 63 , 65 , 81 ];
print (findMaximumElement(heap, n));
|
C#
using System;
class GFG
{
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;
}
public static void Main()
{
int n = 10;
int [] heap = { 10, 25, 23, 45, 50,
30, 35, 63, 65, 81 };
Console.Write(findMaximumElement(heap, n));
}
}
|
Javascript
<script>
function findMaximumElement(heap , n) {
var maximumElement = heap[0];
for (i = 1; i < n; ++i) {
maximumElement = Math.max(maximumElement,
heap[i]);
}
return maximumElement;
}
var n = 10;
var heap = [10, 25, 23, 45, 50,
30, 35, 63, 65, 81];
document.write(findMaximumElement(heap, n));
</script>
|
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 higher value. So we can narrow down our search space to only leaf nodes. In a min 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++14
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int n = 10;
int heap[] = { 10, 25, 23, 45, 50, 30, 35, 63, 65, 81 };
cout << findMaximumElement(heap, n);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
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;
}
public static void main(String args[])
{
int n = 10 ;
int heap[] = { 10 , 25 , 23 , 45 , 50 , 30 , 35 , 63 , 65 , 81 };
System.out.println(findMaximumElement(heap, n));
}
}
|
Python 3
def findMaximumElement(heap, n):
maximumElement = heap[n / / 2 ]
for i in range ( 1 + n / / 2 , n):
maximumElement = max (maximumElement,
heap[i])
return maximumElement
n = 10
heap = [ 10 , 25 , 23 , 45 , 50 ,
30 , 35 , 63 , 65 , 81 ]
print (findMaximumElement(heap, n))
|
C#
using System;
class GFG
{
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;
}
public static void Main()
{
int n = 10;
int [] heap = { 10, 25, 23, 45, 50,
30, 35, 63, 65, 81 };
Console.WriteLine(findMaximumElement(heap, n));
}
}
|
Javascript
<script>
function findMaximumElement(heap , n) {
var maximumElement = heap[n / 2];
for (i = 1 + n / 2; i < n; ++i)
maximumElement = Math.max(maximumElement, heap[i]);
return maximumElement;
}
var n = 10;
var heap = [ 10, 25, 23, 45, 50, 30, 35, 63, 65, 81 ];
document.write(findMaximumElement(heap, n));
</script>
|
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