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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // C++ implementation of above approach #include using namespace std;    // Function to find the // maximum element in a // min heap int findMaximumElement(int heap[], int n) {     int maximumElement = heap;        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; }

Java

 // 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;            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

Python3

 # Python3 implementation of above approach     # Function to find the maximum element  # in a min heap  def findMaximumElement(heap, n):        maximumElement = heap;         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

C#

 // 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;        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

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++

 // C++ implementation of above approach #include 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; }

Java

 // 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));     } }

Python 3

 # 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

C#

 // 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

Output:

81

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