Find the maximum elements in the first and the second halves of the Array

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, Given an array arr[] of N integers. The task is to find the largest elements in the first half and the second half of the array. Note that if the size of the array is odd then the middle element will be included in both halves.
Examples:

Input: arr[] = {1, 12, 14, 5}
Output: 12, 14
First half is {1, 12} and the second half is {14, 5}.
Input: arr[] = {1, 2, 3, 4, 5}
Output: 3, 5

Approach: Calculate the middle index of the array as mid = N / 2. Now the first halve elements will be present in the subarray arr[0…mid-1] and arr[mid…N-1] if N is even.
If N is odd then the halves are arr[0…mid] and arr[mid…N-1]
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print largest element in
// first half and second half of an array
void findMax(int arr[], int n)
{
 
    // To store the maximum element
    // in the first half
    int maxFirst = INT_MIN;
 
    // Middle index of the array
    int mid = n / 2;
 
    // Calculate the maximum element
    // in the first half
    for (int i = 0; i < mid; i++)
        maxFirst = max(maxFirst, arr[i]);
 
    // If the size of array is odd then
    // the middle element will be included
    // in both the halves
    if (n % 2 == 1)
        maxFirst = max(maxFirst, arr[mid]);
 
    // To store the maximum element
    // in the second half
    int maxSecond = INT_MIN;
 
    // Calculate the maximum element
    // int the second half
    for (int i = mid; i < n; i++)
        maxSecond = max(maxSecond, arr[i]);
 
    // Print the found maximums
    cout << maxFirst << ", " << maxSecond;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 12, 14, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findMax(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
    static void findMax(int []arr, int n) 
    { 
     
        // To store the maximum element 
        // in the first half 
        int maxFirst = Integer.MIN_VALUE; 
     
        // Middle index of the array 
        int mid = n / 2; 
     
        // Calculate the maximum element 
        // in the first half 
        for (int i = 0; i < mid; i++) 
        {
            maxFirst = Math.max(maxFirst, arr[i]); 
        }
     
        // If the size of array is odd then 
        // the middle element will be included 
        // in both the halves 
        if (n % 2 == 1) 
        {
            maxFirst = Math.max(maxFirst, arr[mid]); 
        }
         
        // To store the maximum element 
        // in the second half 
        int maxSecond = Integer.MIN_VALUE; 
     
        // Calculate the maximum element 
        // int the second half 
        for (int i = mid; i < n; i++) 
        {
            maxSecond = Math.max(maxSecond, arr[i]); 
        }
         
        // Print the found maximums 
        System.out.print(maxFirst + ", " + maxSecond);
        // cout << maxFirst << ", " << maxSecond; 
    } 
     
    // Driver Code
    public static void main(String[] args)
    {
        int []arr = { 1, 12, 14, 5 }; 
        int n = arr.length; 
     
        findMax(arr, n); 
    } 
}
 
// This code is contributed by anuj_67..

Python3

# Python3 implementation of the approach
import sys
 
# Function to print largest element in
# first half and second half of an array
def findMax(arr, n) :
 
    # To store the maximum element
    # in the first half
    maxFirst = -sys.maxsize - 1
 
    # Middle index of the array
    mid = n // 2;
 
    # Calculate the maximum element
    # in the first half
    for i in range(0, mid):
        maxFirst = max(maxFirst, arr[i])
 
    # If the size of array is odd then
    # the middle element will be included
    # in both the halves
    if (n % 2 == 1):
        maxFirst = max(maxFirst, arr[mid])
 
    # To store the maximum element
    # in the second half
    maxSecond = -sys.maxsize - 1
 
    # Calculate the maximum element
    # int the second half
    for i in range(mid, n):
        maxSecond = max(maxSecond, arr[i])
 
    # Print the found maximums
    print(maxFirst, ",", maxSecond)
 
# Driver code
arr = [1, 12, 14, 5 ]
n = len(arr)
 
findMax(arr, n)
 
# This code is contributed by ihritik

C#

// C# implementation of the approach
using System;
 
class GFG
{
    static void findMax(int []arr, int n) 
    { 
     
        // To store the maximum element 
        // in the first half 
        int maxFirst = int.MinValue; 
     
        // Middle index of the array 
        int mid = n / 2; 
     
        // Calculate the maximum element 
        // in the first half 
        for (int i = 0; i < mid; i++) 
        {
            maxFirst = Math.Max(maxFirst, arr[i]); 
        }
     
        // If the size of array is odd then 
        // the middle element will be included 
        // in both the halves 
        if (n % 2 == 1) 
        {
            maxFirst = Math.Max(maxFirst, arr[mid]); 
        }
         
        // To store the maximum element 
        // in the second half 
        int maxSecond = int.MinValue; 
     
        // Calculate the maximum element 
        // int the second half 
        for (int i = mid; i < n; i++) 
        {
            maxSecond = Math.Max(maxSecond, arr[i]); 
        }
         
        // Print the found maximums 
        Console.WriteLine(maxFirst + ", " + maxSecond);
        // cout << maxFirst << ", " << maxSecond; 
    } 
     
    // Driver Code
    public static void Main()
    {
        int []arr = { 1, 12, 14, 5 }; 
        int n = arr.Length; 
     
        findMax(arr, n); 
    } 
}
 
// This code is contributed by nidhiva

Javascript

// javascript implementation of the approach
    function findMax(arr, n)
    {
      
        // To store the maximum element
        // in the first half
         
        var maxFirst = Number.MIN_VALUE
      
        // Middle index of the array
        var mid = n / 2;
      
        // Calculate the maximum element
        // in the first half
        for (var i = 0; i < mid; i++)
        {
            maxFirst = Math.max(maxFirst, arr[i]);
        }
      
        // If the size of array is odd then
        // the middle element will be included
        // in both the halves
        if (n % 2 == 1)
        {
            maxFirst = Math.max(maxFirst, arr[mid]);
        }
          
        // To store the maximum element
        // in the second half
        var maxSecond = Number.MIN_VALUE
      
        // Calculate the maximum element
        // int the second half
        for (var i = mid; i < n; i++)
        {
            maxSecond = Math.max(maxSecond, arr[i]);
        }
          
        // Print the found maximums
        document.write(maxFirst + ", " + maxSecond);
    }
      
    // Driver Code
        var arr = [ 1, 12, 14, 5 ];
        var n = arr.length;
      
        findMax(arr, n);
 
 // This code is contributed by bunnyram19.

Output:

12, 14

Time Complexity: O(n), since the loop runs from 0 to (mid – 1), and then from mid to (n – 1).

Auxiliary Space: O(1), since no extra space has been taken.

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