在数组的前半部分和后半部分中找到最大元素

给定N个整数的数组arr [] 。任务是在数组的前一半和后一半中找到最大的元素。请注意，如果数组的大小为奇数，则中间元素将同时包含在这两个部分中。
例子：

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 

方法：计算数组的中间索引为mid = N / 2 。现在，如果N为偶数，则前半部分元素将出现在子数组arr [0 … mid-1]和arr [mid … N-1]中。
如果N为奇数，则两半分别为arr [0…mid]和arr [mid … N-1]
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

输出：

12, 14

时间复杂度： O(n)

辅助空间： O(1)