, 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++ 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 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 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# 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 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.
|
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.