, 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 arrayvoid 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 codeint main(){ int arr[] = { 1, 12, 14, 5 }; int n = sizeof(arr) / sizeof(arr[0]); findMax(arr, n); return 0;} |
Java
// Java implementation of the approachimport 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 approachimport sys# Function to print largest element in# first half and second half of an arraydef 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 codearr = [1, 12, 14, 5 ]n = len(arr)findMax(arr, n)# This code is contributed by ihritik |
C#
// C# implementation of the approachusing 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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
