Given an array arr[] of size N, the task is to find the largest element in the given array.
Examples:
Input: arr[] = {10, 20, 4}
Output: 20
Explanation: Among 10, 20 and 4, 20 is the largest.
Input : arr[] = {20, 10, 20, 4, 100}
Output : 100
Iterative Approach to find the largest element of Array:
The simplest approach is to solve this problem is to traverse the whole list and find the maximum among them.
- Create a local variable max and initiate it to arr[0] to store the maximum among the list
- Iterate over the array
- Compare arr[i] with max.
- If arr[i] > max, update max = arr[i].
- Increment i once.
- After the iteration is over, return max as the required answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int largest(int arr[], int n)
{
int i;
int max = arr[0];
for (i = 1; i < n; i++)
if (arr[i] > max)
max = arr[i];
return max;
}
int main()
{
int arr[] = { 10, 324, 45, 90, 9808 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Largest in given array is " << largest(arr, n);
return 0;
}
|
C
#include <stdio.h>
int largest(int arr[], int n)
{
int i;
int max = arr[0];
for (i = 1; i < n; i++)
if (arr[i] > max)
max = arr[i];
return max;
}
int main()
{
int arr[] = { 10, 324, 45, 90, 9808 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Largest in given array is %d", largest(arr, n));
return 0;
}
|
Java
import java.io.*;
class Test {
static int arr[] = { 10, 324, 45, 90, 9808 };
static int largest()
{
int i;
int max = arr[0];
for (i = 1; i < arr.length; i++)
if (arr[i] > max)
max = arr[i];
return max;
}
public static void main(String[] args)
{
System.out.println("Largest in given array is "
+ largest());
}
}
|
Python3
def largest(arr, n):
mx = arr[0]
for i in range(1, n):
if arr[i] > mx:
mx = arr[i]
return mx
if __name__ == '__main__':
arr = [10, 324, 45, 90, 9808]
n = len(arr)
Ans = largest(arr, n)
print("Largest in given array is", Ans)
|
C#
using System;
class GFG {
static int[] arr = { 10, 324, 45, 90, 9808 };
static int largest()
{
int i;
int max = arr[0];
for (i = 1; i < arr.Length; i++)
if (arr[i] > max)
max = arr[i];
return max;
}
public static void Main()
{
Console.WriteLine("Largest in given "
+ "array is " + largest());
}
}
|
Javascript
<script>
function largest(arr) {
let i;
let max = arr[0];
for (i = 1; i < arr.length; i++) {
if (arr[i] > max)
max = arr[i];
}
return max;
}
let arr = [10, 324, 45, 90, 9808];
document.write("Largest in given array is " + largest(arr));
</script>
|
PHP
<?php
function largest($arr, $n)
{
$i;
$max = $arr[0];
for ($i = 1; $i < $n; $i++)
if ($arr[$i] > $max)
$max = $arr[$i];
return $max;
}
$arr= array(10, 324, 45, 90, 9808);
$n = sizeof($arr);
echo "Largest in given array is "
, largest($arr, $n);
?>
|
OutputLargest in given array is 9808
Time complexity: O(N), to traverse the Array completely.
Auxiliary Space: O(1), as only an extra variable is created, which will take O(1) space.
Recursive Approach to find the maximum of Array:
The idea is similar to the iterative approach. Here the traversal of the array is done recursively instead of an iterative loop.
- Set an integer i = 0 to denote the current index being searched.
- Check if i is the last index, return arr[i].
- Increment i and call the recursive function for the new value of i.
- Compare the maximum value returned from the recursion function with arr[i].
- Return the max between these two from the current recursion call.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int largest(int arr[], int n, int i)
{
if (i == n - 1) {
return arr[i];
}
int recMax = largest(arr, n, i + 1);
return max(recMax, arr[i]);
}
int main()
{
int arr[] = { 10, 324, 45, 90, 9808 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Largest in given array is "
<< largest(arr, n, 0);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int largest(int arr[], int n, int i)
{
if (i == n - 1) {
return arr[i];
}
int recMax = largest(arr, n, i + 1);
return Math.max(recMax, arr[i]);
}
public static void main(String[] args)
{
int arr[] = { 10, 324, 45, 90, 9808 };
int n = arr.length;
System.out.println("Largest in given array is "
+ largest(arr, n, 0));
}
}
|
Python3
import math
def largest(arr, n, i):
if (i == n - 1):
return arr[i]
recMax = largest(arr, n, i + 1)
return max(recMax, arr[i])
if __name__ == '__main__':
arr = [10, 324, 45, 90, 9808]
n = len(arr)
print("Largest in given array is " + str(largest(arr, n, 0)))
|
C#
using System;
public class GFG {
public static int largest(int[] arr, int n, int i)
{
if (i == n - 1) {
return arr[i];
}
var recMax = GFG.largest(arr, n, i + 1);
return Math.Max(recMax, arr[i]);
}
public static void Main(String[] args)
{
int[] arr = { 10, 324, 45, 90, 9808 };
var n = arr.Length;
Console.WriteLine(
"Largest in given array is "
+ GFG.largest(arr, n, 0).ToString());
}
}
|
Javascript
function largest(arr, n, i)
{
if (i == n - 1) {
return arr[i];
}
let recMax = largest(arr, n, i + 1);
return Math.max(recMax, arr[i]);
}
let arr = [ 10, 324, 45, 90, 9808 ];
let n = arr.length;
console.log("Largest in given array is", largest(arr, n, 0));
|
PHP
<?php
function largest($arr, $n, $i)
{
if ($i == $n - 1) {
return $arr[$i];
}
$recMax = largest($arr, $n, $i + 1);
return max($recMax, $arr[$i]);
}
$arr = [ 10, 324, 45, 90, 9808 ];
$n = count($arr);
echo "Largest in given array is " , largest($arr, $n, 0) ;
?>
|
OutputLargest in given array is 9808
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), for recursive calls
Find the maximum of Array using Library Function:
Most of the languages have a relevant max() type in-built function to find the maximum element, such as std::max_element in C++. We can use this function to directly find the maximum element.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int largest(int arr[], int n)
{
return *max_element(arr, arr + n);
}
int main()
{
int arr[] = { 10, 324, 45, 90, 9808 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << largest(arr, n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static int largest(int[] arr, int n)
{
Arrays.sort(arr);
return arr[n - 1];
}
static public void main(String[] args)
{
int[] arr = { 10, 324, 45, 90, 9808 };
int n = arr.length;
System.out.println(largest(arr, n));
}
}
|
Python3
def largest(arr, n):
return max(arr)
if __name__ == '__main__':
arr = [10, 324, 45, 90, 9808]
n = len(arr)
print(largest(arr, n))
|
C#
using System;
using System.Linq;
public class GFG {
static int Largest(int[] arr) {
return arr.Max();
}
static public void Main()
{
int[] arr = { 10, 324, 45, 90, 9808 };
Console.WriteLine(Largest(arr));
}
}
|
Javascript
<script>
function largest(arr, n)
{
arr.sort();
return arr[n-1];
}
let arr = [10, 324, 45, 90, 9808];
let n = arr.length;
document.write(largest(arr, n));
</script>
|
PHP
<?php
function largest( $arr, $n)
{
return max($arr);
}
$arr = array(10, 324, 45, 90, 9808);
$n = count($arr);
echo largest($arr, $n);
?>
|
Time complexity: O(N), since the in-built max_element() function takes O(N) time.
Auxiliary Space: O(1), as only an extra variable is created, which will take O(1) space.
Refer below article for more methods: Program to find the minimum (or maximum) element of an array