Given an array of n integers, find the third largest element. All the elements in the array are distinct integers.
Example :
Input: arr[] = {1, 14, 2, 16, 10, 20}
Output: The third Largest element is 14
Explanation: Largest element is 20, second largest element is 16
and third largest element is 14
Input: arr[] = {19, -10, 20, 14, 2, 16, 10}
Output: The third Largest element is 16
Explanation: Largest element is 20, second largest element is 19
and third largest element is 16
Naive Approach: The task is to first find the largest element, followed by the second-largest element and then excluding them both find the third-largest element. The basic idea is to iterate the array twice and mark the maximum and second maximum element and then excluding them both find the third maximum element, i.e the maximum element excluding the maximum and second maximum.
Algorithm:
- First, iterate through the array and find maximum.
- Store this as first maximum along with its index.
- Now traverse the whole array finding the second max, excluding the maximum element.
- Finally traverse the array the third time and find the third largest element i.e., excluding the maximum and second maximum.
C
#include <limits.h>
#include <stdio.h>
void thirdLargest( int arr[], int arr_size)
{
if (arr_size < 3) {
printf ( " Invalid Input " );
return ;
}
int first = arr[0];
for ( int i = 1; i < arr_size; i++)
if (arr[i] > first)
first = arr[i];
int second = INT_MIN;
for ( int i = 0; i < arr_size; i++)
if (arr[i] > second && arr[i] < first)
second = arr[i];
int third = INT_MIN;
for ( int i = 0; i < arr_size; i++)
if (arr[i] > third && arr[i] < second)
third = arr[i];
printf ( "The third Largest element is %d\n" , third);
}
int main()
{
int arr[] = { 12, 13, 1, 10, 34, 16 };
int n = sizeof (arr) / sizeof (arr[0]);
thirdLargest(arr, n);
return 0;
}
|
C++
#include <bits/stdc++.h>
void thirdLargest( int arr[], int arr_size)
{
if (arr_size < 3) {
printf ( " Invalid Input " );
return ;
}
int first = arr[0];
for ( int i = 1; i < arr_size; i++)
if (arr[i] > first)
first = arr[i];
int second = INT_MIN;
for ( int i = 0; i < arr_size; i++)
if (arr[i] > second && arr[i] < first)
second = arr[i];
int third = INT_MIN;
for ( int i = 0; i < arr_size; i++)
if (arr[i] > third && arr[i] < second)
third = arr[i];
printf ( "The third Largest element is %d\n" , third);
}
int main()
{
int arr[] = { 12, 13, 1, 10, 34, 16 };
int n = sizeof (arr) / sizeof (arr[0]);
thirdLargest(arr, n);
return 0;
}
|
Java
class GFG
{
static void thirdLargest( int arr[],
int arr_size)
{
if (arr_size < 3 )
{
System.out.printf( " Invalid Input " );
return ;
}
int first = arr[ 0 ];
for ( int i = 1 ;
i < arr_size ; i++)
if (arr[i] > first)
first = arr[i];
int second = Integer.MIN_VALUE;
for ( int i = 0 ;
i < arr_size ; i++)
if (arr[i] > second &&
arr[i] < first)
second = arr[i];
int third = Integer.MIN_VALUE;
for ( int i = 0 ;
i < arr_size ; i++)
if (arr[i] > third &&
arr[i] < second)
third = arr[i];
System.out.printf( "The third Largest " +
"element is %d\n" , third);
}
public static void main(String []args)
{
int arr[] = { 12 , 13 , 1 ,
10 , 34 , 16 };
int n = arr.length;
thirdLargest(arr, n);
}
}
|
Python3
import sys
def thirdLargest(arr, arr_size):
if (arr_size < 3 ):
print ( " Invalid Input " )
return
first = arr[ 0 ]
for i in range ( 1 , arr_size):
if (arr[i] > first):
first = arr[i]
second = - sys.maxsize
for i in range ( 0 , arr_size):
if (arr[i] > second and
arr[i] < first):
second = arr[i]
third = - sys.maxsize
for i in range ( 0 , arr_size):
if (arr[i] > third and
arr[i] < second):
third = arr[i]
print ( "The Third Largest" ,
"element is" , third)
arr = [ 12 , 13 , 1 ,
10 , 34 , 16 ]
n = len (arr)
thirdLargest(arr, n)
|
Javascript
<script>
function thirdLargest(arr, arr_size)
{
if (arr_size < 3)
{
document.write( " Invalid Input " );
return ;
}
let first = arr[0];
for (let i = 1;
i < arr_size ; i++)
if (arr[i] > first)
first = arr[i];
let second = Number.MIN_VALUE;
for (let i = 0;
i < arr_size ; i++)
if (arr[i] > second &&
arr[i] < first)
second = arr[i];
let third = Number.MIN_VALUE;
for (let i = 0;
i < arr_size ; i++)
if (arr[i] > third &&
arr[i] < second)
third = arr[i];
document.write( "The third Largest " +
"element is " , third);
}
let arr = [12, 13, 1,
10, 34, 16];
let n = arr.length;
thirdLargest(arr, n);
</script>
|
C#
using System;
class GFG
{
static void thirdLargest( int []arr,
int arr_size)
{
if (arr_size < 3)
{
Console.Write( " Invalid Input " );
return ;
}
int first = arr[0];
for ( int i = 1;
i < arr_size ; i++)
if (arr[i] > first)
first = arr[i];
int second = - int .MaxValue;
for ( int i = 0;
i < arr_size ; i++)
if (arr[i] > second &&
arr[i] < first)
second = arr[i];
int third = - int .MaxValue;
for ( int i = 0;
i < arr_size ; i++)
if (arr[i] > third &&
arr[i] < second)
third = arr[i];
Console.Write( "The third Largest " +
"element is " + third);
}
public static void Main()
{
int []arr = {12, 13, 1,
10, 34, 16};
int n = arr.Length;
thirdLargest(arr, n);
}
}
|
PHP
<?php
function thirdLargest( $arr , $arr_size )
{
if ( $arr_size < 3)
{
echo " Invalid Input " ;
return ;
}
$first = $arr [0];
for ( $i = 1; $i < $arr_size ; $i ++)
if ( $arr [ $i ] > $first )
$first = $arr [ $i ];
$second = PHP_INT_MIN;
for ( $i = 0; $i < $arr_size ; $i ++)
if ( $arr [ $i ] > $second &&
$arr [ $i ] < $first )
$second = $arr [ $i ];
$third = PHP_INT_MIN;
for ( $i = 0; $i < $arr_size ; $i ++)
if ( $arr [ $i ] > $third &&
$arr [ $i ] < $second )
$third = $arr [ $i ];
echo "The third Largest element is " ,
$third , "\n" ;
}
$arr = array (12, 13, 1,
10, 34, 16);
$n = sizeof( $arr );
thirdLargest( $arr , $n );
?>
|
Output
The third Largest element is 13
- Complexity Analysis:
- Time Complexity: O(n).
As the array is iterated thrice and is done in a constant time
- Space complexity: O(1).
No extra space is needed as the indices can be stored in constant space.
Efficient Approach: The problem deals with finding the third largest element in the array in a single traversal. The problem can be cracked by taking help of a similar problem- finding the second maximum element. So the idea is to traverse the array from start to end and to keep track of the three largest elements up to that index (stored in variables). So after traversing the whole array, the variables would have stored the indices (or value) of the three largest elements of the array.
Algorithm:
- Create three variables, first, second, third, to store indices of three largest elements of the array. (Initially all of them are initialized to a minimum value).
- Move along the input array from start to the end.
- For every index check if the element is larger than first or not. Update the value of first, if the element is larger, and assign the value of first to second and second to third. So the largest element gets updated and the elements previously stored as largest becomes second largest, and the second largest element becomes third largest.
- Else if the element is larger than the second, then update the value of second,and the second largest element becomes third largest.
- If the previous two conditions fail, but the element is larger than the third, then update the third.
- Print the value of third after traversing the array from start to end
C++
#include <bits/stdc++.h>
void thirdLargest( int arr[], int arr_size)
{
if (arr_size < 3)
{
printf ( " Invalid Input " );
return ;
}
int first = arr[0], second = INT_MIN, third = INT_MIN;
for ( int i = 1; i < arr_size ; i ++)
{
if (arr[i] > first)
{
third = second;
second = first;
first = arr[i];
}
else if (arr[i] > second)
{
third = second;
second = arr[i];
}
else if (arr[i] > third)
third = arr[i];
}
printf ( "The third Largest element is %d\n" , third);
}
int main()
{
int arr[] = {12, 13, 1, 10, 34, 16};
int n = sizeof (arr)/ sizeof (arr[0]);
thirdLargest(arr, n);
return 0;
}
|
Java
class GFG {
static void thirdLargest( int arr[], int arr_size) {
if (arr_size < 3 ) {
System.out.printf( " Invalid Input " );
return ;
}
int first = arr[ 0 ], second = Integer.MIN_VALUE,
third = Integer.MIN_VALUE;
for ( int i = 1 ; i < arr_size; i++) {
if (arr[i] > first) {
third = second;
second = first;
first = arr[i];
}
else if (arr[i] > second) {
third = second;
second = arr[i];
}
else if (arr[i] > third) {
third = arr[i];
}
}
System.out.printf( "The third Largest element is %d\n" , third);
}
public static void main(String []args) {
int arr[] = { 12 , 13 , 1 , 10 , 34 , 16 };
int n = arr.length;
thirdLargest(arr, n);
}
}
|
Python3
import sys
def thirdLargest(arr, arr_size):
if (arr_size < 3 ):
print ( " Invalid Input " )
return
first = arr[ 0 ]
second = - sys.maxsize
third = - sys.maxsize
for i in range ( 1 , arr_size):
if (arr[i] > first):
third = second
second = first
first = arr[i]
elif (arr[i] > second):
third = second
second = arr[i]
elif (arr[i] > third):
third = arr[i]
print ( "The third Largest" ,
"element is" , third)
arr = [ 12 , 13 , 1 ,
10 , 34 , 16 ]
n = len (arr)
thirdLargest(arr, n)
|
Javascript
<script>
function thirdLargest(arr, arr_size)
{
if (arr_size < 3)
{
document.write( " Invalid Input " );
return ;
}
var first = arr[0], second = -1000000000, third = -1000000000;
for ( var i = 1; i < arr_size ; i ++)
{
if (arr[i] > first)
{
third = second;
second = first;
first = arr[i];
}
else if (arr[i] > second)
{
third = second;
second = arr[i];
}
else if (arr[i] > third)
third = arr[i];
}
document.write( "The third Largest element is " + third);
}
var arr = [12, 13, 1, 10, 34, 16];
var n = arr.length;
thirdLargest(arr, n);
</script>
|
C#
using System;
class GFG {
static void thirdLargest( int [] arr, int arr_size)
{
if (arr_size < 3)
{
Console.Write( " Invalid Input " );
return ;
}
int first = arr[0], second = int .MinValue,
third = int .MinValue;
for ( int i = 1; i < arr_size; i++)
{
if (arr[i] > first) {
third = second;
second = first;
first = arr[i];
}
else if (arr[i] > second) {
third = second;
second = arr[i];
}
else if (arr[i] > third) {
third = arr[i];
}
}
Console.Write( "The third Largest element is " + third);
}
public static void Main() {
int [] arr = {12, 13, 1, 10, 34, 16};
int n = arr.Length;
thirdLargest(arr, n);
}
}
|
PHP
<?php
function thirdLargest( $arr , $arr_size )
{
if ( $arr_size < 3)
{
echo " Invalid Input " ;
return ;
}
$first = $arr [0];
$second = PHP_INT_MIN;
$third = PHP_INT_MIN;
for ( $i = 1; $i < $arr_size ; $i ++)
{
if ( $arr [ $i ] > $first )
{
$third = $second ;
$second = $first ;
$first = $arr [ $i ];
}
else if ( $arr [ $i ] > $second )
{
$third = $second ;
$second = $arr [ $i ];
}
else if ( $arr [ $i ] > $third )
$third = $arr [ $i ];
}
echo "The third Largest element is " ,
$third ;
}
$arr = array (12, 13, 1,
10, 34, 16);
$n = sizeof( $arr );
thirdLargest( $arr , $n );
?>
|
Output
The third Largest element is 13
- Complexity Analysis:
- Time Complexity: O(n).
As the array is iterated once and is done in a constant time
- Space complexity: O(1).
No extra space is needed as the indices can be stored in constant space.
Another method: Using Slicing
Here, we are using slicing method, to extract 3rd largest number.
First we will sort the array then we will extract the 3rd largest element using slicing.
C++
#include <bits/stdc++.h>
using namespace std;
int thirdLargest( int arr[], int arr_size)
{
if (arr_size < 3)
{
cout<< " Invalid Input" <<endl;
return 0;
}
else
{
sort(arr,arr+arr_size);
return arr[arr_size-3];
}
}
int main()
{
int arr[] = {12, 13, 1, 10, 34, 16};
int n = sizeof (arr)/ sizeof (arr[0]);
cout<<thirdLargest(arr, n)<<endl;
return 0;
}
|
Java
import java.util.Arrays;
class Main {
public static int thirdLargest( int [] arr)
{
int arr_size = arr.length;
if (arr_size < 3 ) {
System.out.println( "Invalid Input" );
return 0 ;
}
else {
Arrays.sort(arr);
return arr[arr_size - 3 ];
}
}
public static void main(String[] args)
{
int [] arr = { 12 , 13 , 1 , 10 , 34 , 16 };
System.out.println(thirdLargest(arr));
}
}
|
Python3
def thirdLargest(arr, arr_size):
if (arr_size < 3 ):
print ( " Invalid Input " )
return
else :
arr = sorted (arr)
return arr[ - 3 ]
arr = [ 12 , 13 , 1 , 10 , 34 , 16 ]
arr_size = len (arr)
print (thirdLargest(arr, arr_size))
|
C#
using System;
class MainClass {
public static int thirdLargest( int [] arr)
{
int arr_size = arr.Length;
if (arr_size < 3) {
Console.WriteLine( "Invalid Input" );
return 0;
}
else {
Array.Sort(arr);
return arr[arr_size - 3];
}
}
public static void Main( string [] args)
{
int [] arr = { 12, 13, 1, 10, 34, 16 };
Console.WriteLine(thirdLargest(arr));
}
}
|
Javascript
function thirdLargest(arr) {
let arr_size = arr.length;
if (arr_size < 3) {
console.log( "Invalid Input" );
return 0;
}
else
{
arr.sort( function (a, b) {
return a - b;
});
return arr[arr_size - 3];
}
}
let arr = [12, 13, 1, 10, 34, 16];
console.log(thirdLargest(arr));
|
Time complexity: O(nlogn). where n is the length of the array.
Auxiliary space: O(1)
?list=PLqM7alHXFySEQDk2MDfbwEdjd2svVJH9p
Exercise Problem: Extend the above solution to find the third largest when array may have duplicates. For example, if the input array is {10, 5, 15, 5, 15, 10, 1, 1}, then output should be 5. The extended solution should also work in one traversal.
Related Articles:
- Find the smallest and second smallest element in an array
- k largest(or smallest) elements in an array
- K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst Case Linear Time)
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