Given an array of n integers, the task is to find the third largest element.
Examples :
Input: arr[] = [2, 4, 1, 3, 5]
Output: 3
Explanation: The third largest element in the array [2, 4, 1, 3, 5] is 3.
Input: arr[] = [10, 2]
Output: -1
Explanation: There are less than three elements in the array, so the third largest element cannot be determined.Input: arr[] = [5, 5, 5]
Output: 5
Explanation: In the array [5, 5, 5], the third largest element can be considered 5, as there are no other distinct elements.
Try It Yourself
Table of Content
[Naive Approach] Using Sorting - O(n * log n) Time and O(1) Space
The idea is to sort the array and return the third largest element in the array which will be present at (n-3)'th index.
// C++ program to find the third largest
// element in an array.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int thirdLargest(vector<int> &arr) {
int n = arr.size();
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Sort the array
sort(arr.begin(), arr.end());
// Return the third largest element
return arr[n-3];
}
int main() {
vector<int> arr = {2, 4, 1, 3, 5};
cout << thirdLargest(arr) << endl;
return 0;
}
// Java program to find the third largest
// element in an array.
import java.util.Arrays;
class GfG {
static int thirdLargest(int[] arr) {
int n = arr.length;
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Sort the array
Arrays.sort(arr);
// Return the third largest element
return arr[n - 3];
}
public static void main(String[] args) {
int[] arr = {2, 4, 1, 3, 5};
System.out.println(thirdLargest(arr));
}
}
def thirdLargest(arr):
n = len(arr)
if n < 3:
return -1
# Sort the array
arr.sort()
# Return the third largest element
return arr[n - 3]
if __name__ == "__main__":
arr = [2, 4, 1, 3, 5]
print(thirdLargest(arr))
// C# program to find the third largest
// element in an array.
using System;
class GfG {
static int thirdLargest(int[] arr) {
int n = arr.Length;
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Sort the array
Array.Sort(arr);
// Return the third largest element
return arr[n - 3];
}
static void Main() {
int[] arr = {2, 4, 1, 3, 5};
Console.WriteLine(thirdLargest(arr));
}
}
// JavaScript program to find the third largest
// element in an array.
// Function to find the third largest element
function thirdLargest(arr) {
let n = arr.length;
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Sort the array
arr.sort((a, b) => a - b);
// Return the third largest element
return arr[n - 3];
}
let arr = [2, 4, 1, 3, 5];
console.log(thirdLargest(arr));
Output
3
[Expected Approach - 1] Using Three Loops - O(n) Time and O(1) Space
The 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.
Step by step approach:
- 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.
#include <iostream>
#include <vector>
using namespace std;
int thirdLargest(vector<int> &arr) {
int n = arr.size();
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Pass 1: Find the first maximum element and remember its index
int first = -1;
int first_idx = -1;
for (int i = 0; i < n; i++) {
if (arr[i] > first) {
first = arr[i];
first_idx = i;
}
}
// Pass 2: Find the second max element by skipping the EXACT index of the first
int second = -1;
int second_idx = -1;
for (int i = 0; i < n; i++) {
if (i == first_idx) continue;
if (arr[i] > second) {
second = arr[i];
second_idx = i;
}
}
// Pass 3: Find the third largest element by skipping the indices of first and second
int third = -1;
for (int i = 0; i < n; i++) {
if (i == first_idx || i == second_idx) continue;
if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
int main() {
vector<int> arr = {2, 4, 1, 3, 5};
cout << thirdLargest(arr) << endl;
return 0;
}
class GfG {
static int thirdLargest(int[] arr) {
int n = arr.length;
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Pass 1: Find the first maximum element and remember its index
int first = -1;
int first_idx = -1;
for (int i = 0; i < n; i++) {
if (arr[i] > first) {
first = arr[i];
first_idx = i;
}
}
// Pass 2: Find the second max element by skipping the EXACT index of the first
int second = -1;
int second_idx = -1;
for (int i = 0; i < n; i++) {
if (i == first_idx) continue;
if (arr[i] > second) {
second = arr[i];
second_idx = i;
}
}
// Pass 3: Find the third largest element by skipping the indices of first and second
int third = -1;
for (int i = 0; i < n; i++) {
if (i == first_idx || i == second_idx) continue;
if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
public static void main(String[] args) {
int[] arr = {2, 4, 1, 3, 5};
System.out.println(thirdLargest(arr));
}
}
def thirdLargest(arr):
n = len(arr)
if n < 3:
return -1
# Find the first maximum element.
first = -1
first_idx = float('-inf')
for i in range(n):
if arr[i] > first:
first = arr[i]
first_idx = i
# Find the second max element.
second = -1
second_idx = float('-inf')
for i in range(n):
if i == first_idx:
continue
if arr[i] > second:
second = arr[i]
second_idx = i
# Find the third largest element.
third = -1
third = float('-inf')
for i in range(n):
if i == first_idx or i == second_idx:
continue
if arr[i] > third:
third = arr[i]
# Return the third largest element
return third
if __name__ == "__main__":
arr = [2, 4, 1, 3, 5]
print(thirdLargest(arr))
using System;
class GfG {
static int thirdLargest(int[] arr) {
int n = arr.Length;
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Pass 1: Find the first maximum element and remember its index
int first = -1;
int first_idx = -1;
for (int i = 0; i < n; i++) {
if (arr[i] > first) {
first = arr[i];
first_idx = i;
}
}
// Pass 2: Find the second max element by skipping the EXACT index of the first
int second = -1;
int second_idx = -1;
for (int i = 0; i < n; i++) {
if (i == first_idx) continue;
if (arr[i] > second) {
second = arr[i];
second_idx = i;
}
}
// Pass 3: Find the third largest element by skipping the indices of first and second
int third = -1;
for (int i = 0; i < n; i++) {
if (i == first_idx || i == second_idx) continue;
if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
static void Main() {
int[] arr = {2, 4, 1, 3, 5};
Console.WriteLine(thirdLargest(arr));
}
}
function thirdLargest(arr) {
let n = arr.length;
// If the array has less than 3 elements, return -1
if (n < 3) {
return -1;
}
// Pass 1: Find the first maximum element and remember its index
let first = -1;
let first_idx = -1;
for (let i = 0; i < n; i++) {
if (arr[i] > first) {
first = arr[i];
first_idx = i;
}
}
// Pass 2: Find the second max element by skipping the EXACT index of the first
let second = -1;
let second_idx = -1;
for (let i = 0; i < n; i++) {
if (i === first_idx) continue;
if (arr[i] > second) {
second = arr[i];
second_idx = i;
}
}
// Pass 3: Find the third largest element by skipping the indices of first and second
let third = -1;
for (let i = 0; i < n; i++) {
if (i === first_idx || i === second_idx) continue;
if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
// Driver code
let arr = [2, 4, 1, 3, 5];
console.log(thirdLargest(arr));
Output
3
[Expected Approach - 2] Using Single Loop - O(n) Time and O(1) Space
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.
Step by step approach:
- 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.
#include <climits>
#include <iostream>
#include <vector>
using namespace std;
int thirdLargest(vector<int> &arr)
{
int n = arr.size();
int first = INT_MIN, second = INT_MIN, third = INT_MIN;
// If length of array is less than 3, return -1
if (n < 3)
{
return -1;
}
for (int i = 0; i < n; i++)
{
// If arr[i] is greater than first,
// set third to second, second to
// first and first to arr[i].
if (arr[i] > first)
{
third = second;
second = first;
first = arr[i];
}
// If arr[i] is greater than second,
// set third to second and second
// to arr[i].
else if (arr[i] > second)
{
third = second;
second = arr[i];
}
// If arr[i] is greater than third,
// set third to arr[i].
else if (arr[i] > third)
{
third = arr[i];
}
}
// Return the third largest element
return third;
}
int main()
{
vector<int> arr = {2, 4, 1, 3, 5};
cout << thirdLargest(arr) << endl;
return 0;
}
class GfG {
static int thirdLargest(int[] arr) {
int n = arr.length;
int first = Integer.MIN_VALUE, second = Integer.MIN_VALUE,
third = Integer.MIN_VALUE;
// If length of array is less than 3, return -1
if( n < 3 ) {
return -1;
}
for (int i = 0; i < n; i++) {
// If arr[i] is greater than first,
// set third to second, second to
// first and first to arr[i].
if (arr[i] > first) {
third = second;
second = first;
first = arr[i];
}
// If arr[i] is greater than second,
// set third to second and second
// to arr[i].
else if (arr[i] > second) {
third = second;
second = arr[i];
}
// If arr[i] is greater than third,
// set third to arr[i].
else if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
public static void main(String[] args) {
int[] arr = {2, 4, 1, 3, 5};
System.out.println(thirdLargest(arr));
}
}
def thirdLargest(arr):
n = len(arr)
first, second, third = float('-inf'), float('-inf'), float('-inf')
if n < 3:
return -1
for i in range(n):
# If arr[i] is greater than first,
# set third to second, second to
# first and first to arr[i].
if arr[i] > first:
third = second
second = first
first = arr[i]
# If arr[i] is greater than second,
# set third to second and second
# to arr[i].
elif arr[i] > second:
third = second
second = arr[i]
# If arr[i] is greater than third,
# set third to arr[i].
elif arr[i] > third:
third = arr[i]
# Return the third largest element
return third
if __name__ == "__main__":
arr = [2, 4, 1, 3, 5]
print(thirdLargest(arr))
using System;
class GfG {
static int thirdLargest(int[] arr) {
int n = arr.Length;
int first = int.MinValue, second = int.MinValue,
third = int.MinValue;
// If length of array is less than 3, return -1
if (n < 3) {
return -1;
}
for (int i = 0; i < n; i++) {
// If arr[i] is greater than first,
// set third to second, second to
// first and first to arr[i].
if (arr[i] > first) {
third = second;
second = first;
first = arr[i];
}
// If arr[i] is greater than second,
// set third to second and second
// to arr[i].
else if (arr[i] > second) {
third = second;
second = arr[i];
}
// If arr[i] is greater than third,
// set third to arr[i].
else if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
static void Main() {
int[] arr = {2, 4, 1, 3, 5};
Console.WriteLine(thirdLargest(arr));
}
}
function thirdLargest(arr) {
let n = arr.length;
let first = -Infinity, second = -Infinity,
third = -Infinity;
// If length of array is less than 3, return -1
if (n < 3) {
return -1;
}
for (let i = 0; i < n; i++) {
// If arr[i] is greater than first,
// set third to second, second to
// first and first to arr[i].
if (arr[i] > first) {
third = second;
second = first;
first = arr[i];
}
// If arr[i] is greater than second,
// set third to second and second
// to arr[i].
else if (arr[i] > second) {
third = second;
second = arr[i];
}
// If arr[i] is greater than third,
// set third to arr[i].
else if (arr[i] > third) {
third = arr[i];
}
}
// Return the third largest element
return third;
}
let arr = [2, 4, 1, 3, 5];
console.log(thirdLargest(arr));
Output
3
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)