Second Largest Element in an Array
Last Updated :
10 Feb, 2025
Improve
Try it on GfG Practice 
Given an array of positive integers arr[] of size n, the task is to find second largest distinct element in the array.
Note: If the second largest element does not exist, return -1.
Examples:
Input: arr[] = [12, 35, 1, 10, 34, 1]
Output: 34
Explanation: The largest element of the array is 35 and the second largest element is 34.Input: arr[] = [10, 5, 10]
Output: 5
Explanation: The largest element of the array is 10 and the second largest element is 5.Input: arr[] = [10, 10, 10]
Output: -1
Explanation: The largest element of the array is 10 there is no second largest element.
Table of Content
[Naive Approach] Using Sorting
The idea is to sort the array in non-decreasing order. Now, we know that the largest element will be at index n – 1. So, starting from index (n – 2), traverse the remaining array in reverse order. As soon as we encounter an element which is not equal to the largest element, return it as the second largest element in the array. If all the elements are equal to the largest element, return -1.
// C++ program to find second largest element in an array
// using Sorting
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// function to find the second largest element
int getSecondLargest(vector<int> &arr) {
int n = arr.size();
// Sort the array in non-decreasing order
sort(arr.begin(), arr.end());
// start from second last element as last element is the largest
for (int i = n - 2; i >= 0; i--) {
// return the first element which is not equal to the
// largest element
if (arr[i] != arr[n - 1]) {
return arr[i];
}
}
// If no second largest element was found, return -1
return -1;
}
int main() {
vector<int> arr = { 12, 35, 1, 10, 34, 1 };
cout<<getSecondLargest(arr);
return 0;
}
// C program to find second largest element in an array using Sorting
#include <stdio.h>
// function to compare two integers for qsort
int compare(const void *a, const void *b) {
return (*(int *)a - *(int *)b);
}
// function to find the second largest element
int getSecondLargest(int arr[], int n) {
// Sort the array in non-decreasing order
qsort(arr, n, sizeof(int), compare);
// start from second last element as last element is the largest
for (int i = n - 2; i >= 0; i--) {
// return the first element which is not equal to the
// largest element
if (arr[i] != arr[n - 1]) {
return arr[i];
}
}
// If no second largest element was found, return -1
return -1;
}
int main() {
int arr[] = { 12, 35, 1, 10, 34, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", getSecondLargest(arr, n));
return 0;
}
// Java program to find second largest element in an array
// using Sorting
import java.util.Arrays;
class GfG {
// function to find the second largest element
static int getSecondLargest(int[] arr) {
int n = arr.length;
// Sort the array in non-decreasing order
Arrays.sort(arr);
// start from second last element as last element is the largest
for (int i = n - 2; i >= 0; i--) {
// return the first element which is not equal to the
// largest element
if (arr[i] != arr[n - 1]) {
return arr[i];
}
}
// If no second largest element was found, return -1
return -1;
}
public static void main(String[] args) {
int[] arr = { 12, 35, 1, 10, 34, 1 };
System.out.println(getSecondLargest(arr));
}
}
# Python program to find second largest element in an array
# using Sorting
def getSecondLargest(arr):
n = len(arr)
# Sort the array in non-decreasing order
arr.sort()
# start from second last element as last element is the largest
for i in range(n - 2, -1, -1):
# return the first element which is not equal to the
# largest element
if arr[i] != arr[n - 1]:
return arr[i]
# If no second largest element was found, return -1
return -1
if __name__ == "__main__":
arr = [12, 35, 1, 10, 34, 1]
print(getSecondLargest(arr))
// C# program to find second largest element in an array
// using Sorting
using System;
class GfG {
// function to find the second largest element
public static int getSecondLargest(int[] arr) {
int n = arr.Length;
// Sort the array in non-decreasing order
Array.Sort(arr);
// start from second last element as last element is the largest
for (int i = n - 2; i >= 0; i--) {
// return the first element which is not equal to the
// largest element
if (arr[i] != arr[n - 1]) {
return arr[i];
}
}
// If no second largest element was found, return -1
return -1;
}
static void Main() {
int[] arr = { 12, 35, 1, 10, 34, 1 };
Console.WriteLine(getSecondLargest(arr));
}
}
// JavaScript program to find second largest element in an array
// using Sorting
// function to find the second largest element
function getSecondLargest(arr) {
let n = arr.length;
// Sort the array in non-decreasing order
arr.sort((a, b) => a - b);
// start from second last element as last element is the largest
for (let i = n - 2; i >= 0; i--) {
// return the first element which is not equal to the
// largest element
if (arr[i] !== arr[n - 1]) {
return arr[i];
}
}
// If no second largest element was found, return -1
return -1;
}
const arr = [12, 35, 1, 10, 34, 1];
console.log(getSecondLargest(arr));
Output
34
Time Complexity: O(n*log(n)), as sorting the array takes O(n*log(n)) time and traversing the array can take O(n) time in the worst case, so total time complexity = (n*log(n) + n) = O(n*log(n)).
Auxiliary space: O(1), as no extra space is required.
[Better Approach] Two Pass Search
The approach is to traverse the array twice. In the first traversal, find the maximum element. In the second traversal, find the maximum element ignoring the one we found in the first traversal.
Working:
// C++ program to find the second largest element in the array
// using two traversals
#include <iostream>
#include <vector>
using namespace std;
// function to find the second largest element in the array
int getSecondLargest(vector<int> &arr) {
int n = arr.size();
int largest = -1, secondLargest = -1;
// finding the largest element
for (int i = 0; i < n; i++) {
if (arr[i] > largest)
largest = arr[i];
}
// finding the second largest element
for (int i = 0; i < n; i++) {
// Update second largest if the current element is greater
// than second largest and not equal to the largest
if (arr[i] > secondLargest && arr[i] != largest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
int main() {
vector<int> arr = {12, 35, 1, 10, 34, 1};
cout << getSecondLargest(arr);
return 0;
}
// C program to find the second largest element in the array
// using two traversals
#include <stdio.h>
// Function to find the second largest element in the array
int getSecondLargest(int arr[], int n) {
int largest = -1, secondLargest = -1;
// Finding the largest element
for (int i = 0; i < n; i++) {
if (arr[i] > largest)
largest = arr[i];
}
// Finding the second largest element
for (int i = 0; i < n; i++) {
// Update second largest if the current element is greater
// than second largest and not equal to the largest
if (arr[i] > secondLargest && arr[i] != largest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
int main() {
int arr[] = {12, 35, 1, 10, 34, 1};
int n = sizeof(arr)/sizeof(arr[0]);
printf("%d\n", getSecondLargest(arr, n));
return 0;
}
// Java program to find the second largest element in the array
// using two traversals
import java.util.Arrays;
class GfG {
// Function to find the second largest element in the array
static int getSecondLargest(int[] arr) {
int n = arr.length;
int largest = -1, secondLargest = -1;
// Finding the largest element
for (int i = 0; i < n; i++) {
if (arr[i] > largest)
largest = arr[i];
}
// Finding the second largest element
for (int i = 0; i < n; i++) {
// Update second largest if the current element is greater
// than second largest and not equal to the largest
if (arr[i] > secondLargest && arr[i] != largest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
public static void main(String[] args) {
int[] arr = {12, 35, 1, 10, 34, 1};
System.out.println(getSecondLargest(arr));
}
}
# Python program to find the second largest element in the array
# using two traversals
# Function to find the second largest element in the array
def getSecondLargest(arr):
n = len(arr)
largest = -1
secondLargest = -1
# Finding the largest element
for i in range(n):
if arr[i] > largest:
largest = arr[i]
# Finding the second largest element
for i in range(n):
# Update second largest if the current element is greater
# than second largest and not equal to the largest
if arr[i] > secondLargest and arr[i] != largest:
secondLargest = arr[i]
return secondLargest
if __name__ == "__main__":
arr = [12, 35, 1, 10, 34, 1]
print(getSecondLargest(arr))
// C# program to find the second largest element in the array
// using two traversals
using System;
class GfG {
// Function to find the second largest element in the array
static int getSecondLargest(int[] arr) {
int n = arr.Length;
int largest = -1, secondLargest = -1;
// Finding the largest element
for (int i = 0; i < n; i++) {
if (arr[i] > largest)
largest = arr[i];
}
// Finding the second largest element
for (int i = 0; i < n; i++) {
// Update second largest if the current element is greater
// than second largest and not equal to the largest
if (arr[i] > secondLargest && arr[i] != largest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
static void Main() {
int[] arr = { 12, 35, 1, 10, 34, 1 };
Console.WriteLine(getSecondLargest(arr));
}
}
// JavaScript program to find the second largest element in the array
// using two traversals
function getSecondLargest(arr) {
let n = arr.length;
let largest = -1, secondLargest = -1;
// Finding the largest element
for (let i = 0; i < n; i++) {
if (arr[i] > largest)
largest = arr[i];
}
// Finding the second largest element
for (let i = 0; i < n; i++) {
// Update second largest if the current element is greater
// than second largest and not equal to the largest
if (arr[i] > secondLargest && arr[i] !== largest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
let arr = [12, 35, 1, 10, 34, 1];
console.log(getSecondLargest(arr));
Output
34
Time Complexity: O(2*n) = O(n), as we are traversing the array two times.
Auxiliary space: O(1), as no extra space is required.
[Expected Approach] One Pass Search
The idea is to keep track of the largest and second largest element while traversing the array. Initialize largest and secondLargest with -1. Now, for any index i,
- If arr[i] > largest, update secondLargest with largest and largest with arr[i].
- Else If arr[i] < largest and arr[i] > secondLargest, update secondLargest with arr[i].
Working:
// C++ program to find the second largest element in the array
// using one traversal
#include <iostream>
#include <vector>
using namespace std;
// function to find the second largest element in the array
int getSecondLargest(vector<int> &arr) {
int n = arr.size();
int largest = -1, secondLargest = -1;
// finding the second largest element
for (int i = 0; i < n; i++) {
// If arr[i] > largest, update second largest with
// largest and largest with arr[i]
if(arr[i] > largest) {
secondLargest = largest;
largest = arr[i];
}
// If arr[i] < largest and arr[i] > second largest,
// update second largest with arr[i]
else if(arr[i] < largest && arr[i] > secondLargest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
int main() {
vector<int> arr = {12, 35, 1, 10, 34, 1};
cout << getSecondLargest(arr);
return 0;
}
// C program to find the second largest element in the array
// using one traversal
#include <stdio.h>
// function to find the second largest element in the array
int getSecondLargest(int arr[], int n) {
int largest = -1, secondLargest = -1;
// finding the second largest element
for (int i = 0; i < n; i++) {
// If arr[i] > largest, update second largest with
// largest and largest with arr[i]
if(arr[i] > largest) {
secondLargest = largest;
largest = arr[i];
}
// If arr[i] < largest and arr[i] > second largest,
// update second largest with arr[i]
else if(arr[i] < largest && arr[i] > secondLargest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
int main() {
int arr[] = {12, 35, 1, 10, 34, 1};
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", getSecondLargest(arr, n));
return 0;
}
// Java program to find the second largest element in the array
// using one traversal
import java.util.Arrays;
class GfG {
// function to find the second largest element in the array
static int getSecondLargest(int[] arr) {
int n = arr.length;
int largest = -1, secondLargest = -1;
// finding the second largest element
for (int i = 0; i < n; i++) {
// If arr[i] > largest, update second largest with
// largest and largest with arr[i]
if(arr[i] > largest) {
secondLargest = largest;
largest = arr[i];
}
// If arr[i] < largest and arr[i] > second largest,
// update second largest with arr[i]
else if(arr[i] < largest && arr[i] > secondLargest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
public static void main(String[] args) {
int[] arr = {12, 35, 1, 10, 34, 1};
System.out.println(getSecondLargest(arr));
}
}
# Python program to find the second largest element in the array
# using one traversal
# function to find the second largest element in the array
def getSecondLargest(arr):
n = len(arr)
largest = -1
secondLargest = -1
# finding the second largest element
for i in range(n):
# If arr[i] > largest, update second largest with
# largest and largest with arr[i]
if arr[i] > largest:
secondLargest = largest
largest = arr[i]
# If arr[i] < largest and arr[i] > second largest,
# update second largest with arr[i]
elif arr[i] < largest and arr[i] > secondLargest:
secondLargest = arr[i]
return secondLargest
if __name__ == "__main__":
arr = [12, 35, 1, 10, 34, 1]
print(getSecondLargest(arr))
// C# program to find the second largest element in the array
// using one traversal
using System;
class GfG {
// function to find the second largest element in the array
static int getSecondLargest(int[] arr) {
int n = arr.Length;
int largest = -1, secondLargest = -1;
// finding the second largest element
for (int i = 0; i < n; i++) {
// If arr[i] > largest, update second largest with
// largest and largest with arr[i]
if (arr[i] > largest) {
secondLargest = largest;
largest = arr[i];
}
// If arr[i] < largest and arr[i] > second largest,
// update second largest with arr[i]
else if (arr[i] < largest && arr[i] > secondLargest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
static void Main() {
int[] arr = { 12, 35, 1, 10, 34, 1 };
Console.WriteLine(getSecondLargest(arr));
}
}
// JavaScript program to find the second largest element in the array
// using one traversal
// function to find the second largest element in the array
function getSecondLargest(arr) {
const n = arr.length;
let largest = -1, secondLargest = -1;
// finding the second largest element
for (let i = 0; i < n; i++) {
// If arr[i] > largest, update second largest with
// largest and largest with arr[i]
if (arr[i] > largest) {
secondLargest = largest;
largest = arr[i];
}
// If arr[i] < largest and arr[i] > second largest,
// update second largest with arr[i]
else if (arr[i] < largest && arr[i] > secondLargest) {
secondLargest = arr[i];
}
}
return secondLargest;
}
const arr = [12, 35, 1, 10, 34, 1];
console.log(getSecondLargest(arr));
Output
34
Time Complexity: O(n), as we are traversing the array only once.
Auxiliary space: O(1)
Related Article: Smallest and second smallest element in an array
