Skip to content
Related Articles

Related Articles

Improve Article

Find Second largest element in an array

  • Difficulty Level : Easy
  • Last Updated : 16 Jul, 2021

Given an array of integers, our task is to write a program that efficiently finds the second largest element present in the array. 
Example:

Input: arr[] = {12, 35, 1, 10, 34, 1}
Output: The second largest element is 34.
Explanation: The largest element of the 
array is 35 and the second 
largest element is 34

Input: arr[] = {10, 5, 10}
Output: The second largest element is 5.
Explanation: The largest element of 
the array is 10 and the second 
largest element is 5

Input: arr[] = {10, 10, 10}
Output: The second largest does not exist.
Explanation: Largest element of the array 
is 10 there is no second largest element

Simple Solution 
Approach: The idea is to sort the array in descending order and then return the second element which is not equal to the largest element from the sorted array.

C++14




// C++ program to find second largest
// element in an array
 
#include <bits/stdc++.h>
using namespace std;
 
/* Function to print the second largest elements */
void print2largest(int arr[], int arr_size)
{
    int i, first, second;
 
    /* There should be atleast two elements */
    if (arr_size < 2) {
        printf(" Invalid Input ");
        return;
    }
 
    // sort the array
    sort(arr, arr + arr_size);
 
    // start from second last element
    // as the largest element is at last
    for (i = arr_size - 2; i >= 0; i--) {
        // if the element is not
        // equal to largest element
        if (arr[i] != arr[arr_size - 1]) {
            printf("The second largest element is %d\n", arr[i]);
            return;
        }
    }
 
    printf("There is no second largest element\n");
}
 
/* Driver program to test above function */
int main()
{
    int arr[] = { 12, 35, 1, 10, 34, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    print2largest(arr, n);
    return 0;
}

Java




// Java program to find second largest
// element in an array
import java.util.*;
class GFG{
 
// Function to print the
// second largest elements
static void print2largest(int arr[],
                          int arr_size)
{
  int i, first, second;
 
  // There should be
  // atleast two elements
  if (arr_size < 2)
  {
    System.out.printf(" Invalid Input ");
    return;
  }
 
  // Sort the array
  Arrays.sort(arr);
 
  // Start from second last element
  // as the largest element is at last
  for (i = arr_size - 2; i >= 0; i--)
  {
    // If the element is not
    // equal to largest element
    if (arr[i] != arr[arr_size - 1])
    {
      System.out.printf("The second largest " +
                        "element is %d\n", arr[i]);
      return;
    }
  }
 
  System.out.printf("There is no second " +
                    "largest element\n");
}
 
// Driver code
public static void main(String[] args)
{
  int arr[] = {12, 35, 1, 10, 34, 1};
  int n = arr.length;
  print2largest(arr, n);
}
}
 
// This code is contributed by gauravrajput1

Python3




# Python3 program to find second
# largest element in an array
 
# Function to print the
# second largest elements
def print2largest(arr,
                  arr_size):
  
  # There should be
  # atleast two elements
  if (arr_size < 2):  
    print(" Invalid Input ")
    return
  
  # Sort the array
  arr.sort
  
  # Start from second last
  # element as the largest
  # element is at last
  for i in range(arr_size-2,
                 -1, -1):
  
    # If the element is not
    # equal to largest element
    if (arr[i] != arr[arr_size - 1]) :
     
      print("The second largest element is",
            arr[i])
      return
  
  print("There is no second largest element")
 
# Driver code
arr = [12, 35, 1, 10, 34, 1]
n = len(arr)
print2largest(arr, n)
 
# This code is contributed by divyeshrabadiya07

C#




// C# program to find second largest
// element in an array
using System;
 
class GFG{
 
// Function to print the
// second largest elements
static void print2largest(int []arr,
                          int arr_size)
{
  int i;
 
  // There should be
  // atleast two elements
  if (arr_size < 2)
  {
    Console.Write(" Invalid Input ");
    return;
  }
 
  // Sort the array
  Array.Sort(arr);
 
  // Start from second last element
  // as the largest element is at last
  for(i = arr_size - 2; i >= 0; i--)
  {
       
    // If the element is not
    // equal to largest element
    if (arr[i] != arr[arr_size - 1])
    {
      Console.Write("The second largest " +
                    "element is {0}\n", arr[i]);
      return;
    }
  }
 
  Console.Write("There is no second " +
                "largest element\n");
}
 
// Driver code
public static void Main(String[] args)
{
  int []arr = { 12, 35, 1, 10, 34, 1 };
  int n = arr.Length;
   
  print2largest(arr, n);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// Javascript program to find second largest
// element in an array
  
    // Function to print the second largest elements
    function print2largest(arr, arr_size) {
        let i, first, second;
  
        // There should be atleast two elements
        if (arr_size < 2) {
            document.write(" Invalid Input ");
            return;
        }
  
        // sort the array
        arr.sort();
  
        // start from second last element
        // as the largest element is at last
        for (i = arr_size - 2; i >= 0; i--) {
            // if the element is not
            // equal to largest element
            if (arr[i] != arr[arr_size - 1]) {
                document.write("The second largest element is " + arr[i]);
                return;
            }
        }
  
        document.write("There is no second largest element<br>");
    }
  
    // Driver program to test above function
 
    let arr= [ 12, 35, 1, 10, 34, 1 ];
    let n = arr.length;
    print2largest(arr, n);
 
     
// This code is contributed by Surbhi Tyagi
 
</script>

Output:

The second largest element is 34

Complexity Analysis:

  • Time Complexity: O(n log n). 
    Time required to sort the array is O(n log n).
  • Auxiliary space: O(1). 
    As no extra space is required.

Better Solution: 
Approach: The approach is to traverse the array twice. 



In the first traversal find the maximum element. 

In the second traversal find the greatest element in the remaining excluding the previous greatest.

C++14




// C++ program to find the second largest element in the array
#include <iostream>
using namespace std;
 
 
int secondLargest(int arr[], int n) {
    int largest = 0, secondLargest = -1;
 
    // finding the largest element in the array
    for (int i = 1; i < n; i++) {
        if (arr[i] > arr[largest])
            largest = i;
    }
 
    // finding the largest element in the array excluding
    // the largest element calculated above
    for (int i = 0; i < n; i++) {
        if (arr[i] != arr[largest]) {
            // first change the value of second largest
            // as soon as the next element is found
            if (secondLargest == -1)
                secondLargest = i;
            else if (arr[i] > arr[secondLargest])
                secondLargest = i;
        }
    }
    return secondLargest;
}
 
 
int main() {
    int arr[] = {10, 12, 20, 4};
    int n = sizeof(arr)/sizeof(arr[0]);
    int second_Largest = secondLargest(arr, n);
    if (second_Largest == -1)
        cout << "Second largest didn't exit\n";
    else
        cout << "Second largest : " << arr[second_Largest];
}

Java




// Java program to find second largest
// element in an array
class GFG{
 
// Function to print the second largest elements
static void print2largest(int arr[], int arr_size)
{
    int i, first, second;
 
    // There should be atleast two elements
    if (arr_size < 2)
    {
        System.out.printf(" Invalid Input ");
        return;
    }
 
    int largest = second = Integer.MIN_VALUE;
 
    // Find the largest element
    for(i = 0; i < arr_size; i++)
    {
        largest = Math.max(largest, arr[i]);
    }
 
    // Find the second largest element
    for(i = 0; i < arr_size; i++)
    {
        if (arr[i] != largest)
            second = Math.max(second, arr[i]);
    }
    if (second == Integer.MIN_VALUE)
        System.out.printf("There is no second " +
                          "largest element\n");
    else
        System.out.printf("The second largest " +
                          "element is %d\n", second);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 12, 35, 1, 10, 34, 1 };
    int n = arr.length;
     
    print2largest(arr, n);
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program to find
# second largest element
# in an array
 
# Function to print
# second largest elements
def print2largest(arr, arr_size):
 
    # There should be atleast
    # two elements
    if (arr_size < 2):
        print(" Invalid Input ");
        return;
 
    largest = second = -2454635434;
 
    # Find the largest element
    for i in range(0, arr_size):
        largest = max(largest, arr[i]);
 
    # Find the second largest element
    for i in range(0, arr_size):
        if (arr[i] != largest):
            second = max(second, arr[i]);
 
    if (second == -2454635434):
        print("There is no second " +
              "largest element");
    else:
        print("The second largest " +
              "element is \n", second);
 
# Driver code
if __name__ == '__main__':
   
    arr = [12, 35, 1,
           10, 34, 1];
    n = len(arr);
    print2largest(arr, n);
 
# This code is contributed by shikhasingrajput

C#




// C# program to find second largest
// element in an array
using System;
 
class GFG{
 
// Function to print the second largest elements
static void print2largest(int []arr, int arr_size)
{
    // int first;
    int i, second;
 
    // There should be atleast two elements
    if (arr_size < 2)
    {
        Console.Write(" Invalid Input ");
        return;
    }
 
    int largest = second = int.MinValue;
 
    // Find the largest element
    for(i = 0; i < arr_size; i++)
    {
        largest = Math.Max(largest, arr[i]);
    }
 
    // Find the second largest element
    for(i = 0; i < arr_size; i++)
    {
        if (arr[i] != largest)
            second = Math.Max(second, arr[i]);
    }
     
    if (second == int.MinValue)
        Console.Write("There is no second " +
                      "largest element\n");
    else
        Console.Write("The second largest " +
                      "element is {0}\n", second);
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 12, 35, 1, 10, 34, 1 };
    int n = arr.Length;
     
    print2largest(arr, n);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
  
// Javascript program to find second largest
// element in an array
   
    // Function to print the second largest elements
    function print2largest(arr, arr_size) {
        let i;
        let largest = second = -2454635434;
   
        // There should be atleast two elements
        if (arr_size < 2) {
            document.write(" Invalid Input ");
            return;
        }
   
        // finding the largest element
        for (i = 0;i<arr_size;i++){
            if (arr[i]>largest){
                largest = arr[i];
            }
        }
   
        // Now find the second largest element
        for (i = 0 ;i<arr_size;i++){
            if (arr[i]>second && arr[i]<largest){
                second = arr[i];
            }
        }
  
        if (second == -2454635434){
             
        document.write("There is no second largest element<br>");
        }
        else{
            document.write("The second largest element is " + second);
                return;
            }
        }
     
   
    // Driver program to test above function
  
    let arr= [ 12, 35, 1, 10, 34, 1 ];
    let n = arr.length;
    print2largest(arr, n);
   
</script>

Output:

Second largest : 12

Complexity Analysis:

  • Time Complexity: O(n). 
    Two traversals of the array is needed.
  • Auxiliary space: O(1). 
    As no extra space is required.

Efficient Solution 
Approach: Find the second largest element in a single traversal. 
Below is the complete algorithm for doing this:  

1) Initialize the first as 0(i.e, index of arr[0] element
2) Start traversing the array from array[1],
   a) If the current element in array say arr[i] is greater
      than first. Then update first and second as,
      second = first
      first = arr[i]
   b) If the current element is in between first and second,
      then update second to store the value of current variable as
      second = arr[i]
3) Return the value stored in second.

C




// C program to find second largest
// element in an array
 
#include <limits.h>
#include <stdio.h>
 
/* Function to print the second largest elements */
void print2largest(int arr[], int arr_size)
{
    int i, first, second;
 
    /* There should be atleast two elements */
    if (arr_size < 2) {
        printf(" Invalid Input ");
        return;
    }
 
    first = second = INT_MIN;
    for (i = 0; i < arr_size; i++) {
        /* If current element is greater than first
           then update both first and second */
        if (arr[i] > first) {
            second = first;
            first = arr[i];
        }
 
        /* If arr[i] is in between first and
           second then update second  */
        else if (arr[i] > second && arr[i] != first)
            second = arr[i];
    }
    if (second == INT_MIN)
        printf("There is no second largest element\n");
    else
        printf("The second largest element is %dn", second);
}
 
/* Driver program to test above function */
int main()
{
    int arr[] = { 12, 35, 1, 10, 34, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    print2largest(arr, n);
    return 0;
}

C++




// C++ program to find the second largest element
 
#include <iostream>
using namespace std;
 
// returns the index of second largest
// if second largest didn't exist return -1
int secondLargest(int arr[], int n) {
    int first = 0, second = -1;
    for (int i = 1; i < n; i++) {
        if (arr[i] > arr[first]) {
            second = first;
            first = i;
        }
        else if (arr[i] < arr[first]) {
            if (second == -1 || arr[second] < arr[i])
                second = i;
        }
    }
    return second;
}
 
int main() {
    int arr[] = {10, 12, 20, 4};
    int index = secondLargest(arr, sizeof(arr)/sizeof(arr[0]));
    if (index == -1)
        cout << "Second Largest didn't exist";
    else
        cout << "Second largest : " << arr[index];
}

Java




// JAVA Code for Find Second largest
// element in an array
class GFG {
 
    /* Function to print the second largest
    elements */
    public static void print2largest(int arr[],
                                     int arr_size)
    {
        int i, first, second;
 
        /* There should be atleast two elements */
        if (arr_size < 2) {
            System.out.print(" Invalid Input ");
            return;
        }
 
        first = second = Integer.MIN_VALUE;
        for (i = 0; i < arr_size; i++) {
            /* If current element is greater than
            first then update both first and second */
            if (arr[i] > first) {
                second = first;
                first = arr[i];
            }
 
            /* If arr[i] is in between first and
               second then update second  */
            else if (arr[i] > second && arr[i] != first)
                second = arr[i];
        }
 
        if (second == Integer.MIN_VALUE)
            System.out.print("There is no second largest"
                             + " element\n");
        else
            System.out.print("The second largest element"
                             + " is " + second);
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = { 12, 35, 1, 10, 34, 1 };
        int n = arr.length;
        print2largest(arr, n);
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3




# Python program to
# find second largest
# element in an array
 
 
# Function to print the
# second largest elements
def print2largest(arr, arr_size):
 
    # There should be atleast
        # two elements
    if (arr_size < 2):
     
        print(" Invalid Input ")
        return
     
 
    first = second = -2147483648
    for i in range(arr_size):
     
        # If current element is
                # smaller than first
        # then update both
                # first and second
        if (arr[i] > first):
         
            second = first
            first = arr[i]
         
 
        # If arr[i] is in
                # between first and
        # second then update second
        elif (arr[i] > second and arr[i] != first):
            second = arr[i]
     
    if (second == -2147483648):
        print("There is no second largest element")
    else:
        print("The second largest element is", second)
 
 
# Driver program to test
# above function
arr = [12, 35, 1, 10, 34, 1]
n = len(arr)
 
print2largest(arr, n)
 
# This code is contributed
# by Anant Agarwal.

C#




// C# Code for Find Second largest
// element in an array
using System;
 
class GFG {
 
    // Function to print the
    // second largest elements
    public static void print2largest(int[] arr,
                                     int arr_size)
    {
        int i, first, second;
 
        // There should be atleast two elements
        if (arr_size < 2) {
            Console.WriteLine(" Invalid Input ");
            return;
        }
 
        first = second = int.MinValue;
        for (i = 0; i < arr_size; i++) {
            // If current element is smaller than
            // first then update both first and second
            if (arr[i] > first) {
                second = first;
                first = arr[i];
            }
 
            // If arr[i] is in between first
            // and second then update second
            else if (arr[i] > second && arr[i] != first)
                second = arr[i];
        }
 
        if (second == int.MinValue)
            Console.Write("There is no second largest"
                          + " element\n");
        else
            Console.Write("The second largest element"
                          + " is " + second);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 12, 35, 1, 10, 34, 1 };
        int n = arr.Length;
        print2largest(arr, n);
    }
}
 
// This code is contributed by Parashar.

PHP




<?php
// PHP program to find second largest
// element in an array
 
// Function to print the
// second largest elements
function print2largest($arr, $arr_size)
{
 
    // There should be atleast
    // two elements
    if ($arr_size < 2)
    {
        echo(" Invalid Input ");
        return;
    }
 
    $first = $second = PHP_INT_MIN;
    for ($i = 0; $i < $arr_size ; $i++)
    {
         
        // If current element is
        // smaller than first
        // then update both
        // first and second
        if ($arr[$i] > $first)
        {
            $second = $first;
            $first = $arr[$i];
        }
 
        // If arr[i] is in
        // between first and
        // second then update
        // second
        else if ($arr[$i] > $second &&
                 $arr[$i] != $first)
            $second = $arr[$i];
    }
    if ($second == PHP_INT_MIN)
        echo("There is no second largest element\n");
    else
        echo("The second largest element is " . $second . "\n");
}
 
// Driver Code
$arr = array(12, 35, 1, 10, 34, 1);
$n = sizeof($arr);
print2largest($arr, $n);
 
// This code is contributed by Ajit.
?>

Javascript




<script>
  
// Javascript program to find second largest
// element in an array
   
    // Function to print the second largest elements
    function print2largest(arr, arr_size) {
        let i;
        let largest = second = -2454635434;
   
        // There should be atleast two elements
        if (arr_size < 2) {
            document.write(" Invalid Input ");
            return;
        }
   
        // finding the largest element
         
        for (i = 0 ;i<arr_size;i++){
            if (arr[i]>largest){
                second = largest ;
                largest = arr[i]
            }
 
            else if (arr[i]!=largest && arr[i]>second ){
                second = arr[i];
            }
        }
  
        if (second == -2454635434){
             
        document.write("There is no second largest element<br>");
        }
        else{
            document.write("The second largest element is " + second);
                return;
            }
        }
     
 
    // Driver program to test above function
  
    let arr= [ 12, 35, 1, 10, 34, 1 ];
    let n = arr.length;
    print2largest(arr, n);
     
 // This code is contributed by Shaswat Singh
  
</script>

Output:

Second largest : 12

Complexity Analysis:

  • Time Complexity: O(n). 
    Only one traversal of the array is needed.
  • Auxiliary space: O(1). 
    As no extra space is required.

Related Article
Smallest and second smallest element in an array
This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :