Find Second largest element in an array
Given an array of integers, our task is to write a program that efficiently finds the second-largest element present in the array.
Examples:
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 34Input: 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 5Input: 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
Find Second Largest element using Sorting:
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.
Below is the implementation of the above idea:
// 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)
{
// sort the array in descending order
sort(arr, arr + arr_size, greater<int>());
// start from second element as first
// element is the largest
for (int i = 1; i < arr_size; i++) {
// if the element is not equal to largest element
if (arr[i] != arr[0]) {
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;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// C program to find second largest element in an array
#include <stdio.h>
#include <stdlib.h>
// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
return (*(int*)b - *(int*)a);
}
/* Function to print the second largest elements */
void print2largest(int arr[], int arr_size)
{
// sort the array in descending order
qsort(arr, arr_size, sizeof(int), cmpfunc);
// start from second element as first
// element is the largest
for (int i = 1; i < arr_size; i++) {
// if the element is not equal
// to the largest element
if (arr[i] != arr[0]) {
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;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// 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(Integer arr[], int arr_size)
{
// Sort the array in descending order
Arrays.sort(arr, Collections.reverseOrder());
// Start from second element as first
// element is the largest
for (int i = 1; i < arr_size; i++) {
// If the element is not
// equal to largest element
if (arr[i] != arr[0]) {
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)
{
Integer arr[] = { 12, 35, 1, 10, 34, 1 };
int n = arr.length;
print2largest(arr, n);
}
}
// This code is contributed by gauravrajput1
# Python3 program to find second
# largest element in an array
# Function to print the
# second largest elements
def print2largest(arr, arr_size):
# Sort the array in descending order
arr.sort(reverse=True)
# Start from second last element as first
# element is the largest
for i in range(1, arr_size):
# If the element is not
# equal to largest element
if (arr[i] != arr[0]):
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
using System;
using System.Linq;
class Program
{
// Function to print the second largest element
static void PrintSecondLargest(int[] arr)
{
// Sort the array in descending order
Array.Sort(arr);
Array.Reverse(arr);
int n = arr.Length;
int largest = arr[0];
int secondLargest = -1;
// Find the second largest element
for (int i = 1; i < n; i++)
{
if (arr[i] != largest)
{
secondLargest = arr[i];
break;
}
}
if (secondLargest != -1)
Console.WriteLine($"The second largest element is {secondLargest}");
else
Console.WriteLine("There is no second largest element");
}
// Driver code to test the function
static void Main()
{
int[] arr = { 12, 35, 1, 10, 34, 1 };
PrintSecondLargest(arr);
}
}
// This code is contributed by shivamgupta0987654321
<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
The second largest element is 34
Time Complexity: O(nlogn), where n is the size of input array.
Auxiliary space: O(1), as no extra space is required.
Find Second Largest element by traversing the array twice (Two Pass):
The approach is to traverse the array twice. In the first traversal, find the maximum element. In the second traversal, find the greatest element excluding the previous greatest.
Below is the implementation of the above idea:
// 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[] = { 12, 35, 1, 10, 34, 1 };
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 program to find second largest
// element in an array
import java.io.*;
class GFG {
// Function to print the second largest elements
static void print2largest(int arr[], int arr_size)
{
int i, 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);
}
}
# 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# 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
<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 : 34
Time Complexity: O(n), where n is the size of input array.
Auxiliary space: O(1), as no extra space is required.
Find Second Largest element by traversing the array once (One Pass):
The idea is to keep track of the largest and second largest element while traversing the array. If an element is greater than the largest element, we update the largest as well as the second largest. Else if an element is smaller than largest but greater than second largest, then we update the second largest only.
Below is the complete algorithm for doing this:
- Initialize the first as 0(i.e, index of arr[0] element
- Start traversing the array from array[1],
- If the current element in array say arr[i] is greater than first. Then update first and second as, second = first and first = arr[i]
- If the current element is in between first and second, then update second to store the value of current variable as second = arr[i]
- Return the value stored in second.
Below is the implementation of the above approach:
// 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[] = { 12, 35, 1, 10, 34, 1 };
int index = secondLargest(arr, sizeof(arr)/sizeof(arr[0]));
if (index == -1)
cout << "Second Largest didn't exist";
else
cout << "Second largest : " << arr[index];
}
// 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 %d", 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;
}
// JAVA Code for Find Second largest
// element in an array
import java.io.*;
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.
# 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
# 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
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# 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 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.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.
<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>
<?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 ; $i++)
{
// If current element is
// greater than first
// then update both
// first and second
if ($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 == 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.
?>
Output
Second largest : 34
Time Complexity: O(n), where n is the size of input array.
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.
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