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 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