Hello Everyone,
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
Efficient Solution
Approach: Find the second largest element in a single traversal.
Below is the complete algorithm for doing this:
 Initialize two variables first and second to INT_MIN as first = second = INT_MIN 2) Start traversing the array, 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 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;
}
Output:
The second largest element is 34
Complexity Analysis:

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