Introduction

Finding the second largest element in an array means identifying the second biggest distinct value present in the array.

Given an array of integers arr[] of size n , the task is to find the second largest distinct element in the array.

Note:
If the second largest element does not exist, return: -1

Example

Input: arr[] = [7, 12, 4, 9, 15, 10]
Output: 12
Explanation:
Largest element = 15 
Second largest element = 12
Input: arr[] = [5, 5, 5]
Output: -1
Explanation:
All elements are same
No second largest element exists

 Constraints

  • 1 <= n <= 10^5
  • -10^9 <= arr[i] <= 10^9

Approach 1 : Brute Force (Using Sorting)

Explanation

The simplest way to solve this problem is:

  • Sort the array
  • Traverse from the end
  • Find the first element smaller than the largest element

This approach is easy to understand but sorting increases time complexity.

Steps

  1. Sort the array in ascending order.
  2. Store the largest element.
  3. Traverse from the second last index.
  4. Find the first element different from the largest element.
  5. Return that value as the second largest element.

    Dry Run:

    Input Array: [12, 35, 1, 10, 34, 1]
After Sorting: [1, 1, 10, 12, 34, 35]
Largest Element: 35
Traverse from right to left:
34 is smaller than 35
Therefore, Second Largest = 34
Final Result: 34

Brute Force Code 

Complexity Analysis
Time Complexity: O(n log n)Explanation:
Sorting the array takes O(n log n) time.
Traversing from the end takes O(n) time.
Space Complexity: O(1)
Explanation:
No extra space is used apart from a few variables.

Approach 2 : Optimized Solution (Single Traversal)

Explanation

Instead of sorting the array, we can find both the largest and second largest elements in a single traversal.

We maintain:

  • largest element
  • second largest element

and update them whenever required.

This avoids unnecessary sorting operations and improves efficiency.

Steps

  1. Initialize:
    • largest = -1
    • secondLargest = -1
  2. Traverse the array.
  3. If current element is greater than largest:
    • update secondLargest = largest
    • update largest = current element

  4. Else if current element:

    • is smaller than largest
    • and greater than secondLargest

    update secondLargest.

  5. Return secondLargest.

Dry Run:​

Input Array: [12, 35, 1, 10, 34, 1]Initially:
Largest = -1
Second Largest = -1

Traverse 12:
12 is greater than Largest
Update:
Largest = 12
Second Largest = -1

Traverse 35:
35 is greater than Largest (12)
Update:
Largest = 35
Second Largest = 12

Traverse 1:
1 is smaller than both values
No update required

Traverse 10:
10 is greater than Second Largest (-1)
Update Second Largest = 10

Traverse 34:
34 is smaller than Largest (35)
34 is greater than Second Largest (10)
Update Second Largest = 34

Traverse 1:
No update required
Final Result: Second Largest = 34

Optimized Code 

Complexity Analysis

Time Complexity: O(n)Explanation:
The array is traversed only once.
Each element is processed exactly one time.

Space Complexity: O(1)
Explanation:
Only two extra variables are used:
largest and secondLargest.
No additional data structures are required.

Edge Cases

  • Array contains all same elements
  • Array contains only one element
  • Array contains negative numbers
  • Duplicate largest elements are present

Why This Problem is Important

This problem builds the foundation for:

  • Optimization-based problems
  • Greedy techniques
  • Heap problems
  • Top K element problems
  • Efficient traversal logic

Understanding how to track multiple values efficiently is important in many advanced DSA problems.

Real-World Applications

Finding the second largest element is used in:

  • Ranking systems
  • Sports leaderboards
  • Performance analytics
  • Competitive scoring systems
  • Statistical analysis

Common Mistakes

  • Forgetting duplicate elements
  • Incorrect initialization
  • Updating second largest incorrectly
  • Sorting unnecessarily in optimized solution

Interview Tips

Interviewers often expect:

  • an O(n) traversal solution
  • handling duplicate values properly
  • constant space optimization

Always explain why the single traversal approach is better than sorting.

Related Questions

  • Find Largest Element
  • Top K Frequent Elements
  • Kth Largest Element
  • Third Largest Element
  • Leaders in an Array

Final Takeaway

Finding the second largest element in an array is an important beginner-level DSA problem that teaches efficient traversal, conditional updates, and optimization techniques. Understanding this problem helps build strong foundations for advanced array and heap-based interview problems.