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Second Largest Element in Array

Certification: Basic Level Accuracy: 0% Submissions: 0 Points: 5

Write a JavaScript program to find the second largest element in an array of integers. If there is no second largest element (e.g., all elements are the same), return -1. The array contains at least one element.

Example 1
    
      
  • Input: arr = [3, 1, 4, 1, 5, 9, 2]
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  • Output: 5
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  • Explanation: 
    • The array contains elements [3, 1, 4, 1, 5, 9, 2]. 
    • The largest element in the array is 9. 
    • The second largest element in the array is 5. 
    • Therefore, the output is 5.
    •
Example 2
    
      
  • Input: arr = [5, 5, 5, 5]
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  • Output: -1
    • 
      
  • Explanation: 
    • The array contains elements [5, 5, 5, 5]. 
    • All elements in the array are the same (5). 
    • There is no second largest element since all values are identical. 
    • Therefore, the output is -1.
    •
Constraints
    
      
  • 1 ≤ arr.length ≤ 1000
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  • -1000 ≤ arr[i] ≤ 1000
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  • If no second largest exists, return -1
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  • Time Complexity: O(n) where n is the length of the array
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  • Space Complexity: O(1)
    •
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Hint
Solution

Solution Hints

    
      
  • Initialize two variables to track the largest and second largest elements
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  • Set both variables to negative infinity initially
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  • Iterate through the array once
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  • For each element, compare it with the current largest element
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  • If the element is larger than the largest, update both largest and second largest
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  • If the element is between largest and second largest, update only second largest
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  • Make sure to handle duplicate values correctly by checking for strict inequality
    •

Steps to solve by this approach:

 Step 1: Initialize two variables 'largest' and 'secondLargest' to negative infinity
 Step 2: Iterate through each element in the array using a for loop
 Step 3: For each element, first check if it's greater than the current largest element
 Step 4: If yes, update secondLargest to the previous largest value and set largest to current element
 Step 5: If no, check if the element is greater than secondLargest but less than largest
 Step 6: If this condition is true, update secondLargest to the current element
 Step 7: After the loop, check if secondLargest is still negative infinity and return -1 if true, otherwise return secondLargest

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