Problem Solving

Problem Statement

Given an array of integers, reverse the array so that the first element becomes the last, the second becomes the second last, and so on. The task is to return or print the array in reversed order.

Example:

      Input: arr = [10, 20, 30, 5, 25]
      Output: [25, 5, 30, 20, 10]
      Explanation: The array elements are reversed.
        

Solution Approach

1. Start iterating from the last index of the array to the first index.
2. Print or store each element to get the reversed order.
3. Alternatively, use built-in Kotlin functions like reversedArray() for simplicity.

Your Kotlin Solution

      import kotlinx.coroutines.*

      fun main(){
          val arrInput = arrayOf(10, 20, 30, 5, 25) // expected outcome = 25,5,30,20,10
          for(i in arrInput.size-1 downTo 0){
              print(arrInput[i])
          }
      }
        

Alternate Kotlin Solution Using Built-in Function

      fun main() {
          val arrInput = arrayOf(10, 20, 30, 5, 25)
          val reversedArr = arrInput.reversedArray()
          println(reversedArr.joinToString()) // Output: 25, 5, 30, 20, 10
      }
        

Explanation

• The first solution uses a for loop starting from the last index down to 0.
• Elements are printed in reverse order, effectively reversing the array.
• The built-in reversedArray() function is more concise and returns a new reversed array.

Complexity Analysis

• Time Complexity: O(N) — traversing the array once.
• Space Complexity: O(N) for reversedArray(), O(1) if printing directly.

Problem Statement

You are given an array of integers. Your task is to find the second largest element in the array. The array may contain duplicate elements.

Example:

      Input: arr = [12, 35, 1, 10, 34, 1]
      Output: 34
      Explanation: The largest element is 35, the second largest is 34.
        

Example 2:

      Input: arr = [5, 5, 5, 2]
      Output: 2
      Explanation: The largest element is 5 (repeated), second largest is 2.
        

Solution Approach

1. Initialize two variables: largest and secondLargest with Int.MIN_VALUE.
2. Traverse the array once.
3. For each element:
   • If element is greater than largest, update secondLargest to largest and then update largest to the element.
   • Else if element is greater than secondLargest and not equal to largest, update secondLargest.
4. After traversal, secondLargest contains the second largest element.

Kotlin Solution

      fun findSecondLargest(arr: Array): Int {
          var largest = Int.MIN_VALUE
          var secondLargest = Int.MIN_VALUE

          arr.forEach { num ->
              if (num > largest) {
                  secondLargest = largest
                  largest = num
              } else if (num > secondLargest && num != largest) {
                  secondLargest = num
              }
          }

          return secondLargest
      }

      fun main() {
          val arr = arrayOf(12, 35, 1, 10, 34, 1)
          val secondLargest = findSecondLargest(arr)
          println("Second largest element is: $secondLargest") // Output: 34
      }
        

Step-by-Step Example

For arr = [12, 35, 1, 10, 34, 1]:

1. Start: largest = MIN_VALUE, secondLargest = MIN_VALUE
2. 12 → largest = 12, secondLargest = MIN_VALUE
3. 35 → largest = 35, secondLargest = 12
4. 1 → largest = 35, secondLargest = 12
5. 10 → largest = 35, secondLargest = 12
6. 34 → largest = 35, secondLargest = 34
7. 1 → largest = 35, secondLargest = 34

Result: secondLargest = 34

Explanation

• This approach keeps track of the two largest numbers in a single pass.
• No extra array or filtering is needed.
• Handles duplicates correctly by checking num != largest.

Complexity Analysis

• Time Complexity: O(N) — Only one traversal of the array is needed.
• Space Complexity: O(1) — Only two variables are used.

Edge Cases

• If the array has all elements the same, the second largest might not exist — you may return a special value or handle separately.
• If the array has less than 2 distinct elements, handle accordingly.