SE-1

To reverse a linked list, we change the direction of its next pointers so that each node points to its previous node.

Steps to Reverse the List:
Initialize Three Pointers:

prev = NULL (to store the previous node).
curr = head (to track the current node).
next (to store the next node).
Iterate Through the List:

Save the next node (next = curr.next).
Reverse the link (curr.next = prev).
Move prev and curr forward (prev = curr, curr = next).
Once curr becomes NULL, prev will be the new head of the reversed list.

The problem requires finding the K-th digit of the N-th palindrome in a sequence of palindromes.

Step 1: Understanding the Problem
A palindrome is a number that reads the same forward and backward (e.g., 121, 333, 1221).
Given N, find the N-th palindrome in the sequence.
Extract the K-th digit from this palindrome.
Step 2: Approach to Solve the Problem
Generate the N-th palindrome:

Consider palindromes based on their number of digits.
Generate palindromes systematically in increasing order until reaching the N-th one.
Extract the K-th digit:

Convert the N-th palindrome to a string.
Retrieve the K-th digit (1-based index).

The Maximum Subarray Sum problem requires finding the contiguous subarray with the largest sum in a given array.

Step 1: Understanding the Problem
Given an array arr[] of size N, find the contiguous subarray that has the maximum sum.

Example:
Input: arr = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output: 6
Explanation: The subarray [4, -1, 2, 1] has the maximum sum = 6.

Step 2: Kadane’s Algorithm (Optimal Approach)
Initialize two variables:

maxSum = -∞ (stores the maximum sum found).
currentSum = 0 (stores the sum of the current subarray).
Iterate through the array:

Add the current element to currentSum.
If currentSum is greater than maxSum, update maxSum.
If currentSum becomes negative, reset it to 0 (start a new subarray).
Return maxSum after completing the iteration.