*This summary of the video was created by an AI. It might contain some inaccuracies.*

## 00:00:00 – 00:06:23

The video discusses minimizing the cost to make an array palindromic by finding the optimal palindromic number that minimizes the absolute difference between original numbers and the palindromic number. The key approach involves finding the median of the array and adjusting numbers around it to create palindromes. The process is demonstrated for both even and odd-numbered arrays, showcasing methods to manipulate medians for palindromic numbers. Important terms include palindromic numbers and medians. Ultimately, the video emphasizes efficient strategies for achieving minimal cost in creating palindromic arrays.

### 00:00:00

In this segment of the video, the YouTuber discusses the minimum cost to make an array palindromic. The goal is to find a palindromic number that minimizes the absolute difference between the original numbers in the array and the palindromic number. The optimal solution involves finding the median of the array. If the array has an odd number of elements, the median will be an integer, and if it has an even number of elements, there will be two potential medians. The YouTuber demonstrates how to adjust the medians to create palindromic numbers in the array, ultimately minimizing the overall score.

### 00:03:00

In this segment of the video, the speaker discusses creating palindromes by manipulating numbers around the median. They highlight the process of finding palindromes for both even and odd numbers. Using the example of the array 10, 12, 13, 14, 15, they emphasize the importance of sorting the array to find the median (which is 13 in this case). The speaker then demonstrates operations to achieve palindromic numbers within the array, ultimately providing the answers as 11 and 22. They conclude by determining the minimum between these two values.

### 00:06:00

In this segment of the video, the speaker discusses finding the value of A5, which is determined to be 11 as the minimum. They explain a process for when n is even, involving adding and subtracting 5 and then removing the 0. Ultimately, the final answer is determined to be A1 A2 A3 A4 minimum.