Solving the SPOJ Problem: Finding the Largest Palindromic Sub-Squares

When tackling the SPOJ problem to find the largest palindromic sub-square in a board (2D array) for each position, the solution can be approached by analyzing the sub-squares with the current position as the bottom-right corner. The goal is to determine the maximum palindromic sub-square across all positions in the board.

In this article, we will explore an efficient algorithm to solve the problem. We start by determining the largest palindromic sub-square for each position and then use this information to find the overall maximum. The key is to check for the presence of palindromes horizontally and vertically for every possible length.

Step 1: Identifying Palindromes Horizontally

To identify palindromes horizontally, we use a boolean array b[n][n][n] to store the information. The array b[i][j][k] indicates whether a palindrome of length k ends at position (i, j). The array is computed using the following steps:

For each position (i, j), if the sub-square of length k 1 to n is a palindrome, set b[i][j][k] to 1 For b[i][j][k], if b[i][j][k] is true, then b[i][j-k 1][k-2] must also be true. This is because if there is a palindrome of length k ending at (i, j), there must be a palindrome of length (k-2) ending at (i, j-1). Similarly, if b[i][j][k] is true, then table[i][j-k 1] table[i][j], where table is the original input array.

After this step, we have a complete record of all palindromes of different lengths ending at each position in the horizontal direction.

Step 2: Storing Consecutive Palindromic Rows

To facilitate further analysis, we store the number of consecutive rows containing palindromes of a particular length (k) using an array (h[n][n][n]). The array (h[i][j][k]) is defined as follows:

If b[i][j][k] is true, then h[i][j][k] 1 h[i-1][j][k]. This means that the number of consecutive rows containing palindromes of length k ending at (i, j) is the same as the number of such rows ending at (i-1, j) plus one. If b[i][j][k] is false, then h[i][j][k] 0.

Step 3: Identifying Palindromes Vertically

The process for vertical palindromes is analogous to the horizontal one. We use another boolean array (v[n][n][n]) to store the information about vertical palindromes. The steps to compute this are the same as those for the horizontal palindromes, but with the vertical positions instead of horizontal positions.

Step 4: Determining the Maximum Palindromic Sub-Square

To find the maximum palindromic sub-square for each position, we need to check if both h[i][j][k] and v[i][j][k] are equal to k. If both are true, it means that there are palindromes of length k ending at (i-k, j) to (i, j) horizontally and at (i, j-k) to (i, j) vertically. This confirms the existence of a palindromic sub-square of length k.

The time complexity of this algorithm is O(N^3), as it involves three nested loops to iterate over the board and the lengths of the palindromes.

Conclusion

The described algorithm efficiently solves the SPOJ problem by breaking it down into manageable tasks and using boolean arrays to store intermediate results. By leveraging this efficient approach, we can determine the largest palindromic sub-square for each position in the 2D array. The overall time complexity ensures that the solution scales well with the size of the input board.

For more information or to dive deeper into the implementation details, refer to the following resources:

Palindromes on Wikipedia SPOJ Problems Repository on GitHub GeeksforGeeks: Check Palindrome Substrings in Given Range