I am trying to solve a problem where I am given a nXn square matrix of characters and I want to find out size of the largest palindrome square from this? The largest palindrome square is, a square with all rows and all columns as palindrome.

For eg. Input

```
a g h j k
s d g d j
s e f e n
a d g d h
r y d g s
```

The output will be:

```
3
```

corresponding to the middle square. I am thinking of dynamic programming solution but unable to formulate the recurrence relation. I am thinking the dimensions should be a(i,j,k) where i, j are the bottom-right of rectangle and k be the size of palindrome square. Can someone help me with the recurrence relation for this problem?

EDIT:

n<500, so I believe that I can't go beyond O(n^3).

Answer:

Assuming that you can solve the following problem:

- Ending at cell
`(i, j)`

is there any palindrome with different length horizontally and vertically.

Hint for above problem:

```
boolean[][][]palindrome;//Is there any palindrome ending at (i , j) has length k
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
palindrome[i][j][0] = true;
palindrome[i][j][1] = true;
for(int k = 2; k <= n; k++)
if(data[i][j - k + 1] == data[i][j] && palindrome[i][j - 1][k - 2])
palindrome[i][j][k] = true;
}
}
```

So, we can create two three dimensional arrays `int[n][n][n]col`

and `int[n][n][n]row`

.

For each cell(i, j), we will calculate the total number of palindrome with length k, ending at cell (0, j), (1, j) , ... (i, j) and total number of palindrome with length k, ending at cell (i,0), (i, 1), ... (i, j)

```
for(int k = 1; k <= n; k++)
if(there is palindrome length k horizontally, end at cell (i, j))
row[i][j][k] = 1 + row[i - 1][j][k];
if(there is palindrome length k vertically, end at cell (i, j))
col[i][j][k] = 1 + col[i][j - 1][k];
```

Finally, `if row[i][j][k] >= k && col[i][j][k] >= k`

-> there is an square palindrome length k ending at (i,j).

In total, the time complexity will be **O(n^3)**

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