A Collection of Code Snippets in as Many Programming Languages as Possible

This project is maintained by TheRenegadeCoder

Welcome to the Longest Palindromic Substring in Commodore Basic page! Here, you'll find the source code for this program as well as a description of how the program works.

```
10 DIM M%(99, 99): REM Match array
20 GOSUB 1000: REM Input string
30 IF S$ = "" THEN GOTO 200
40 GOSUB 2000: REM Get longest palindromic substring
50 IF LEN(R$) < 2 THEN GOTO 200
60 PRINT R$
70 END
200 PRINT "Usage: please provide a string that contains at least ";
210 PRINT "one palindrome"
220 END
1000 REM Read input value one character at a time since Commodore BASIC
1001 REM has trouble reading line from stdin properly
1002 REM S$ = string
1003 REM Initialize
1010 S$ = ""
1015 REM Append characters until end of input
1020 GET A$
1030 C = ASC(A$)
1040 IF C = 13 OR C = 255 THEN RETURN: REM end of value or input
1050 S$ = S$ + A$
1060 GOTO 1020
2000 REM Find longest palindromic substring using matching array
2001 REM Source:
2002 REM https://www.geeksforgeeks.org/
2003 REM longest-palindromic-substring-using-dynamic-programming/
2004 REM Inputs: S$ contains string
2005 REM Outputs:
2006 REM - R$ contains longest palindromic substring
2010 REM Indicate all length 1 strings match and everything else does not
2020 N = LEN(S$)
2030 ML% = 1: REM Maximum length
2040 SI% = 1: REM Start index
2050 IF N < 1 THEN GOTO 2420
2060 FOR I = 0 TO N - 1
2070 FOR J = 0 TO N - 1
2080 M%(I, J) = 0
2090 NEXT J
2100 M%(I, I) = 1
2110 NEXT I
2120 REM Convert string to lowercase
2130 T$ = ""
2140 FOR I = 1 TO N
2150 C$ = MID$(S$, I, 1)
2160 IF C$ >= "A" AND C$ <= "Z" THEN C$ = CHR$(ASC(C$) + 32)
2170 T$ = T$ + C$
2180 NEXT I
2190 REM Find all length 2 matches
2200 IF N < 2 THEN GOTO 2420
2210 FOR I = 0 TO N - 2
2220 IF MID$(T$, I + 1, 1) <> MID$(T$, I + 2, 1) THEN GOTO 2270
2230 M%(I, I + 1) = 1
2240 IF ML >= 2 THEN GOTO 2270
2250 ML% = 2
2260 SI% = I + 1
2270 NEXT I
2280 REM Find all length 3 or higher matches
2290 IF N < 3 THEN GOTO 2420
2300 FOR L = 3 TO N
2310 REM Loop through each starting character
2320 FOR I = 0 TO N - L
2330 J = I + L - 1
2340 IF M%(I + 1, J - 1) = 0 THEN GOTO 2400
2350 IF MID$(T$, I + 1, 1) <> MID$(T$, J + 1, 1) THEN GOTO 2400
2360 M%(I, J) = 1
2370 IF L <= ML% THEN GOTO 2400
2380 ML% = L
2390 SI% = I + 1
2400 NEXT I
2410 NEXT L
2420 R$ = MID$(S$, SI%, ML%)
2430 RETURN
```

Longest Palindromic Substring in Commodore Basic was written by:

- rzuckerm

If you see anything you'd like to change or update, please consider contributing.

No 'How to Implement the Solution' section available. Please consider contributing.

No 'How to Run the Solution' section available. Please consider contributing.