How to resolve the algorithm Jewels and stones step by step in the Prolog programming language

Published on 12 May 2024 09:40 PM
#Prolog

How to resolve the algorithm Jewels and stones step by step in the Prolog programming language

Table of Contents

Problem Statement

Create a function which takes two string parameters: 'stones' and 'jewels' and returns an integer. Both strings can contain any number of upper or lower case letters. However, in the case of 'jewels', all letters must be distinct. The function should count (and return) how many 'stones' are 'jewels' or, in other words, how many letters in 'stones' are also letters in 'jewels'.

Note that: So, for example, if passed "aAAbbbb" for 'stones' and "aA" for 'jewels', the function should return 3. This task was inspired by this problem.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Jewels and stones step by step in the Prolog programming language

Source code in the prolog programming language

:- system:set_prolog_flag(double_quotes,codes) .

count_jewels(STONEs0,JEWELs0,COUNT)
:-
findall(X,(member(X,JEWELs0),member(X,STONEs0)),ALLs) ,
length(ALLs,COUNT)
.


count_jewels(Stones, Jewels, N):-
    string_codes(Stones, Scodes),
    string_codes(Jewels, Jcodes),
    msort(Scodes, SScodes),
    sort(Jcodes, SJcodes),
    count_jewels(SScodes, SJcodes, N, 0).

count_jewels([], _, N, N):-!.
count_jewels(_, [], N, N):-!.
count_jewels([C|Stones], [C|Jewels], N, R):-
    !,
    R1 is R + 1,
    count_jewels(Stones, [C|Jewels], N, R1).
count_jewels([S|Stones], [J|Jewels], N, R):-
    J < S,
    !,
    count_jewels([S|Stones], Jewels, N, R).
count_jewels([_|Stones], Jewels, N, R):-
    count_jewels(Stones, Jewels, N, R).

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