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| -rw-r--r-- | challenge-055/luca-ferrari/raku/ch-1.p6 | 51 |
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diff --git a/challenge-055/luca-ferrari/raku/ch-1.p6 b/challenge-055/luca-ferrari/raku/ch-1.p6 new file mode 100644 index 0000000000..82ea461143 --- /dev/null +++ b/challenge-055/luca-ferrari/raku/ch-1.p6 @@ -0,0 +1,51 @@ +#!env raku +# +# You are given a binary number B, +# consisting of N binary digits 0 or 1: s0, s1, …, s(N-1). +# Choose two indices L and R such that 0 ≤ L ≤ R < N +# and flip the digits s(L), s(L+1), …, s(R). +# By flipping, we mean change 0 to 1 and vice-versa. +# +# For example, given the binary number 010, +# the possible flip pair results are listed below: +# +# L=0, R=0 the result binary: 110 +# L=0, R=1 the result binary: 100 +# L=0, R=2 the result binary: 101 +# L=1, R=1 the result binary: 000 +# L=1, R=2 the result binary: 001 +# L=2, R=2 the result binary: 011 +# +# Write a script to find the indices (L,R) +# that results in a binary number with maximum number of 1s. +# If you find more than one maximal pair L,R then print all of them. +# +# Continuing our example, note that we had three pairs +# (L=0, R=0), (L=0, R=2), and (L=2, R=2) that resulted +# in a binary number with two 1s, which was the maximum. +# So we would print all three pairs. + + +sub MAIN( Int:D $number where { $number.chars > 1 && $number ~~ / ^ <[01]>+ $ / } ) { + say "Number $number"; + + my @src = $number.comb; + my $N = @src.elems; + say @src; + say $N; + + my $wanted = $N - 1; + say "Searching for at least $wanted ones..."; + + for 0 ..^ $N -> $L { + my @working = @src; + @working[ $L ] = @working[ $L ] == 1 ?? 0 !! 1; + + for $L ..^ $N -> $R { + @working[ $R ] = @working[ $R ] == 1 ?? 0 !! 1 if ( $L != $R ); + say "Found { @working } with L = $L and R = $R" if @working.grep( * == 1 ).elems >= $wanted; + } + + + } +} |