Solution to challenge 075 task 1 and 2 in Raku by Noud

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diff --git a/challenge-075/noud/raku/ch-1.p6 b/challenge-075/noud/raku/ch-1.p6
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+# You are given a set of coins @C, assuming you have infinite amount of each
+# coin in the set.
+#
+# Write a script to find how many ways you make sum $S using the coins from the
+# set @C.
+# Example:
+#
+# Input:
+#     @C = (1, 2, 4)
+#     $S = 6
+#
+# Output: 6
+# There are 6 possible ways to make sum 6.
+# a) (1, 1, 1, 1, 1, 1)
+# b) (1, 1, 1, 1, 2)
+# c) (1, 1, 2, 2)
+# d) (1, 1, 4)
+# e) (2, 2, 2)
+# f) (2, 4)
+
+sub coins-sum(@C, $S) {
+    if ($S == 0) {
+        return [(),];
+    }
+
+    my @ret = [];
+    for @C -> $c {
+        if ($c <= $S) {
+            for coins-sum(@C, $S - $c) -> @r {
+                if ($c <= @r.min) {
+                    @ret.push([$c, |(@r)]);
+                }
+            }
+        }
+    }
+    return @ret.unique;
+}
+
+coins-sum((1, 2, 4), 6).say;
diff --git a/challenge-075/noud/raku/ch-2.p6 b/challenge-075/noud/raku/ch-2.p6
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+# You are given an array of positive numbers @A.
+#
+# Write a script to find the largest rectangle histogram created by the given
+# array.  BONUS: Try to print the histogram as shown in the example, if
+# possible.
+#
+# Example 1:
+#
+# Input: @A = (2, 1, 4, 5, 3, 7)
+#
+#      7           #
+#      6           #
+#      5       #   #
+#      4     # #   #
+#      3     # # # #
+#      2 #   # # # #
+#      1 # # # # # #
+#      _ _ _ _ _ _ _
+#        2 1 4 5 3 7
+#
+# Looking at the above histogram, the largest rectangle (4 x 3) is formed by
+# columns (4, 5, 3 and 7).
+# Output: 12
+#
+# Example 2:
+#
+# Input: @A = (3, 2, 3, 5, 7, 5)
+#
+#      7         #
+#      6         #
+#      5       # # #
+#      4       # # #
+#      3 #   # # # #
+#      2 # # # # # #
+#      1 # # # # # #
+#      _ _ _ _ _ _ _
+#        3 2 3 5 7 5
+#
+# Looking at the above histogram, the largest rectangle (3 x 5) is formed by
+# columns (5, 7 and 5).
+# Output: 15
+
+# Without the BONUS. ;)
+
+sub largest-rec-hist(@A) {
+    my @largest-sub-rec = [];
+    for 0..(@A.elems - 1) -> $i {
+        for ($i + 1)..(@A.elems - 1) -> $j {
+            @largest-sub-rec.push(($j - $i + 1) * @A[$i..$j].min);
+        }
+    }
+    return @largest-sub-rec.max;
+}
+
+largest-rec-hist((2, 1, 4, 5, 3, 7)).say;
+largest-rec-hist((3, 2, 3, 5, 7, 5)).say;