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+# Task 1 of the Weekly Challenge 342
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-342/#TASK1
+
+say balance-string('a0b1c2'); # 0a1b2c
+say balance-string('abc12'); # 0a1b2c
+say balance-string('0a2b1c3'); # 0a1b2c3
+say balance-string('1a23'); # ''
+say balance-string('ab123'); # 1a2b3
+
+# Extra case with 3 letters and 3 digits
+say balance-string('abc123'); # 1a2b3c
+
+
+sub balance-string($s) {
+    my @digits = $s.comb(/\d/).sort;
+    my @letters = $s.comb(/<[a..z]>/).sort;
+
+    # It's only possible to balance a string if
+    # the number of digits and letters differs by not more than 2.
+    if ((@digits.elems - @letters.elems).abs > 1) {
+        return '';
+    }
+    elsif (@digits.elems == @letters.elems) {
+        # Assuming that alphabetical sort puts digits first
+        return [~] (@digits Z~ @letters);
+    }
+
+    # Now the two cases where the size is different, but the
+    # difference of sizes is always 1 here. So appending the extra character.
+    elsif (@digits.elems > @letters.elems) {
+        return ([~] (@digits Z~ @letters)) ~ @digits[*-1];
+    }
+    else {
+        return ([~] (@letters Z~ @digits)) ~ @letters[*-1];
+    }
+}