Add Awk solution to challenge 001

2 files changed, 48 insertions, 0 deletions

diff --git a/challenge-001/paulo-custodio/awk/ch-1.awk b/challenge-001/paulo-custodio/awk/ch-1.awk
new file mode 100644
index 0000000000..c954b16d73
--- /dev/null
+++ b/challenge-001/paulo-custodio/awk/ch-1.awk
@@ -0,0 +1,31 @@
+#!/usr/bin/gawk
+
+# Challenge 001
+#
+# Challenge #1
+# Write a script to replace the character ‘e’ with ‘E’ in the string
+# ‘Perl Weekly Challenge’. Also print the number of times the character ‘e’
+# is found in the string.
+
+function join(array, start, end, sep,    result, i)
+{
+    if (sep == "")
+       sep = " "
+    else if (sep == SUBSEP) # magic value
+       sep = ""
+    result = array[start]
+    for (i = start + 1; i <= end; i++)
+        result = result sep array[i]
+    return result
+}
+
+function alen(a, i) {
+    for(i in a);
+    return i
+}
+
+BEGIN {
+    text = join(ARGV, 1, alen(ARGV))
+    print gsub(/e/, "E", text) " " text
+    exit 0
+}
diff --git a/challenge-001/paulo-custodio/awk/ch-2.awk b/challenge-001/paulo-custodio/awk/ch-2.awk
new file mode 100644
index 0000000000..f4c9a9fafb
--- /dev/null
+++ b/challenge-001/paulo-custodio/awk/ch-2.awk
@@ -0,0 +1,17 @@
+#!/usr/bin/gawk
+
+# Challenge 001
+#
+# Challenge #2
+# Write a one-liner to solve the FizzBuzz problem and print the numbers 1
+# through 20. However, any number divisible by 3 should be replaced by the word
+# ‘fizz’ and any divisible by 5 by the word ‘buzz’. Those numbers that are both
+# divisible by 3 and 5 become ‘fizzbuzz’.
+
+BEGIN {
+    num = ARGV[1] ? ARGV[1] : 20
+    for (i=1; i<=num; i++) {
+        print (i%15)==0 ? "fizzbuzz" : (i%3)==0 ? "fizz" : (i%5)==0 ? "buzz" : i
+    }
+    exit 0
+}