Solutions to challenge 080, both parts, in tcl

2 files changed, 39 insertions, 0 deletions

diff --git a/challenge-080/shawn-wagner/tcl/ch1.tcl b/challenge-080/shawn-wagner/tcl/ch1.tcl
new file mode 100755
index 0000000000..df578d5e42
--- /dev/null
+++ b/challenge-080/shawn-wagner/tcl/ch1.tcl
@@ -0,0 +1,20 @@
+#!/usr/bin/env tclsh
+
+# Perl one used hashes, so let's take a different approach in the tcl
+# solution; a sorted list.
+
+proc task1 args {
+    set N [lsort -integer [lmap n $args { if {$n <= 0} { continue }; set n }]]
+    set i 1
+    set n 0
+    set len [llength $N]
+    while {$n < $len && $i == [lindex $N $n]} {
+        incr n
+        incr i
+    }
+    puts $i;
+}
+
+task1 5 2 -2 0
+task1 1 8 -1
+task1 2 0 -1
diff --git a/challenge-080/shawn-wagner/tcl/ch2.tcl b/challenge-080/shawn-wagner/tcl/ch2.tcl
new file mode 100755
index 0000000000..0f4837d31f
--- /dev/null
+++ b/challenge-080/shawn-wagner/tcl/ch2.tcl
@@ -0,0 +1,19 @@
+#!/usr/bin/env tclsh
+
+# Same approach as the perl solution
+
+proc task2 args {
+    set candies [llength $args]
+    set len $candies
+    for {set n 0} {$n < $len} {incr n} {
+        if {($n > 0 && [lindex $args $n] > [lindex $args $n-1]) ||
+            ($n < $len - 1 && [lindex $args $n] > [lindex $args $n+1])} {
+            incr candies
+        }
+    }
+    puts $candies
+}
+
+
+task2 1 2 2
+task2 1 4 3 2