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-rwxr-xr-xchallenge-080/shawn-wagner/tcl/ch1.tcl20
-rwxr-xr-xchallenge-080/shawn-wagner/tcl/ch2.tcl19
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