Bash: 2022 day 4 part 2

2022/RESULTS.txt

@@ -63,3 +63,17 @@ aoc.bash: res=2760
 aoc-c: res=2760
 	time: 0:00.00 real, 0.00 user, 0.00 sys
 	context-switch:	0+1, page-faults: 0+88
+
+=========================================
+================= day04 =================
+=========================================
+
++++++++++++++++++ part 1
+aoc.bash: res=444
+	time: 0:00.04 real, 0.04 user, 0.00 sys
+	context-switch:	5+1, page-faults: 0+338
+
++++++++++++++++++ part 2
+aoc.bash: res=801
+	time: 0:00.04 real, 0.03 user, 0.00 sys
+	context-switch:	0+1, page-faults: 0+336

2022/day04/README.org

@@ -67,8 +67,6 @@ pairs.
 
 Your puzzle answer was =444=.
 
-The first half of this puzzle is complete! It provides one gold star: *
-
 ** --- Part Two ---
 It seems like there is still quite a bit of duplicate work planned.
 Instead, the Elves would like to know the number of pairs that /overlap
@@ -87,4 +85,6 @@ So, in this example, the number of overlapping assignment pairs is =4=.
 
 /In how many assignment pairs do the ranges overlap?/
 
-Answer:
+Your puzzle answer was =801=.
+
+Both parts of this puzzle are complete! They provide two gold stars: **

2022/day04/aoc.bash

@@ -22,11 +22,9 @@ parse() {
 
     while IFS=-, read -ra _arr; do
         # arrange the two sections so that the lowest is the first
-        printf "_arr=%s\n" "${_arr[*]}"
         if ((_arr[0] > _arr[2])); then
             ((_tmp=_arr[0], _arr[0]=_arr[2], _arr[2]=_tmp))
             ((_tmp=_arr[1], _arr[1]=_arr[3], _arr[3]=_tmp))
-            printf "\t->%s\n" "${_arr[*]}"
         fi
         sections+=("${_arr[*]}")
     done
@@ -45,6 +43,17 @@ part1() {
     done
 }
 
+part2() {
+    declare -ig res=0
+    local -a _sect
+
+    for line in "${sections[@]}"; do
+        # shellcheck disable=SC2206
+        _sect=($line)
+        (( _sect[2] <= _sect[1] )) && (( res++ ))
+    done
+}
+
 solve() {
     if (($1 == 1)); then
         part1