Use bitmask to save space on Day 3

odin/day03/day03.odin

@@ -70,36 +70,28 @@ sum_rucksack :: proc(sack: string) -> int {
 }
 
 find_common_item :: proc(sacks: []string) -> int {
-	size := 52 * len(sacks)
-	fmt.printf("Sacks: %d; Size: %d\n", len(sacks), size)
+	size := 52
+	counter := make([]int, size) // Dynamic array initialized to all 0
+	defer delete(counter) // Clean up memory when it goes out of scope
 
-	counter := make([]int, size)
-	defer delete(counter)
-
-	for idx in 0..<size {
-		counter[idx] = 0
+	// Compute the max value for a counter
+	max_counter := 0
+	idx : u32 = 0
+	for _ in 0..<len(sacks) {
+		max_counter = max_counter & (1 << idx)
+		idx += 1
 	}
 
-	for str, offset in sacks {
-		fmt.printf("Sack: %s\n", str)
-		for chr, _ in str {
+	// Check every item in every rucksack
+	for sack, offset_ in sacks {
+		offset := u32(offset_)
+		for chr, _ in sack {
 			prio := priority(chr) - 1
-			fmt.printf("Found char: %c (%d)\n", chr, prio + 1)
-			counter[(len(sacks) * prio) + offset] = 1
-		}
-	}
-
-	fmt.println(counter)
-	for prio in 0..<52 {
-		fmt.printf("Checking prio %d\n", prio)
-		matched := true
-		for offset in 0..<len(sacks) {
-			idx := (len(sacks) * prio) + offset
-			fmt.printf("Checking index %d\n", idx)
-			matched = matched && (counter[idx] == 1)
-		}
-		if matched {
-			return prio + 1
+			counter[prio] = counter[prio] & (1 << offset)
+			// If we hit the max value, we can just quit without checking anything else - this is because per the problem there should only be one item that appears in all three rucksacks
+			if counter[prio] == max_counter {
+				return prio + 1
+			}
 		}
 	}
 	return 0
@@ -430,7 +422,6 @@ main :: proc() {
 		upper_bound := idx + step
 		offset := min(len(input), upper_bound)
 		prio := find_common_item(input[idx:offset])
-		fmt.printf("Priority: %d\n", prio)
 		total += prio
 		idx = upper_bound
 	}