update Watching Mooloo

solutions/bronze/usaco-1301.mdx

@@ -9,9 +9,9 @@ author: Chu Minjae
 
 ## Explanation
 
-We solve this problem with a greedy algorithm by first sorting all the days Bessie wants to watch Mooloo.
+We solve this problem with a greedy algorithm.
 
-Next, we iterate through all the days.
+First, we iterate through all the days.
 On the first day, Bessie will always have to buy a new subscription.
 However, on subsequent days, there's two cases:
 1. It's better to extend the last subscription.
@@ -38,21 +38,19 @@ int main() {
 	vector<long long> days(n);
 	for (long long &d : days) { cin >> d; }
 
-	sort(days.begin(), days.end());
-
 	long long last_day = days[0];
 	long long cost = k + 1;  // Start the first subscription
-	for (long long d : days) {
+	for (int i = 1; i < n; i++) {
 		// Should Bessie extend the most recent subscription?
-		if (d - last_day < k + 1) {
-			cost += d - last_day;
+		if (days[i] - last_day < k + 1) {
+			cost += days[i] - last_day;
 		} else {
 			// Or just start a new one entirely?
 			cost += k + 1;
 		}
 
 		// Store the date of the last subscription
-		last_day = d;
+		last_day = days[i];
 	}
 
 	cout << cost << endl;
@@ -78,21 +76,19 @@ public class WatchingMooloo {
 		br.close();
 		for (int i = 0; i < n; i++) { days[i] = Long.parseLong(daysInput[i]); }
 
-		Arrays.sort(days);
-
 		long lastDay = days[0];
 		long cost = k + 1;  // Start the first subscription
-		for (long d : days) {
+		for (int i = 1; i < n; i++) {
 			// Should Bessie extend the most recent subscription?
-			if (d - lastDay < k + 1) {
-				cost += d - lastDay;
+			if (days[i] - lastDay < k + 1) {
+				cost += days[i] - lastDay;
 			} else {
 				// Or just start a new one entirely?
 				cost += k + 1;
 			}
 
 			// Store the date of the last subscription
-			lastDay = d;
+			lastDay = days[i];
 		}
 
 		System.out.println(cost);
@@ -101,4 +97,28 @@ public class WatchingMooloo {
 ```
 
 </JavaSection>
+<PySection>
+
+```py
+n, k = map(int, input().split())
+days = list(map(int, input().split()))
+
+last_day = days[0]
+cost = k + 1  # Start the first subscription
+
+for i in range(1, n):
+	# Should Bessie extend the most recent subscription?
+	if days[i] - last_day < k + 1:
+		cost += days[i] - last_day
+	else:
+		# Or just start a new one entirely?
+		cost += k + 1
+
+	# Store the date of the last subscription
+	last_day = days[i]
+
+print(cost)
+```
+
+</PySection>
 </LanguageSection>