updated Trie problem

Java/Data Structures/Java Dequeue/Solution.java

@@ -27,7 +27,7 @@ public class test {
         for (int i = 0; i < n; i++) {
             /* Remove old value (if necessary) */
             if (i >= m) {
-                int old = deque.remove();
+                int old = deque.removeFirst();
                 if (map.get(old) == 1) {
                     map.remove(old);
                 } else {
@@ -37,7 +37,7 @@ public class test {
             
             /* Add new value */
             int num = scan.nextInt();
-            deque.add(num);
+            deque.addLast(num);
             map.merge(num, 1, Integer::sum);
             
             max = Math.max(max, map.size());

Java/Data Structures/Java Priority Queue/Solution.java

@@ -64,10 +64,9 @@ public class Solution {
 }
 
 class StudentComparator implements Comparator<Student> {
-    double epsilon = 0.001; // since we shouldn't use "==" with doubles
     @Override
     public int compare(Student s1, Student s2) {
-        if (Math.abs(s1.getCgpa() - s2.getCgpa()) > epsilon) {
+        if (Double.compare(s1.getCgpa(), s2.getCgpa()) != 0) {
             return s1.getCgpa() < s2.getCgpa() ? 1 : -1; // descending order
         } else if ( ! s1.getFname().equals(s2.getFname())) {
             return s1.getFname().compareTo(s2.getFname());

Java/Data Structures/Java Subarray/Solution.java

@@ -4,16 +4,15 @@
 
 import java.util.Scanner;
 
-// Our subarray must be contiguous, so we have O(n^2) contiguous subarrays
+// A subarray must be contiguous. There are O(n^2) contiguous subarrays.
 
 //  Time Complexity: O(n^2)
 // Space Complexity: O(1)
 public class Solution {
     public static void main(String[] args) {
-        /* Save input */
         Scanner scan = new Scanner(System.in);
         int size     = scan.nextInt();
-        int [] array = new int[size];        
+        int[] array = new int[size];        
         for (int i = 0; i < size; i++) {
             array[i] = scan.nextInt();
         }
@@ -22,17 +21,16 @@ public class Solution {
         System.out.println(negativeSubarrays(array));
     }
     
-    private static int negativeSubarrays(int [] array) {
+    private static int negativeSubarrays(int[] array) {
         int count = 0;
-        int sum   = 0;
         for (int i = 0; i < array.length; i++) {
+            int sum = 0;
             for (int j = i; j < array.length; j++) {
                 sum += array[j];
                 if (sum < 0) {
                     count++;
                 }
             }
-            sum = 0;
         }
         return count;
     }