chore: fix some comments (#5333)

src/main/java/com/thealgorithms/backtracking/ParenthesesGenerator.java

@@ -19,7 +19,7 @@ public final class ParenthesesGenerator {
      */
     public static List<String> generateParentheses(final int n) {
         if (n < 0) {
-            throw new IllegalArgumentException("The number of pairs of parentheses cannot be nagative");
+            throw new IllegalArgumentException("The number of pairs of parentheses cannot be negative");
         }
         List<String> result = new ArrayList<>();
         generateParenthesesHelper(result, "", 0, 0, n);

src/main/java/com/thealgorithms/datastructures/trees/BSTRecursive.java

@@ -95,7 +95,7 @@ public class BSTRecursive {
     }
 
     /**
-     * Serach recursively if the given value is present in BST or not.
+     * Search recursively if the given value is present in BST or not.
      *
      * @param node the current node to check
      * @param data the value to be checked

src/main/java/com/thealgorithms/misc/PalindromePrime.java

@@ -6,7 +6,7 @@ public final class PalindromePrime {
     private PalindromePrime() {
     }
 
-    public static void main(String[] args) { // Main funtion
+    public static void main(String[] args) { // Main function
         Scanner in = new Scanner(System.in);
         System.out.println("Enter the quantity of First Palindromic Primes you want");
         int n = in.nextInt(); // Input of how many first palindromic prime we want

src/main/java/com/thealgorithms/searches/HowManyTimesRotated.java

@@ -17,7 +17,7 @@ import java.util.Scanner;
     from its initial sorted position.
     Eg. For [2,5,6,8,11,12,15,18], 1 rotation gives [5,6,8,11,12,15,18,2], 2 rotations
    [6,8,11,12,15,18,2,5] and so on. Finding the minimum element will take O(N) time but, we can  use
-   Binary Search to find the mimimum element, we can reduce the complexity to O(log N). If we look
+   Binary Search to find the minimum element, we can reduce the complexity to O(log N). If we look
    at the rotated array, to identify the minimum element (say a[i]), we observe that
    a[i-1]>a[i]<a[i+1].