Update Array Manipulation Notes.md

Grade 10/Computer Science/ICS4U1/Array Manipulation Notes.md

@@ -20,7 +20,7 @@ String array[] = {"s", "d", "f"};
 
 ## String Manipulataion
 
-- ```length()` to find the length of a string
+- `length()` to find the length of a string
 - `stringName.charAt(0)` will output the first character 
 - Each character in a string has a positio. String positions start  at 0
 - `stringName.indexOf("T")` will output the index number that the letter `T` belongs to
@@ -28,16 +28,81 @@ String array[] = {"s", "d", "f"};
 - Eg. `string stringName = "Jim hates math"`
     - Output: `hates math`
 
-- First # is for the index that you want and last number is for the index that you don't wnat
+- First # is for the index that you want and last number is for the index that you don't want
 - `stringName2 = "     i hate math         "`
-- `stringName2.trim(); 
-    - output: i hate math"`
+- `stringName2.trim();` 
+    - Output: `i hate math`
 - `stringName.toLowerCase()` makes every character a lower case letter
 - `stringName.toUpperCase()` makes every character an upper case letter
 
-- concat()
+- `concat()`
 - `stringname1.concat(stringname2);` will put both string into 1 word
-- `stringname.replce("a", "p");` will replace every `'a'` with `'p'`
+- `stringname.replace("a", "p");` will replace every `a` with `p`
 - `stringname1.equals(stringname2);` will return true of false depending on whether the two contain the same value
 - `stringname1.compareTo(stringname2);` is for alphabetical order
-    - if the condition above is < 0, name1 is before name2. >0, name1 is before name2, if =0, its the same string
+    - if the condition above is $`< 0`$, name1 is before name2. $`> 0`$, name1 is after name2, if $`=0`$, its the same string
+
+## Sorting
+```java
+static void ImprovedBubbleSorted(int arr[]) {
+    boolean isSorted = false;
+    int j = 1;
+    while(!isSorted) {
+        isSorted = true;
+        for(int i = 0; i < arr.length - j; i++) {
+            if(arr[i]> arr[i + 1]) {
+                int t = arr[i];
+                arr[i] = arr[i + 1];
+                arr[i + 1] = t;
+
+                isSorted = true;
+            }
+        }
+        j++;
+    }
+}
+```
+
+## Searching
+```java
+boolean found=false;
+System.out.println("Enter the word you want to search");
+String word = sc.next();
+for(int i=0; i<arr.length; i++) {
+    if(arr[i].equals(word)) {
+        found=true;
+    }
+}
+if(found) {
+    System.out.println("The word was found");
+}
+else {
+    System.out.println("The word was not found");
+}
+```
+
+## Searching Continued
+```java
+boolean found=false;
+System.out.println("Enter the word you want to search");
+String word = sc.next();
+int pos=0;
+for(int i=0; i<arr.length; i++) {
+    if(arr[i].equals(word)) {
+        found=true;
+        pos=i;
+    }
+}
+if(found) {
+    System.out.println("The word was found at index: " + pos);
+}
+else {
+    System.out.println("The word was not found");
+}
+```
+
+## Conclusion
+- Arrays are very useful when problem solving when a large numbe of data is involved
+- `Organize` large sets of data
+- Variety of ways you can access information in an array
+