Precourse-2 Done

Exercise_1.java

@@ -1,8 +1,33 @@
-class BinarySearch { 
+// Time Complexity : O(log n), assuming input array is sorted
+// Space Complexity : O(1), no additional space needed
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+class BinarySearch {
     // Returns index of x if it is present in arr[l.. r], else return -1 
     int binarySearch(int arr[], int l, int r, int x) 
-    { 
-        //Write your code here
+    {
+        int length = arr.length;
+        int result = -1;
+        if(length == 0) return result;
+        while(l <= r){
+            // Find the mid of the sorted array
+            int mid = (l+r)/2;
+
+            // return mid index if it equals the target value
+            if(arr[mid] == x){
+                return mid;
+            }
+
+            // search in the second of the array if target > mid value, else search in other half
+            if(x < arr[mid]){
+                r = mid-1;
+            }else {
+                l = mid+1;
+            }
+        }
+
+        return result;
     } 
 
     // Driver method to test above 
@@ -11,7 +36,7 @@ public static void main(String args[])
         BinarySearch ob = new BinarySearch(); 
         int arr[] = { 2, 3, 4, 10, 40 }; 
         int n = arr.length; 
-        int x = 10; 
+        int x = 10;
         int result = ob.binarySearch(arr, 0, n - 1, x); 
         if (result == -1) 
             System.out.println("Element not present"); 

Exercise_2.java

@@ -1,27 +1,58 @@
-class QuickSort 
-{ 
-    /* This function takes last element as pivot, 
-       places the pivot element at its correct 
-       position in sorted array, and places all 
-       smaller (smaller than pivot) to left of 
-       pivot and all greater elements to right 
-       of pivot */
+// Time Complexity : Average case - O(n logn), Worst Case - O(n^2) - based on what we choose as pivot
+// Space Complexity : In place swapping, 0(1)
+// Did this code successfully run on Leetcode :
+// Ran into time limit exceeded error
+// Any problem you faced while coding this : No
+class QuickSort
+{
+    /* A utility function to swap elements at i and j index */
     void swap(int arr[],int i,int j){
-        //Your code here   
+        int temp = arr[i];
+        arr[i] = arr[j];
+        arr[j] = temp;
     }
-
+
+    /* This function takes last element as pivot,
+       places the pivot element at its correct
+       position in sorted array, and places all
+       smaller (smaller than pivot) to left of
+       pivot and all greater elements to right
+       of pivot */
     int partition(int arr[], int low, int high) 
-    { 
-   	//Write code here for Partition and Swap 
+    {
+        // select last element in the array for comparison
+        int lastIndex = high;
+        while (low < high) {
+            // find first larger value(than the last value) from left
+            while(low < high && arr[low] <= arr[lastIndex]){
+                low++;
+            }
+            // find first smaller value(than the last value) from right
+            while(low < high && arr[high] >= arr[lastIndex]){
+                high--;
+            }
+            if(low >= high) break;
+            // swap the larger and smaller value, so that all smaller values
+            // are on left and larger values are on right
+            swap(arr, low, high);
+        }
+
+        return low;
     } 
     /* The main function that implements QuickSort() 
       arr[] --> Array to be sorted, 
       low  --> Starting index, 
-      high  --> Ending index */
+      high  --> Ending index
+      Find the pivot, swap the pivot value with last element, so that array is divided into 2 halves,
+      sort the left side of the pivot and the right side of the pivot*/
     void sort(int arr[], int low, int high) 
-    {  
-            // Recursively sort elements before 
-            // partition and after partition 
+    {
+        if(low >= high) return;
+
+        int pivotIndex = partition(arr, low, high);
+        swap(arr, pivotIndex, high);
+        sort(arr, low, pivotIndex-1);
+        sort(arr, pivotIndex+1, high);
     } 
 
     /* A utility function to print array of size n */
@@ -36,8 +67,8 @@ static void printArray(int arr[])
     // Driver program 
     public static void main(String args[]) 
     { 
-        int arr[] = {10, 7, 8, 9, 1, 5}; 
-        int n = arr.length; 
+        int arr[] = {10, 7, 8, 9, 1, 5};
+        int n = arr.length;
 
         QuickSort ob = new QuickSort(); 
         ob.sort(arr, 0, n-1); 

Exercise_3.java

@@ -1,53 +1,77 @@
-class LinkedList 
-{ 
+// Time Complexity : O(N) - Need to go through all linked list nodes at least once
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+class LinkedList
+{
     Node head; // head of linked list 
-  
+
     /* Linked list node */
-    class Node 
-    { 
-        int data; 
-        Node next; 
-        Node(int d) 
-        { 
-            data = d; 
-            next = null; 
-        } 
-    } 
-  
+    class Node
+    {
+        int data;
+        Node next;
+        Node(int d)
+        {
+            data = d;
+            next = null;
+        }
+    }
+
     /* Function to print middle of linked list */
    //Complete this function
-    void printMiddle() 
-    { 
-        //Write your code here
-	//Implement using Fast and slow pointers
-    } 
-
-    public void push(int new_data) 
-    { 
-        Node new_node = new Node(new_data); 
-        new_node.next = head; 
-        head = new_node; 
-    } 
+    void printMiddle()
+    {
+        if (head == null) {
+            System.out.println("List is empty!!");
+            return;
+        }
+
+        if(head.next == null){
+            System.out.println(head.data);
+            return;
+        }
+
+        // Keep 2 points, slow moves one node at a time, fast moves 2 nodes at a time,
+        // By the time fast pointer reaches the end of the Linked list, slow pointer covers only half distance and
+        // reaches middle.
+        Node slow = head;
+        Node fast = slow.next.next;
+        while(fast != null){
+            fast = fast.next;
+            if(fast == null) break;
+            fast = fast.next;
+            slow = slow.next;
+        }
+        System.out.println(slow.next.data);
+    }
+
+    public void push(int new_data)
+    {
+        Node new_node = new Node(new_data);
+        new_node.next = head;
+        head = new_node;
+    }
+
+    public void printList()
+    {
+        Node tnode = head;
+        while (tnode != null)
+        {
+            System.out.print(tnode.data+"->");
+            tnode = tnode.next;
+        }
+        System.out.println("NULL");
+    }
 
-    public void printList() 
-    { 
-        Node tnode = head; 
-        while (tnode != null) 
-        { 
-            System.out.print(tnode.data+"->"); 
-            tnode = tnode.next; 
-        } 
-        System.out.println("NULL"); 
-    } 
-
-    public static void main(String [] args) 
-    { 
-        LinkedList llist = new LinkedList(); 
-        for (int i=15; i>0; --i) 
-        { 
-            llist.push(i); 
-            llist.printList(); 
-            llist.printMiddle(); 
-        } 
-    } 
+    public static void main(String [] args)
+    {
+        LinkedList llist = new LinkedList();
+        for (int i=15; i>0; --i)
+        {
+            llist.push(i);
+            llist.printList();
+            llist.printMiddle();
+        }
+    }
 } 

Exercise_4.java

@@ -1,19 +1,63 @@
-class MergeSort 
+// Time Complexity : Average case - O(n logn),
+// Space Complexity : Additional array created in the recursive call which can have max "n" elements, O(n)
+// Did this code successfully run on Leetcode :
+// Yes
+// Any problem you faced while coding this : No
+
+class MergeSort
 { 
     // Merges two subarrays of arr[]. 
     // First subarray is arr[l..m] 
     // Second subarray is arr[m+1..r] 
     void merge(int arr[], int l, int m, int r) 
-    {  
-       //Your code here  
+    {   int arrALen = m-l+1;
+        int arrBLen = r-m;
+        int[] arrA = new int[arrALen];
+        int[] arrB = new int[arrBLen];
+
+        // copy the array into 2 sub arrays
+        int startIndex = l;
+        for(int i = 0; i < arrALen; i++){
+            arrA[i] = arr[startIndex++];
+        }
+
+        startIndex = m+1;
+        for(int i = 0; i < arrBLen; i++){
+            arrB[i] = arr[startIndex++];
+        }
+        int i = 0, j = 0, index = l;
+
+        // compare the sorted array and combine
+        while(i < arrALen && j < arrBLen){
+            if(arrA[i] <= arrB[j]){
+                arr[index++] = arrA[i++];
+            }else{
+                arr[index++] = arrB[j++];
+            }
+        }
+
+        while(i < arrALen){
+            arr[index++] = arrA[i++];
+        }
+
+        while(j < arrBLen){
+            arr[index++] = arrB[j++];
+        }
+
     } 
 
     // Main function that sorts arr[l..r] using 
     // merge() 
     void sort(int arr[], int l, int r) 
     { 
-	//Write your code here
-        //Call mergeSort from here 
+	    if(l>=r) return;
+        int mid = (l+r)/2;
+        // Divide the array from the middle
+        sort(arr, l, mid);
+        sort(arr, mid+1, r);
+
+        // merge the array comparing sub arrays
+        merge(arr, l, mid, r);
     } 
 
     /* A utility function to print array of size n */