Completed Precourse2

Exercise_1.java

@@ -1,8 +1,28 @@
-class BinarySearch { 
+// Time Complexity : O(log n)
+// Space Complexity : O(n) just for the array
+// Your code here along with comments explaining your approach
+/*
+* I try to find the mid element iteratively until my left value gets less than or equal to right index
+* If I find my target element dorectly at mid index, I return it, if not, I compare the value of target
+* with value of mid to see and move my left, right indices to narrow the search criteria. Return -1 if
+* target not found.
+* */
+
+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
+        while(l <= r) {
+            int mid = l + (r - l) / 2;
+            if(x == arr[mid])
+                return mid;
+            else if(arr[mid] < x)
+                l = mid + 1;
+            else
+                r = mid - 1;
+        }
+        return -1;
     } 
 
     // Driver method to test above 
@@ -11,7 +31,7 @@ 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

@@ -1,4 +1,14 @@
-class QuickSort 
+// Time Complexity : O(nlog n) best and average case , O(n^2) worst case
+// Space Complexity : O(n) worst case
+// Your code here along with comments explaining your approach
+/*
+ * The idea is to find a specific partition index and sort using it for first and second halfs recursively
+ * But, to choose this partition, I need to understand the order of the elements. So, I assume a pivot
+ * element first, then try to compare iteratively each element of the array with pivot and make sure to
+ * place/swap all those smaller elements to left side of the array in comparison with pivot. then, I return
+ * that least index element which acts as a partition of smaller sorted elements till now and the rest.
+ * */
+class QuickSort
 { 
     /* This function takes last element as pivot, 
        places the pivot element at its correct 
@@ -7,12 +17,26 @@ class QuickSort
        pivot and all greater elements to right 
        of pivot */
     void swap(int arr[],int i,int j){
-        //Your code here   
+        //Your code here
+        int temp = arr[i];
+        arr[i] = arr[j];
+        arr[j] = temp;
     }
 
     int partition(int arr[], int low, int high) 
     { 
-   	//Write code here for Partition and Swap 
+   	//Write code here for Partition and Swap
+        int i = low - 1;
+        int pivot = arr[high];
+
+        for(int j = low ; j <= high - 1 ; j++) {
+            if(arr[j] < pivot) {
+                i++;
+                swap(arr, i, j);
+            }
+        }
+        swap(arr, i + 1, high);
+        return i + 1;
     } 
     /* The main function that implements QuickSort() 
       arr[] --> Array to be sorted, 
@@ -21,7 +45,12 @@ int partition(int arr[], int low, int high)
     void sort(int arr[], int low, int high) 
     {  
             // Recursively sort elements before 
-            // partition and after partition 
+            // partition and after partition
+        if(low < high) {
+            int pi = partition(arr, low, high);
+            sort(arr, low, pi - 1);
+            sort(arr, pi + 1, high);
+        }
     } 
 
     /* A utility function to print array of size n */

Exercise_3.java

@@ -1,4 +1,12 @@
-class LinkedList 
+// Time Complexity : O(n)
+// Space Complexity : O(1)
+
+// Your code here along with comments explaining your approach
+/*
+I take 2 pointers fast and slow and make sure to move fast pointer quickly and slow pointer just behind
+it to fetch the exact middle element.
+ */
+class LinkedList
 { 
     Node head; // head of linked list 
 
@@ -20,6 +28,16 @@ void printMiddle()
     { 
         //Write your code here
 	//Implement using Fast and slow pointers
+        if(head == null) {
+            System.out.println("List is empty");
+        }
+        Node slow = head;
+        Node fast = head;
+        while(fast != null && fast.next != null) {
+                fast = fast.next.next;
+                slow = slow.next;
+        }
+        System.out.println("Middle elmement is " + slow.data);
     } 
 
     public void push(int new_data) 

Exercise_4.java

@@ -1,19 +1,57 @@
-class MergeSort 
+// Time Complexity : O(nlog n)
+// Space Complexity : O(n) worst case
+// Your code here along with comments explaining your approach
+/*
+ * The idea is to split and sort left and right sub arrays recursively first and then eventually merge
+ *  using 2 pointers making sure to pick smaller element each time, placing it in main array.
+ * */
+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  
+       //Your code here
+        int n1 = m - l + 1;
+        int n2 = r - m;
+
+        int[] N1 = new int[n1];
+        int[] N2 = new int[n2];
+
+        for(int i = 0 ; i < n1 ; i++)
+            N1[i] = arr[l + i];
+        for(int j = 0 ; j < n2 ; j++)
+            N2[j] = arr[m + 1 + j];
+
+        int i = 0, j = 0 , k = l;
+
+        while(i < n1 && j < n2) {
+            if(N1[i] < N2[j]) {
+                arr[k++] = N1[i++];
+            }
+            else
+                arr[k++] = N2[j++];
+        }
+        while(i < n1)
+            arr[k++] = N1[i++];
+        while(j < n2)
+            arr[k++] = N2[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 
+        //Call mergeSort from here
+        if(l < r) {
+            int m = l + (r - l) / 2;
+            sort(arr, l, m);
+            sort(arr, m + 1, r);
+
+            merge(arr, l , m , r);
+        }
     } 
 
     /* A utility function to print array of size n */

Exercise_5.java

@@ -1,20 +1,66 @@
-class IterativeQuickSort { 
+// Time Complexity : O(nlog n)
+// Space Complexity : O(n)
+// Your code here along with comments explaining your approach
+/*
+ * The idea is to use the same partition logic, but instead of recursion, use a stack logic to store (l,h) pairs of
+ * subarrays.Each iteration pops a subarray, partitions it and pushes left/right subarrays
+ * Instead of extra variable, I used XOR logic to swap
+ * */
+class IterativeQuickSort {
     void swap(int arr[], int i, int j) 
     { 
-	//Try swapping without extra variable 
+	//Try swapping without extra variable
+        if(i != j) {
+            arr[i] = arr[i] ^ arr[j];
+            arr[j] = arr[i] ^ arr[j];
+            arr[i] = arr[i] ^ arr[j];
+        }
     } 
 
     /* This function is same in both iterative and 
        recursive*/
     int partition(int arr[], int l, int h) 
     { 
         //Compare elements and swap.
+        int i = l - 1;
+        int pivot = arr[h];
+
+        for(int j = l ; j < h ; j++) {
+            if(arr[j] <= pivot) {
+                i++;
+                swap(arr, i , j);
+            }
+        }
+        swap(arr, i + 1, h);
+        return i + 1;
     } 
 
     // Sorts arr[l..h] using iterative QuickSort 
     void QuickSort(int arr[], int l, int h) 
     { 
         //Try using Stack Data Structure to remove recursion.
+        int[] stack = new int[h - l + 1];
+        int top = -1;
+
+        stack[++top] = l;
+        stack[++top] = h;
+
+        while(top >= 0) {
+            h = stack[top--];
+            l = stack[top--];
+
+            int p = partition(arr, l, h);
+            if(l < p - 1) {
+                stack[++top] = l;
+                stack[++top] = p - 1;
+            }
+            if(p + 1 < h) {
+                stack[++top] = p + 1;
+                stack[++top] = h;
+            }
+        }
+
+
     } 
 
     // A utility function to print contents of arr