improve UniqueArrayElement.java

Basic/UniqueArrayElement.java

@@ -11,9 +11,27 @@ class UniqueArrayElement{
         for(int i=0;i<n;i++){
             arr[i] = sc.nextInt();
         }
-        System.out.println("The unique element in the array is "+findUnique(arr));
+        sc.close();
+        // is there any way to show the time taken by each function to find the unique element in the array?
+        // yes, we can use the System.currentTimeMillis() function to get the current time in milliseconds
+        // we can use this function to get the time before and after the function call
+        // then we can subtract the two times to get the time taken by the function to execute
+        // can you rewrite the above code to show the time taken by each function to find the unique element in the array?
+        // write the code to show the time taken by each function to find the unique element in the array
+        long start = System.currentTimeMillis();
+        System.out.println("The unique element in the array is "+findUniqueUsingXOR(arr));
+        long end = System.currentTimeMillis();
+        System.out.println("Time taken by findUnique function is "+(end-start)+" milliseconds");
+        start = System.currentTimeMillis();
+        System.out.println("The unique element in the array is "+findUniqueUsingMapping(arr));
+        end = System.currentTimeMillis();
+        System.out.println("Time taken by findUniqueUsingMapping function is "+(end-start)+" milliseconds");
+        start = System.currentTimeMillis();
+        System.out.println("The unique element in the array is "+findUniqueBruteforce(arr));
+        end = System.currentTimeMillis();
+        System.out.println("Time taken by findUniqueBruteforce function is "+(end-start)+" milliseconds");
     }
-    public static int findUnique(int[] arr){
+    public static int findUniqueUsingXOR(int[] arr){
         int res = 0;
         for(int i=0;i<arr.length;i++){
             res = res^arr[i];
@@ -33,4 +51,51 @@ class UniqueArrayElement{
     // res = 7^3 = 4
     // return 4
 
+    // write another function using hasing technique to find the unique element in the array
+    // write using mapping technique to find the unique element in the array
+    // DO NOT USE XOR
+    // make another array of size of the largest element in the array and then map the elements of the array to the new array
+    // then find the element which is mapped to 1
+    // return that element
+    // return -1 if no such element is found
+    // write the code for the above logic
+    static int findUniqueUsingMapping(int[] arr){
+        int max = Integer.MIN_VALUE;
+        for(int i=0;i<arr.length;i++){
+            if(arr[i]>max){
+                max = arr[i];
+            }
+        }
+        int[] map = new int[max+1];
+        for(int i=0;i<arr.length;i++){
+            map[arr[i]]++;
+        }
+        for(int i=0;i<map.length;i++){
+            if(map[i]==1){
+                return i;
+            }
+        }
+        return -1;
+    }
+
+    // write a bruteforce method to find the unique element in the array
+    // compare each element with all the other elements in the array
+    // if the element is not found in the array then return that element
+    // return -1 if no such element is found
+    static int findUniqueBruteforce(int[] arr){
+        for(int i=0;i<arr.length;i++){
+            boolean found = false;
+            for(int j=0;j<arr.length;j++){
+                if(i!=j && arr[i]==arr[j]){
+                    found = true;
+                    break;
+                }
+            }
+            if(!found){
+                return arr[i];
+            }
+        }
+        return -1;
+    }
+
 }