Solution of finding first and last occurrence of number

Author: MohammedHassan07

LeetCode/Occurrence.java

@@ -0,0 +1,62 @@
+import java.util.Arrays;
+
+public class Occurrence {           // Question: Find the first and last occurrence of the target number in array in O(log n) time complexity.
+
+    public static void main(String[] args) {
+
+        int[] arr = {1, 1, 2, 3, 5, 6, 8, 8, 10, 11, 12, 13, 13};       // Initializing and declaring the sorted array
+
+//        int[] arr = new int[20];      // Since array is initialized with zero it will print [0 19]
+
+        System.out.println(Arrays.toString(occurrence(arr, 4)));    // Calling the function and printing the return value
+    }
+
+    public static int[] occurrence(int[] arr, int target) {
+
+        int[] pos = {-1, -1};
+
+        pos[0] = searchElement(arr, target, true);
+
+        if(pos[0] != -1){       // If the target is not present in left side then no need to check the target in right side of the array
+                                // because if target present then the index value will not be -1
+                                // This check makes the program much faster
+
+            pos[1] = searchElement(arr, target, false);
+        }
+
+        return pos;
+    }
+
+    // Function to find the target element
+    public static int searchElement(int[] arr, int target, boolean check) {
+
+        int start = 0;
+        int end = arr.length - 1;
+        int occur = -1;
+
+        while(start <= end) {
+
+            int midElement = start + (end - start) / 2;     // Finding the middle element
+
+            if(target < arr[midElement]) {
+
+                end = midElement - 1;
+            }else if(target > arr[midElement]) {
+
+                start = midElement + 1;
+            }else {
+
+                occur = midElement;
+
+                if(check) {     // if target is present in array it might be possible that the more target elements are at left side and right side of the
+                                // array so we check that, is target present in left side
+                    end = midElement - 1;
+                }else {
+
+                    start = midElement + 1;     // Here we check that, is target present at right side
+                }
+            }
+        }
+        return occur;
+    }
+}