Solution of Search in rotated sorted array 2 in java with explanation

Author: AkashSingh19

81.Search in a rotated sorted array II.java

@@ -0,0 +1,31 @@
+  public boolean search(int[] nums, int target) {
+        int start = 0, end = nums.length - 1, mid = -1;
+        while(start <= end) {
+            mid = (start + end) / 2;
+            if (nums[mid] == target) {
+                return true;
+            }
+            //If we know for sure right side is sorted or left side is unsorted
+            if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
+                if (target > nums[mid] && target <= nums[end]) {
+                    start = mid + 1;
+                } else {
+                    end = mid - 1;
+                }
+            //If we know for sure left side is sorted or right side is unsorted
+            } else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
+                if (target < nums[mid] && target >= nums[start]) {
+                    end = mid - 1;
+                } else {
+                    start = mid + 1;
+                }
+            //If we get here, that means nums[start] == nums[mid] == nums[end], then shifting out
+            //any of the two sides won't change the result but can help remove duplicate from
+            //consideration, here we just use end-- but left++ works too
+            } else {
+                end--;
+            }
+        }
+        
+        return false;
+    }