Merge pull request #3067 from viet2307/java/leetcode-0540

Create 0540-single-element-in-a-sorted-array.java

Author: sujal-goswami

java/0540-single-element-in-a-sorted-array.java

@@ -0,0 +1,47 @@
+/*
+ * Time complexity: O(n)
+ * Space complexity: O(1)
+ */
+
+class Solution {
+    public int singleNonDuplicate(int[] nums) {
+        int n = nums.length;
+        int left = 0, right = n-1;
+
+        // if nums only has one element
+        if(n < 2) return nums[0];
+        
+        // if the answer lies at either end of the array
+        if(nums[left+1] != nums[left]) return nums[left];
+        if(nums[right-1] != nums[right]) return nums[right];
+
+        // when you see requirement of O(log n) time and O(1) space
+        // most of the time you gonna need binary search
+        // either that or some hacky method using math or bitwise
+        while(left < right) {
+            int mid = left + (right-left)/2;
+            int curr = nums[mid];
+            // we already check either end so mid will never gonna be out of array's bound
+            int prev = nums[mid-1];
+            int next = nums[mid+1];
+
+            // if mid lands on the result, just return it
+            if(prev != curr && curr != next) return curr;
+
+            // because there are no others criteria
+            // remember to check the array's index to see if you miss anything
+            // you can see that the even index should always be the start of the duplication
+            // so if the sequence is broken: left of odd index is still a duplicate, the sequence
+            // was broken before that index -> right = mid and vice versa
+            if(mid % 2 == 0) {
+                if(curr != prev) left = mid+1;
+                else right = mid;
+                continue;
+            }
+
+            if(curr == prev) left = mid+1;
+            else right = mid;
+        }
+        return nums[left];
+    }
+}