search-insert-position.py

Author: Chris Wu

problems/search-insert-position.py

@@ -0,0 +1,49 @@
+"""
+We use the concept of binary search to find the index.
+`l` and `r` indicates the range we are going to search.
+
+First, check if the target is out of range. [0]
+Second, check if the target is the same at `l` or `r`. [1]
+Third, we use a pivot to navigate our `l` and `r`. [2]
+    * If the value on the pivot is larger then the target, we search the left-half.
+    * If the value on the pivot is smaller then the target, we search the right-half.
+Repeat those three steps.
+
+Last, If we could not find the target, the `l` and `r` will collapse (`l==r`) and it will be the value that is closet to the target. [4]
+
+One thing to keep in mind, if we decided to insert to the right of the index `k`, the output will be `k+1`.
+If we decided to insert to the left of the index `k`, the output will be `k`, because we shift all the others to the right.
+
+Time complexity is O(LogN), space complexity is O(1).
+"""
+class Solution(object):
+    def searchInsert(self, nums, target):
+        if nums is None or len(nums)==0: return 0
+        l = 0
+        r = len(nums)-1
+
+        while l<r:
+            #[0]
+            if nums[l]>target: return l
+            if nums[r]<target: return r+1
+
+            #[1]
+            if nums[l]==target: return l
+            if nums[r]==target: return r
+
+            #[2]
+            p = (l+r)/2
+            if nums[p]==target:
+                return p
+            elif nums[p]>target:
+                r = p-1
+            else:
+                l = p+1
+
+        #[4]
+        if nums[l]==target:
+            return l
+        elif nums[l]>target:
+            return l
+        else:
+            return l+1