dynamic programming for counting-bits

Author: sean424

counting-bits/bhyun-kim.py

@@ -2,7 +2,7 @@
 338. Counting Bits
 https://leetcode.com/problems/counting-bits/description/
 
-Solution:
+Solution 1:
     - Convert the number to binary string
     - Sum the number of 1s in the binary string
     - Append the sum to the output list
@@ -14,12 +14,7 @@
 
 Space complexity: O(n)
     - The output list has n elements
-"""
-
-
-from typing import List
-
-
+    
 class Solution:
     def countBits(self, n: int) -> List[int]:
         output = []
@@ -29,3 +24,42 @@ def countBits(self, n: int) -> List[int]:
             output.append(_sum)
 
         return output
+"""
+
+"""
+Solution 2
+    We can solve this problem with dynamic programming. 
+    1. Initialize output with n elements 
+    2. The first element is 0 because iteration starts from zero. 
+    3. Iterate from 1 to n+1 
+    4. The last digit of each number is 0 for even number 1 for odd number 
+        So add (i & 1) to the output 
+    5. The digits except the last one can be found when the number is divided by two. 
+        Instead for division by two, we can use one step of bit shift to the right. 
+
+    0 = 00000
+    1 = 00001
+    2 = 00010
+    3 = 00011
+    4 = 00100
+    5 = 00101
+    6 = 00110
+    7 = 00111
+
+Time complexity: O(n)
+    - The for loop runs n times
+
+Space complexity: O(n)
+    - The output list has n elements
+"""
+
+from typing import List
+
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        output = [0] * (n+1)
+
+        for i in range(1, n+1):
+            output[i] = output[i >> 1] + (i & 1)
+
+        return output