🩹 fixed java > counting bits > Approach 3

Author: htadg

java/counting-bits.md

@@ -1,9 +1,23 @@
 # [Leetcode 338: Counting Bits](https://leetcode.com/problems/counting-bits/)
 
 ## Approaches
-- [Approach 1: Naive Approach](#approach-1-naive-approach)
-- [Approach 2: DP with Last Significant Bit](#approach-2-dp-with-last-significant-bit)
-- [Approach 3: DP with Highest Power of Two](#approach-3-dp-with-highest-power-of-two)
+- [Leetcode 338: Counting Bits](#leetcode-338-counting-bits)
+  - [Approaches](#approaches)
+    - [Approach 1: Naive Approach](#approach-1-naive-approach)
+      - [Intuition](#intuition)
+      - [Code](#code)
+      - [Time Complexity](#time-complexity)
+      - [Space Complexity](#space-complexity)
+    - [Approach 2: DP with Last Significant Bit](#approach-2-dp-with-last-significant-bit)
+      - [Intuition](#intuition-1)
+      - [Code](#code-1)
+      - [Time Complexity](#time-complexity-1)
+      - [Space Complexity](#space-complexity-1)
+    - [Approach 3: DP with Highest Power of Two](#approach-3-dp-with-highest-power-of-two)
+      - [Intuition](#intuition-2)
+      - [Code](#code-2)
+      - [Time Complexity](#time-complexity-2)
+      - [Space Complexity](#space-complexity-2)
 
 ---
 
@@ -81,9 +95,9 @@ public int[] countBits(int n) {
     for (int i = 1; i <= n; i++) {
         if (i == pow) {
             pow *= 2; // Move to the next power of two
-            x = 1;
+            x = i;
         }
-        res[i] = res[x++] + 1; // Use the result from a smaller index plus one
+        res[i] = res[i-x] + 1; // Use the result from a smaller index plus one
     }
     return res;
 }