Solved LeetCode 338 using dynamic programming by deriving each bit count from its half and least significant bit with runtime = 2ms.

Author: rajit2004

src/LeetCode/DynamicProgramming/LeetCode_338_CountingBits.java

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+package LeetCode.DynamicProgramming;
+
+public class LeetCode_338_CountingBits {
+    public static void main(String[] args) {
+
+        int n = 5;
+
+        int[] answer = countBits(n);
+
+        for (int bit : answer) {
+            System.out.print(bit + " ");
+        }
+    }
+
+    /*
+        Dynamic Programming Approach :
+
+        Let: bits[i] = Number of set bits in i
+
+        Observation: Dividing a number by 2 removes its least significant bit.
+
+        Therefore, bits[i] = bits[i / 2] + (i % 2)
+
+        where: bits[i / 2] gives the number of set bits after removing the last bit and (i % 2) tells whether the removed bit was 0 or 1.
+     */
+    static int[] countBits(int n) {
+
+        // bits[i] stores the number of set bits in i.
+        int[] bits = new int[n + 1];
+
+        // bits[0] is already 0.
+
+        // Compute the answer for every number from 1 to n.
+        for (int i = 1; i <= n; i++) {
+
+//                Remove the last bit by dividing by 2. Add 1 if the removed bit was 1, otherwise add 0.
+            bits[i] = bits[i / 2] + (i % 2);
+        }
+
+        return bits;
+    }
+}
+
+/*
+---------------------------------------------------------
+Complexity Analysis
+---------------------------------------------------------
+
+Let: n = input number
+
+---------------------------------------------------------
+
+Time Complexity: O(n)
+
+Reason: Each number from 1 to n is processed exactly once.
+
+Overall: O(n)
+
+---------------------------------------------------------
+
+Space Complexity: O(n)
+
+Reason: An array of size (n + 1) is used to store the number of set bits for every integer.
+
+Overall: O(n)
+
+---------------------------------------------------------
+
+Key Observation:
+
+The number of set bits in a number can be derived from the result of half of that number.
+
+Since: bits[i] = bits[i / 2] + (i % 2) every answer is built using a previously computed value, making Dynamic Programming possible.
+
+---------------------------------------------------------
+*/