finds total set bits in given number in O(N) time

Author: NirmalSilwal

Leetcode/DailyChallenges/March2022/Day1CountingBitsLC338.java

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+package LeetcodeDaily.March2022;
+
+import java.util.Arrays;
+
+public class Day1CountingBitsLC338 {
+
+	// O(N) time | O(N) space
+	public static int[] countBits(int n) {
+
+		int[] countOfOnes = new int[n + 1];
+		countOfOnes[0] = 0;
+
+		for (int i = 1; i <= n; i++) {
+
+			if (i % 2 == 0) {
+				// for even number, LSB will be zero, so doing right shift of number by 1 bit
+				// will not lose set one bit(1)
+				countOfOnes[i] = countOfOnes[i / 2];
+			} else {
+				// for odd number, LSB will always be one, so doing right shift of number by 1
+				// bit will also lose 1 bit
+				countOfOnes[i] = 1 + countOfOnes[i / 2];
+			}
+		}
+
+		return countOfOnes;
+	}
+
+	public static void main(String[] args) {
+		System.out.println(Arrays.toString(countBits(15)));
+	}
+
+}