4092 - Binary Digit Score (Easy)Copy Anchor Link

Problem LinkCopy Anchor Link

https://leetcode.com/problems/binary-digit-score

Problem StatementCopy Anchor Link

You are given two integers a and b.

Let s1 and s2 be the number of set bits in the binary representations of a and b, respectively.

The score of these integers is defined as follows:

• If s1 >= s2, the score is s1 * s2.
• Otherwise, the score is s2 / s1 using integer division.

Return an integer representing the score of a and b.

A set bit refers to a bit in the binary representation of a number that has a value of 1.

Example 1:

Input: a = 14, b = 6

Output: 6

Explanation:

• The binary representation of a = 14 is "1110", so s1 = 3.
• The binary representation of b = 6 is "110", so s2 = 2.
• Since s1 > s2, the score of a and b is s1 * s2 = 3 * 2 = 6.

Example 2:

Input: a = 5, b = 15

Output: 2

Explanation:

• The binary representation of a = 5 is "101", so s1 = 2.
• The binary representation of b = 15 is "1111", so s2 = 4.
• Since s1 < s2, the score of a and b is s2 / s1 = 4 / 2 = 2.

Example 3:

Input: a = 10, b = 12

Output: 4

Explanation:

• The binary representation of a = 10 is "1010", so s1 = 2.
• The binary representation of b = 12 is "1100", so s2 = 2.
• Since s1 == s2, the score of a and b is s1 * s2 = 2 * 2 = 4.

Constraints:

• 1 <= a, b <= 100

Approach 1: Bit CountingCopy Anchor Link

Count the number of 1s in the binary representations of $a$ and $b$. Then compare the two counts: multiply them if the first count is greater than or equal to the second; otherwise, divide the second count by the first using integer division.

The time complexity is $O(1)$ and the space complexity is $O(1)$.

class Solution:
    def calculateScore(self, a: int, b: int) -> int:
        s1, s2 = a.bit_count(), b.bit_count()
        return s1 * s2 if s1 >= s2 else s2 // s1