Here is what each function in your code does: 1. **`bool...

August 31, 2025 at 03:26 AM

// Math Algorithms bool isPowerOfTwo(long n) { return (n > 0) && ((n & (n - 1)) == 0); } long lowerPowerOfTwo(const long &n) { long currentPower = 1; while (currentPower <= n) { currentPower *= 2; } return currentPower / 2; } int findMedian(std::vector<int> numbers) { std::sort(numbers.begin(), numbers.end()); int n = numbers.size(); if (n % 2 == 0) { return (numbers[n / 2 - 1] + numbers[n / 2]) / 2; } else { return numbers[n / 2]; } } Int intSize(int x){ // 1 = 1; 12 = 2; 123 = 3; 1234 = 4 int y = x, size = 0; while(y >= 1){ size++; y /= 10; } return size; }

Here is what each function in your code does:

  1. bool isPowerOfTwo(long n):

    • This function determines if the integer n is a power of two.
    • It works by checking two conditions:
      1. n > 0: Ensures the number is positive.
      2. (n & (n - 1)) == 0: Uses bitwise AND to verify that n has only a single bit set (a property of powers of two).
    • Example:
      • isPowerOfTwo(4)true (4 is (2^2)).
      • isPowerOfTwo(6)false.
  2. long lowerPowerOfTwo(const long &n):

    • This function finds the largest power of two less than or equal to the input n.
    • It iteratively doubles currentPower (starting at 1) until it exceeds n, then returns the previous value (currentPower / 2).
    • Example:
      • lowerPowerOfTwo(15)8 (largest power of two ≤ 15).
      • lowerPowerOfTwo(20)16.
  3. int findMedian(std::vector<int> numbers):

    • This function calculates the median of a vector of integers.
    • Steps:
      1. Sorts the vector in ascending order.
      2. Finds the median using the size of the vector:
        • If the size is odd, returns the middle element.
        • If the size is even, returns the average of the two middle elements.
    • Example:
      • findMedian({1, 3, 2})2 (sorted: {1, 2, 3}, median is 2).
      • findMedian({4, 2, 1, 3})2.5 (sorted: {1, 2, 3, 4}, median is (2 + 3) / 2).
  4. int intSize(int x):

    • This function calculates the number of digits in a positive integer x.
    • It does so by repeatedly dividing x by 10 and incrementing size until x becomes less than 1.
    • Example:
      • intSize(1)1 (1 has 1 digit).
      • intSize(1234)4 (1234 has 4 digits).
      • Note: Negative numbers are not explicitly handled, so for negatives, the behavior may not be as intended.
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