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3314. Construct the Minimum Bitwise Array I 👍

  • Time: $O(\Sigma \log\texttt{nums[i]}) \approx O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> minBitwiseArray(vector<int>& nums) {
    vector<int> ans;

    for (const int num : nums)
      ans.push_back(num == 2 ? -1 : num - getLeadingOneOfLastGroupOfOnes(num));

    return ans;
  }

 private:
  // Returns the leading one of the last group of 1s in the binary
  // representation of num. For example, if num = 0b10111, the leading one of
  // the last group of 1s is 0b100.
  int getLeadingOneOfLastGroupOfOnes(int num) {
    int leadingOne = 1;
    while ((num & leadingOne) > 0)
      leadingOne <<= 1;
    return leadingOne >> 1;
  }
};

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class Solution {
  public int[] minBitwiseArray(List<Integer> nums) {
    int[] ans = new int[nums.size()];

    for (int i = 0; i < nums.size(); ++i)
      ans[i] = nums.get(i) == 2 ? -1 : nums.get(i) - getLeadingOneOfLastGroupOfOnes(nums.get(i));

    return ans;
  }

  // Returns the leading one of the last group of 1s in the binary
  // representation of num. For example, if num = 0b10111, the leading one of
  // the last group of 1s is 0b100.
  private int getLeadingOneOfLastGroupOfOnes(int num) {
    int leadingOne = 1;
    while ((num & leadingOne) > 0)
      leadingOne <<= 1;
    return leadingOne >> 1;
  }
}

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class Solution:
  def minBitwiseArray(self, nums: list[int]) -> list[int]:
    return [-1 if num == 2 else num - self._getLeadingOneOfLastGroupOfOnes(num)
            for num in nums]

  def _getLeadingOneOfLastGroupOfOnes(self, num: int) -> int:
    """
    Returns the leading one of the last group of 1s in the binary
    representation of num. For example, if num = 0b10111, the leading one of
    the last group of 1s is 0b100.
    """
    leadingOne = 1
    while (num & leadingOne) > 0:
      leadingOne <<= 1
    return leadingOne >> 1