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2571. Minimum Operations to Reduce an Integer to 0

  • Time: $O(\log n)$
  • Space: $O(1)$
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class Solution {
 public:
  int minOperations(int n) {
    // The strategy is that when the end of n is
    //   1. consecutive 1s, add 1 (2^0).
    //   2. single 1, substract 1 (2^0).
    //   3. 0, substract 2^k to omit the last 1. Equivalently, n >> 1.
    //
    // e.g.
    //
    //         n = 0b101
    // n -= 2^0 -> 0b100
    // n -= 2^2 -> 0b0
    //         n = 0b1011
    // n += 2^0 -> 0b1100
    // n -= 2^2 -> 0b1000
    // n -= 2^3 -> 0b0
    int ans = 0;

    while (n > 0)
      if ((n & 3) == 3) {
        ++n;
        ++ans;
      } else if ((n & 1) == 1) {
        --n;
        ++ans;
      } else {
        n >>= 1;
      }

    return ans;
  }
};

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class Solution {
  public int minOperations(int n) {
    // The strategy is that when the end of n is
    //   1. consecutive 1s, add 1 (2^0).
    //   2. single 1, substract 1 (2^0).
    //   3. 0, substract 2^k to omit the last 1. Equivalently, n >> 1.
    //
    // e.g.
    //
    //         n = 0b101
    // n -= 2^0 -> 0b100
    // n -= 2^2 -> 0b0
    //         n = 0b1011
    // n += 2^0 -> 0b1100
    // n -= 2^2 -> 0b1000
    // n -= 2^3 -> 0b0
    int ans = 0;

    while (n > 0)
      if ((n & 3) == 3) {
        ++n;
        ++ans;
      } else if ((n & 1) == 1) {
        --n;
        ++ans;
      } else {
        n >>= 1;
      }

    return ans;
  }
}

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class Solution:
  def minOperations(self, n: int) -> int:
    # The strategy is that when the end of n is
    #   1. consecutive 1s, add 1 (2^0).
    #   2. single 1, substract 1 (2^0).
    #   3. 0, substract 2^k to omit the last 1. Equivalently, n >> 1.
    #
    # e.g.
    #
    #         n = 0b101
    # n -= 2^0 -> 0b100
    # n -= 2^2 -> 0b0
    #         n = 0b1011
    # n += 2^0 -> 0b1100
    # n -= 2^2 -> 0b1000
    # n -= 2^3 -> 0b0
    ans = 0

    while n > 0:
      if (n & 3) == 3:
        n += 1
        ans += 1
      elif (n & 1) == 1:
        n -= 1
        ans += 1
      else:
        n >>= 1

    return ans