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995. Minimum Number of K Consecutive Bit Flips 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int minKBitFlips(vector<int>& nums, int k) {
    int ans = 0;
    int flippedTime = 0;

    for (int i = 0; i < nums.size(); ++i) {
      if (i >= k && nums[i - k] == 2)
        --flippedTime;
      if (flippedTime % 2 == nums[i]) {
        if (i + k > nums.size())
          return -1;
        ++ans;
        ++flippedTime;
        nums[i] = 2;
      }
    }

    return ans;
  }
};

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class Solution {
  public int minKBitFlips(int[] nums, int k) {
    int ans = 0;
    int flippedTime = 0;

    for (int i = 0; i < nums.length; ++i) {
      if (i >= k && nums[i - k] == 2)
        --flippedTime;
      if (flippedTime % 2 == nums[i]) {
        if (i + k > nums.length)
          return -1;
        ++ans;
        ++flippedTime;
        nums[i] = 2;
      }
    }

    return ans;
  }
}

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class Solution:
  def minKBitFlips(self, nums: list[int], k: int) -> int:
    ans = 0
    flippedTime = 0

    for i, num in enumerate(nums):
      if i >= k and nums[i - k] == 2:
        flippedTime -= 1
      if flippedTime % 2 == num:
        if i + k > len(nums):
          return -1
        ans += 1
        flippedTime += 1
        nums[i] = 2

    return ans