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487. Max Consecutive Ones II 👍

Approach 1: Sliding Window

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

    for (int l = 0, r = 0; r < nums.size(); ++r) {
      if (nums[r] == 0)
        ++zeros;
      while (zeros == 2)
        if (nums[l++] == 0)
          --zeros;
      ans = max(ans, r - l + 1);
    }

    return ans;
  }
};

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class Solution {
  public int findMaxConsecutiveOnes(int[] nums) {
    int ans = 0;
    int zeros = 0;

    for (int l = 0, r = 0; r < nums.length; ++r) {
      if (nums[r] == 0)
        ++zeros;
      while (zeros == 2)
        if (nums[l++] == 0)
          --zeros;
      ans = Math.max(ans, r - l + 1);
    }

    return ans;
  }
}

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class Solution:
  def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
    ans = 0
    zeros = 0

    l = 0
    for r, num in enumerate(nums):
      if num == 0:
        zeros += 1
      while zeros == 2:
        if nums[l] == 0:
          zeros -= 1
        l += 1
      ans = max(ans, r - l + 1)

    return ans

Approach 2: Follow up: Generalize to k maxZeros

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int findMaxConsecutiveOnes(vector<int>& nums) {
    constexpr int maxZeros = 1;
    int ans = 0;
    queue<int> q;  // Store indices of zero

    for (int l = 0, r = 0; r < nums.size(); ++r) {
      if (nums[r] == 0)
        q.push(r);
      if (q.size() > maxZeros)
        l = q.front() + 1, q.pop();
      ans = max(ans, r - l + 1);
    }

    return ans;
  }
};

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class Solution {
  public int findMaxConsecutiveOnes(int[] nums) {
    final int maxZeros = 1;
    int ans = 0;
    // Store indices of zero
    Queue<Integer> q = new ArrayDeque<>();

    for (int l = 0, r = 0; r < nums.length; ++r) {
      if (nums[r] == 0)
        q.offer(r);
      if (q.size() > maxZeros)
        l = q.poll() + 1;
      ans = Math.max(ans, r - l + 1);
    }

    return ans;
  }
}

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class Solution:
  def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
    maxZeros = 1
    ans = 0
    q = collections.deque()  # Store indices of zero

    l = 0
    for r, num in enumerate(nums):
      if num == 0:
        q.append(r)
      if len(q) > maxZeros:
        l = q.popleft() + 1
      ans = max(ans, r - l + 1)

    return ans

Approach 3: Follow up: $\texttt{maxZeros == 1}$

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int findMaxConsecutiveOnes(vector<int>& nums) {
    int ans = 0;
    int lastZeroIndex = -1;

    for (int l = 0, r = 0; r < nums.size(); ++r) {
      if (nums[r] == 0) {
        l = lastZeroIndex + 1;
        lastZeroIndex = r;
      }
      ans = max(ans, r - l + 1);
    }

    return ans;
  }
};

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class Solution {
  public int findMaxConsecutiveOnes(int[] nums) {
    int ans = 0;
    int lastZeroIndex = -1;

    for (int l = 0, r = 0; r < nums.length; ++r) {
      if (nums[r] == 0) {
        l = lastZeroIndex + 1;
        lastZeroIndex = r;
      }
      ans = Math.max(ans, r - l + 1);
    }

    return ans;
  }
}

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class Solution:
  def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
    ans = 0
    lastZeroIndex = -1

    l = 0
    for r, num in enumerate(nums):
      if num == 0:
        l = lastZeroIndex + 1
        lastZeroIndex = r
      ans = max(ans, r - l + 1)

    return ans