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Question

Formatted question description: https://leetcode.ca/all/487.html

 487. Max Consecutive Ones II
 Given a binary array, find the maximum number of consecutive 1s in this array if you can flip at most one 0.

 Example 1:
 Input: [1,0,1,1,0]
 Output: 4
 Explanation: Flip the first zero will get the the maximum number of consecutive 1s.
 After flipping, the maximum number of consecutive 1s is 4.

 Note:

 The input array will only contain 0 and 1.
 The length of input array is a positive integer and will not exceed 10,000

 Follow up:
 What if the input numbers come in one by one as an infinite stream? In other words, you can't store all numbers coming from the stream as it's too large to hold in memory. Could you solve it efficiently?

Algorithm

What if the question says that it can be flipped k times?

It is best to use a general solution to deal with this kind of problem. A window [left,right] can be maintained to hold at least k zeros. When it encounters 0, it accumulates the number of zero, and then judges if the number of 0 is greater than k at this time, then shifts the left boundary left to the right, and if the removed nums[left] is 0, then zero decrements by 1. If it is not greater than k, use the number of digits in the window to update res.

Follow up

The above method cannot be used in the case of follow up, because nums[left] needs to access the previous number. We can save all the positions of 0 we encountered in a queue, so that we know where to move when we need to move left.

Code

• Java
• C++
• Python

• 
  public class Max_Consecutive_Ones_II {
  
      public class Solution {
  
          public int findMaxConsecutiveOnes(int[] nums) {
  
              int res = 0, zero = 0, left = 0, k = 1;
  
              for (int right = 0; right < nums.length; ++right) {
                  if (nums[right] == 0) ++zero;
                  while (zero > k) {
                      if (nums[left++] == 0) --zero;
                  }
                  res = Math.max(res, right - left + 1);
              }
  
              return res;
          }
      }
  
      public class Solution_followup {
  
          public int findMaxConsecutiveOnes(int[] nums) {
  
              int res = 0, left = 0, k = 1;
              Queue<Integer> q = new LinkedList<>();
  
              for (int right = 0; right < nums.length; ++right) {
                  if (nums[right] == 0) q.offer(right);
                  if (q.size() > k) {
                      left = q.poll() + 1;
                  }
                  res = Math.max(res, right - left + 1);
              }
              return res;
  
          }
      }
  }
  
  

• // OJ: https://leetcode.com/problems/max-consecutive-ones-ii/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int findMaxConsecutiveOnes(vector<int>& A) {
          int zero = 0, i = 0, j = 0, N = A.size(), ans = 0;
          while (j < N) {
              zero += A[j++] == 0;
              while (zero > 1) zero -= A[i++] == 0;
              ans = max(ans, j - i);
          }
          return ans;
      }
  };
  

• class Solution:
      def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
          l = r = 0
          k = 1
          while r < len(nums):
              if nums[r] == 0:
                  k -= 1
              if k < 0:
                  if nums[l] == 0:
                      k += 1
                  l += 1
              r += 1
          return r - l
  
  ############
  
  class Solution(object):
    def __init__(self):
      self.ans = 0
      self.count = 0
      self.lastCount = 0
  
    def findMaxConsecutiveOnes(self, nums):
      """
      :type nums: List[int]
      :rtype: int
      """
      for num in nums:
        self.readNum(num)  # stream the input
      return self.ans
  
    def readNum(self, num):
      """
      :type nums: int
      """
      if num == 1:
        self.count += 1
      else:
        self.count = self.count - self.lastCount + 1
        self.lastCount = self.count
      self.ans = max(self.ans, self.count)
  
  

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