Massive Algorithms: LeetCode 487

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?
https://discuss.leetcode.com/topic/75445/java-clean-solution-easily-extensible-to-flipping-k-zero-and-follow-up-handled

The idea is to keep a window [l, h] that contains at most k zero

The following solution does not handle follow-up, because nums[l] will need to access previous input stream
Time: O(n) Space: O(1)

    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0, zero = 0, k = 1; // flip at most k zero
        for (int l = 0, h = 0; h < nums.length; h++) {
            if (nums[h] == 0)                                           
                zero++;
            while (zero > k)
                if (nums[l++] == 0)
                    zero--;                                     
            max = Math.max(max, h - l + 1);
        }                                                               
        return max;             
    }

Now let's deal with follow-up, we need to store up to k indexes of zero within the window [l, h] so that we know where to move lnext when the window contains more than k zero. If the input stream is infinite, then the output could be extremely large because there could be super long consecutive ones. In that case we can use BigInteger for all indexes. For simplicity, here we will use int
Time: O(n) Space: O(k)

    public int findMaxConsecutiveOnes(int[] nums) {                 
        int max = 0, k = 1; // flip at most k zero
        Queue<Integer> zeroIndex = new LinkedList<>(); 
        for (int l = 0, h = 0; h < nums.length; h++) {
            if (nums[h] == 0)
                zeroIndex.offer(h);
            if (zeroIndex.size() > k)                                   
                l = zeroIndex.poll() + 1;
            max = Math.max(max, h - l + 1);
        }
        return max;                     
    }

Note that setting k = 0 will give a solution to the earlier version Max Consecutive Ones

For k = 1 we can apply the same idea to simplify the solution. Here q stores the index of zero within the window [l, h] so its role is similar to Queue in the above solution

    public int findMaxConsecutiveOnes(int[] nums) {
        int max = 0, q = -1;
        for (int l = 0, h = 0; h < nums.length; h++) {
            if (nums[h] == 0) {
                l = q + 1;
                q = h;
            }
            max = Math.max(max, h - l + 1);
        }                                                               
        return max;             
    }

X.
https://discuss.leetcode.com/topic/75457/sliding-window-2-pointers-o-n-time-1-pass-o-1-space

    public int findMaxConsecutiveOnes(int[] nums) {
        int j=0;
        int len=0;
        int zero=0;
        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                zero++;
            }
            
            while(zero>1){
                if(nums[j]==0){
                    zero--;
                }
                j++;
            }
            len=Math.max(i-j+1,len);
        }
        
        return len;
    }

https://discuss.leetcode.com/topic/75426/concise-o-n-solution-slight-modification-of-two-pointers-and-1-liner-solution-for-fun/2

Based on the above solution, for "Max Consecutive Ones II", we simply keep track of 2 low pointers:

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        nums.append(0)
        lo1, lo2 = 0, 0
        ret = 0
        for hi,n in enumerate(nums):
            if n==0:
                ret = max(ret, hi-lo1)
                lo1, lo2 = lo2, hi+1
        return ret

https://discuss.leetcode.com/topic/75439/java-concise-o-n-time-o-1-space
Use variable index to record the position of zero, and initiate the index to -1. Once we find a zero, refresh the count and store the new index.

        int max = 0, count = 0, index = -1;
        for(int i = 0;i< nums.length; i++){
            if(nums[i] == 1){
                count++;
            }else{
                count = i - index;
                index = i;
            }
            max = Math.max(max,count);
        }
        return max;
    }

zero right can be considered as 1 accumulated. If the gap between 1 and 1 is only 1(zero), zeroleft can be a temp storage to store how many 1 has been calculated, and then move on accumulate. But if there are 2 or more 0 gap, after the 1st 0, left will only be used to the flipped single 1.

public int findMaxConsecutiveOnes(int[] nums) {
     int maxConsecutive = 0, zeroLeft = 0, zeroRight = 0;
     for (int i=0;i<nums.length;i++) {
         zeroRight++;
         if (nums[i] == 0) {
             zeroLeft = zeroRight;
             zeroRight = 0;
         }
         maxConsecutive = Math.max(maxConsecutive, zeroLeft+zeroRight); 
     }
     return maxConsecutive;
}

线性遍历+计数器

统计恰好相隔1个'0'的两个连续1子数组的最大长度之和

def findMaxConsecutiveOnes(self, nums): """ :type nums: List[int] :rtype: int """ if len(nums) == sum(nums): return sum(nums) a = b = c = 0 for n in nums: if n == 1: c += 1 else: b, c = c, 0 a = max(a, b + c + 1) return a

X. https://discuss.leetcode.com/topic/75435/java-dp-o-n-solution

The idea is to use an extra array dp to store number of 1 to its left and right. Then the answer is the MAXof dp[i] + 1 of all nums[i] == 0.

    public int findMaxConsecutiveOnes(int[] nums) {
        int result = 0, n = nums.length, count = 0;
        int[] dp = new int[n];
        
        for (int i = 0; i < n; i++) {
            dp[i] = count;
            if (nums[i] == 1) {
         count++;
         result = Math.max(count, result);
            }
            else count = 0;
        }
        
        count = 0;
        for (int i = n - 1; i >= 0; i--) {
            dp[i] += count;
            if (nums[i] == 1) count++;
            else count = 0;
        }
        
        for (int i = 0; i < n; i++) {
            if (nums[i] == 0) {
         result = Math.max(result, dp[i] + 1);
            }
        }
        
        return result;
    }

LeetCode 487 - Max Consecutive Ones II

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