Max Consecutive Ones

Given a binary array nums, return the maximum number of consecutive 1's in the array.

Problem Statement

When you're given a problem like this, the first thing is to understand the problem statement and the constraints. In this case, we are given a binary array, where the elements are either 0 or 1, and we are asked to return the maximum number of consecutive 1's in the array.

Examples

Input: nums = [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3.

Clarifying Questions

1. Is the binary array a list of integers or boolean values?
2. Are there any specific constraints on the size of the array, such as maximum length?
3. Should I return the length of the maximum sequence or the sequence itself?
4. Can the input array contain non-binary values? If so, how should I handle them?
5. Should I return 0 if there are no 1's in the array?
6. Can the array have multiple sequences of the same maximum length, and should I return one of them or all of them?
7. Is the input array guaranteed to be non-empty, or should I include special cases for an empty array?
8. What is the expected input format for the binary array? Can it be assumed that it will always be a valid binary array with only 0's and 1's?
9. Is the output required to be the length of the longest contiguous sequence of 1's or is it required to be the number of 1's in the longest contiguous sequence of 1's?
10. Is there a requirement for the solution to have a specific time complexity or can any efficient solution be acceptable?

Solutions

There are several approaches to solving the problem of finding the maximum number of consecutive 1's in a binary array:

Here are the code implementations in various programming languages:

Approach I: Brute force

This involves using nested loops to check for consecutive 1's at each position in the binary array. This approach has a time complexity of O(n^2), where n is the length of the binary array.

Algorithm:

1. Initialize a variable maxCount to 0, to keep track of the maximum number of consecutive 1's
2. For each element i in the binary array nums:
   1. Initialize a variable count to 0
   2. For each element j starting from i until the end of the binary array nums:
      1. If nums[j] is 1, increment count
      2. If nums[j] is 0, break out of the inner loop
   3. Set maxCount to the maximum of maxCount and count
3. Return maxCount

        
       
          public int FindMaxConsecutiveOnes(int[] nums) {
            int maxCount = 0;
            for (int i = 0; i < nums.Length; i++) {
                int count = 0;
                for (int j = i; j < nums.Length; j++) {
                    if (nums[j] == 1) {
                        count++;
                    } else {
                        break;
                    }
                }
                maxCount = Math.Max(maxCount, count);
            }
            return maxCount;
        }
        
        
      

        
       
      def findMaxConsecutiveOnes(nums: List[int]) -> int:
          maxCount = 0
          for i in range(len(nums)):
              count = 0
              for j in range(i, len(nums)):
                  if nums[j] == 1:
                      count += 1
                  else:
                      break
              maxCount = max(maxCount, count)
          return maxCount
      
          

        
      

        
       
          public int findMaxConsecutiveOnes(int[] nums) {
            int maxCount = 0;
            for (int i = 0; i < nums.length; i++) {
                int count = 0;
                for (int j = i; j < nums.length; j++) {
                    if (nums[j] == 1) {
                        count++;
                    } else {
                        break;
                    }
                }
                maxCount = Math.max(maxCount, count);
            }
            return maxCount;
        }
        
        
      

        
          int findMaxConsecutiveOnes(vector<int>& nums) {
            int maxCount = 0;
            for (int i = 0; i < nums.size(); i++) {
                int count = 0;
                for (int j = i; j < nums.size(); j++) {
                    if (nums[j] == 1) {
                        count++;
                    } else {
                        break;
                    }
                }
                maxCount = max(maxCount, count);
            }
            return maxCount;
        }
        
      

Time Complexity: O(n^2), where n is the length of the input array. This is because for each element in the array, we are checking all of the elements that come after it.

Space Complexity: O(1), because we only use a few variables to keep track of the maximum count and the current count.

Approach II: Single pass

This approach involves keeping track of the count of consecutive 1's as we traverse the binary array in a single pass. This way, we can achieve a time complexity of O(n), where n is the length of the binary array.

Algorithm:

1. Initialize two variables maxCount to 0 and count to 0, to keep track of the maximum number of consecutive 1's and the current number of consecutive 1's, respectively
2. For each element i in the binary array nums:
   1. If nums[i] is 1, increment count
   2. If nums[i] is 0, reset count to 0
   3. Set maxCount to the maximum of maxCount and count
3. Return maxCount

        
       
          public int FindMaxConsecutiveOnes(int[] nums)
          {
              int maxCount = 0;
              int count = 0;
              foreach (int num in nums)
              {
                  if (num == 1)
                      count++;
                  else
                  {
                      maxCount = Math.Max(maxCount, count);
                      count = 0;
                  }
              }
              return Math.Max(maxCount, count);
          }
          
        
        
      

        
       
      def findMaxConsecutiveOnes(nums: List[int]) -> int:
          maxCount = 0
          count = 0
          for num in nums:
              if num == 1:
                  count += 1
              else:
                  maxCount = max(maxCount, count)
                  count = 0
          return max(maxCount, count)
      
        
      

        
       
          public int findMaxConsecutiveOnes(int[] nums) {
            int maxCount = 0;
            int count = 0;
            for (int num : nums) {
                if (num == 1)
                    count++;
                else {
                    maxCount = Math.max(maxCount, count);
                    count = 0;
                }
            }
            return Math.max(maxCount, count);
        }
        
        
        
      

        
        
        int findMaxConsecutiveOnes(vector<int>& nums) {
          int maxCount = 0;
          int count = 0;
          for (int num : nums) {
              if (num == 1)
                  count++;
              else {
                  maxCount = max(maxCount, count);
                  count = 0;
              }
          }
          return max(maxCount, count);
      }
      
        
      

Time Complexity: O(n), where n is the length of the input array. This is because we are only looping through the array once.

Space Complexity: O(1), because we only use a few variables to keep track of the maximum count and the current count.

Approach III: Bit manipulation

This approach involves using bit manipulation techniques to count the number of consecutive 1's in the binary representation of the input. This can be an efficient solution for very large binary arrays but is less intuitive and harder to understand.

Algorithm:

1. Initialize two variables maxCount to 0 and count to 0, to keep track of the maximum number of consecutive 1's and the current number of consecutive 1's, respectively
2. For each element i in the binary array nums:
   1. Convert nums[i] to its binary representation
   2. Use bit manipulation techniques to count the number of consecutive 1's in the binary representation of nums[i]
   3. Set count to the maximum of count and the number of consecutive 1's found in the binary representation
   4. Set maxCount to the maximum of maxCount and count
3. Return maxCount

        
       
          public int FindMaxConsecutiveOnes(int[] nums)
          {
              int count = 0;
              int maxCount = 0;
              for (int i = 0; i < nums.Length; i++)
              {
                  if ((nums[i] & 1) == 1) // check if the least significant bit is 1
                  {
                      count++;
                  }
                  else
                  {
                      maxCount = Math.Max(maxCount, count);
                      count = 0;
                  }
              }
              return Math.Max(maxCount, count);
          }
          
        
        
      

        
       
      def findMaxConsecutiveOnes(nums: List[int]) -> int:
          count = 0
          maxCount = 0
          for num in nums:
              if num & 1: # check if the least significant bit is 1
                  count += 1
              else:
                  maxCount = max(maxCount, count)
                  count = 0
          return max(maxCount, count)
      
      
        
      

        
          public int findMaxConsecutiveOnes(int[] nums) {
            int count = 0;
            int maxCount = 0;
            for (int num : nums) {
                if ((num & 1) == 1) // check if the least significant bit is 1
                    count++;
                else {
                    maxCount = Math.max(maxCount, count);
                    count = 0;
                }
            }
            return Math.max(maxCount, count);
        }
        
        
      

        
        
    int findMaxConsecutiveOnes(vector<int>& nums)
        int count = 0;
        int maxCount = 0;
        for (int num : nums) {
            if (num & 1) // check if the least significant bit is 1
                count++;
            else {
                maxCount = max(maxCount, count);
                count = 0;
            }
        }
        return max(maxCount, count);
    }
    
      
        
      

Time Complexity: O(n), because we iterate through the array once, and the bitwise operations are constant time, so the time complexity is O(n).

Space Complexity: O(1), because we only need a constant amount of extra space to store the count and maxCount variables, so the space complexity is O(1).

Approach IV: Dynamic programming

This involves using dynamic programming techniques to keep track of the maximum number of consecutive 1's up to each position in the binary array. This approach has a time complexity of O(n), where n is the length of the binary array and is more memory efficient than the optimized single pass approach.

Algorithm:

1. Initialize an array dp of the same length as the binary array nums, to keep track of the maximum number of consecutive 1's up to each position
2. For each element i in the binary array nums:
   1. If nums[i] is 1, set dp[i] to dp[i-1] + 1
   2. If nums[i] is 0, set dp[i] to 0
   3. Set maxCount to the maximum of maxCount and dp[i]
3. Return maxCount

        
       
          public int FindMaxConsecutiveOnes(int[] nums)
          {
              int[] dp = new int[nums.Length];
              int maxCount = 0;
              dp[0] = nums[0];
              if (nums[0] == 1) maxCount = 1;
              for (int i = 1; i < nums.Length; i++)
              {
                  if (nums[i] == 1)
                      dp[i] = dp[i - 1] + 1;
                  if (nums[i] == 0 || i == nums.Length - 1)
                      maxCount = Math.Max(maxCount, dp[i]);
              }
              return maxCount;
          }
          
        
        
      

        
       
      def findMaxConsecutiveOnes(nums: List[int]) -> int:
          dp = [0] * len(nums)
          maxCount = dp[0] = nums[0]
          for i in range(1, len(nums)):
              if nums[i] == 1:
                  dp[i] = dp[i - 1] + 1
              if nums[i] == 0 or i == len(nums) - 1:
                  maxCount = max(maxCount, dp[i])
          return maxCount
      
        
      

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

        
        
    int findMaxConsecutiveOnes(vector<int>& nums){
      vector<int> dp(nums.size(), 0);
      int maxCount = dp[0] = nums[0];
      for (int i = 1; i < nums.size(); i++) {
          if (nums[i] == 1)
              dp[i] = dp[i - 1] + 1;
          if (nums[i] == 0 || i == nums.size() - 1)
              maxCount = max(maxCount, dp[i]);
      }
      return maxCount;
  }
  
      
        
      

Time Complexity: O(n), where n is the length of the input array. This is because we are only looping through the array once.

Space Complexity: O(n), because we are using an array dp of length n to store the maximum consecutive 1's ending at each index.