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This approach involves iterating through the array and counting sequences of 1s. If a 0 is encountered, the count is reset to 0. We keep track of the maximum count during the iteration.
Time Complexity: O(n), where n is the number of elements in the array, as we make a single pass.
Space Complexity: O(1) since no extra space proportional to input size is used.
1def findMaxConsecutiveOnes(nums):
2 maxCount = 0
3 currentCount = 0
4 for num in nums:
5 if num == 1:
6 currentCount += 1
7 maxCount = max(maxCount, currentCount)
8 else:
9 currentCount = 0
10 return maxCount
11
12nums = [1, 1, 0, 1, 1, 1]
13print("The maximum number of consecutive 1s is:", findMaxConsecutiveOnes(nums))This Python code iterates through the list nums, updating counts similarly to previous implementations. It's straightforward with the use of Python's built-in max function.
This approach uses a variation of the sliding window technique. The idea is to maintain a window of the current sequence of 1s and adjust the window whenever a 0 is encountered.
Time Complexity: O(n)
Space Complexity: O(1)
1#include <iostream>
2#include <vector>
using namespace std;
int findMaxConsecutiveOnes(vector<int>& nums) {
int left = 0, right = 0, maxCount = 0;
while (right < nums.size()) {
if (nums[right] == 0) {
left = right + 1;
}
maxCount = max(maxCount, right - left + 1);
++right;
}
return maxCount;
}
int main() {
vector<int> nums = {1, 0, 1, 1, 0, 1};
cout << "The maximum number of consecutive 1s is: " << findMaxConsecutiveOnes(nums) << endl;
return 0;
}This C++ solution expands on the sliding window technique, ensuring that left is moved to keep the window focusing on 1s by adjusting based on zeroes found at right.