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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.
1using System;
2
3public class MaxConsecutiveOnes {
4 public static int FindMaxConsecutiveOnes(int[] nums) {
5 int maxCount = 0, currentCount = 0;
6 foreach (int num in nums) {
7 if (num == 1) {
8 currentCount++;
9 maxCount = Math.Max(maxCount, currentCount);
10 } else {
11 currentCount = 0;
12 }
13 }
14 return maxCount;
15 }
16
17 public static void Main() {
18 int[] nums = {1, 1, 0, 1, 1, 1};
19 Console.WriteLine("The maximum number of consecutive 1s is: " + FindMaxConsecutiveOnes(nums));
20 }
21}
22This C# method uses a foreach loop to process the input array. The max consecutive count is determined using logic similar to that in the provided C# code, leveraging the static method Math.Max for comparison.
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.