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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.
1#include <iostream>
2#include <vector>
3
4int findMaxConsecutiveOnes(std::vector<int>& nums) {
5 int maxCount = 0, currentCount = 0;
6 for(int num : nums) {
7 if(num == 1) {
8 currentCount++;
9 maxCount = std::max(maxCount, currentCount);
10 } else {
11 currentCount = 0;
12 }
13 }
14 return maxCount;
15}
16
17int main() {
18 std::vector<int> nums = {1, 1, 0, 1, 1, 1};
19 std::cout << "The maximum number of consecutive 1s is: " << findMaxConsecutiveOnes(nums) << std::endl;
20 return 0;
21}This C++ solution uses a for-each loop for cleanly iterating over each element in the vector. The logic remains similar: incrementing the count for each 1 and resetting on encountering a 0.
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)
1using System;
2
public class MaxConsecutiveOnes {
public static int FindMaxConsecutiveOnes(int[] nums) {
int left = 0, maxCount = 0;
for (int right = 0; right < nums.Length; right++) {
if (nums[right] == 0) {
left = right + 1;
}
maxCount = Math.Max(maxCount, right - left + 1);
}
return maxCount;
}
public static void Main() {
int[] nums = {1, 0, 1, 1, 0, 1};
Console.WriteLine("The maximum number of consecutive 1s is: " + FindMaxConsecutiveOnes(nums));
}
}
This C# solution applies the sliding window approach by dynamically adjusting the potential sequence boundaries with respect to encounters with zeros.