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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 <stdio.h>
2
3int findMaxConsecutiveOnes(int* nums, int numsSize) {
4 int maxCount = 0, currentCount = 0;
5 for(int i = 0; i < numsSize; i++) {
6 if(nums[i] == 1) {
7 currentCount++;
8 if(currentCount > maxCount) {
9 maxCount = currentCount;
10 }
11 } else {
12 currentCount = 0;
13 }
14 }
15 return maxCount;
16}
17
18int main() {
19 int nums[] = {1, 1, 0, 1, 1, 1};
20 int size = sizeof(nums) / sizeof(nums[0]);
21 int result = findMaxConsecutiveOnes(nums, size);
22 printf("The maximum number of consecutive 1s is: %d\n", result);
23 return 0;
24}The function findMaxConsecutiveOnes iterates over the input array nums. It uses a counter to track consecutive ones. When a one is encountered, the counter is incremented. If a zero is encountered, the counter is reset. Throughout the iteration, the maximum sequence length is updated.
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.