Sponsored
Sponsored
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.
1public class MaxConsecutiveOnes {
2 public static int findMaxConsecutiveOnes(int[] nums) {
3 int maxCount = 0, currentCount = 0;
4 for (int num : nums) {
5 if (num == 1) {
6 currentCount++;
7 maxCount = Math.max(maxCount, currentCount);
8 } else {
9 currentCount = 0;
10 }
11 }
12 return maxCount;
13 }
14
15 public static void main(String[] args) {
16 int[] nums = {1, 1, 0, 1, 1, 1};
17 System.out.println("The maximum number of consecutive 1s is: " + findMaxConsecutiveOnes(nums));
18 }
19}
20This Java program employs a for-each loop to iterate over the array. The principles of counting and resetting are maintained as described.
This Python solution applies the same technique using the right index iterating over the list, dynamically adjusting the left boundary when it hits zero.