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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.
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.
Here, Java implements the sliding window by smoothly sliding the left after encountering zeros and considering the right index for effective window measurement.