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This approach involves iterating through the array while maintaining a variable to track the length of the current continuous increasing subsequence. As we encounter each element, we check if it is larger than the previous one to decide if we should continue the current subsequence or start a new one. We also maintain a global maximum to store the maximum length found during our iteration.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(1), as we use a constant amount of space.
1public class LCIS {
2 public static int findLengthOfLCIS(int[] nums) {
3 if (nums.length == 0) return 0;
4 int maxLen = 1, currLen = 1;
5
6 for (int i = 1; i < nums.length; i++) {
7 if (nums[i] > nums[i - 1]) {
8 currLen++;
9 maxLen = Math.max(maxLen, currLen);
10 } else {
11 currLen = 1;
12 }
13 }
14
15 return maxLen;
16 }
17
18 public static void main(String[] args) {
19 int[] nums = {1, 3, 5, 4, 7};
20 System.out.println(findLengthOfLCIS(nums));
21 }
22}The Java code implements the same technique: iterating through the array, updating currLen when the current element is greater, and resetting it otherwise. The max length is updated accordingly.
The greedy approach leverages a single pass through the array to determine the maximum length of an increasing subsequence. This method is optimal because it immediately processes each element only once, updating the sequence lengths on the fly, and computing the maximum without revisiting any part of the array.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(1), as no additional space other than variables is utilized.
This Python code implements a greedy check to only keep adjusts on a basic condition, replicating rolling continuity steps that keep sync throughout iterations in a lazy kind of sequence.