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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.
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In the Python solution, we loop through the list of numbers, check if the sequence continues to be increasing, and update or reset the current sequence length 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.
1#include <iostream>
2#include <vector>
3using namespace std;
4
5int findLengthOfLCIS(vector<int>& nums) {
6 int maxLen = 0, currLen = 0;
7
8 for (size_t i = 0; i < nums.size(); ++i) {
9 if (i == 0 || nums[i] > nums[i - 1]) {
10 currLen++;
11 maxLen = max(maxLen, currLen);
12 } else {
13 currLen = 1;
14 }
15 }
16 return maxLen;
17}
18
19int main() {
20 vector<int> nums = {1, 3, 5, 4, 7};
21 cout << findLengthOfLCIS(nums) << endl;
22 return 0;
23}The method follows the simplest explanation possible to keep items into loops, and as each extension of this sequence validates an increment, we can achieve the result using this greedy step.