Sponsored
Sponsored
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.
1using System;
2
3class LCIS {
4 public static int FindLengthOfLCIS(int[] nums) {
5 if (nums.Length == 0) return 0;
6 int maxLen = 1, currLen = 1;
7
8 for (int i = 1; i < nums.Length; i++) {
9 if (nums[i] > nums[i - 1]) {
10 currLen++;
11 maxLen = Math.Max(maxLen, currLen);
12 } else {
13 currLen = 1;
14 }
15 }
16
17 return maxLen;
18 }
19
20 static void Main() {
21 int[] nums = {1, 3, 5, 4, 7};
22 Console.WriteLine(FindLengthOfLCIS(nums));
23 }
24}The C# solution also uses a simple loop to determine the maximum length of continuous increasing subsequence by leveraging the conditional logic and variable updates.
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.
#include <vector>
using namespace std;
int findLengthOfLCIS(vector<int>& nums) {
int maxLen = 0, currLen = 0;
for (size_t i = 0; i < nums.size(); ++i) {
if (i == 0 || nums[i] > nums[i - 1]) {
currLen++;
maxLen = max(maxLen, currLen);
} else {
currLen = 1;
}
}
return maxLen;
}
int main() {
vector<int> nums = {1, 3, 5, 4, 7};
cout << findLengthOfLCIS(nums) << endl;
return 0;
}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.