Longest Continuous Increasing Subsequence

• LeetCode: Longest Continuous Increasing Subsequence

• Problem:

Given an unsorted array of integers, find the length of longest continuous increasing subsequence.

• Note:

Length of the array will not exceed 10,000.

• Examples:

Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. 
Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4.

Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1.

• Analysis:

It's equal to the longest increasing subarry.

• Code - Java:

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int len = nums.length;
        int left = 0;
        int result = 0;
        for (int i = 1; i < len; i++) {
            if (nums[i - 1] >= nums[i]) {
                result = Math.max(i - left, result);
                left = i;
            }
        }
        result = Math.max(len - left, result);
        return result;
    }
}

• Code - C++:

public:
    int findLengthOfLCIS(vector<int>& nums) {
        int size = nums.size();
        int left = 0;
        int result = 0;
        for (int i = 1; i < size; i++) {
            if (nums[i - 1] >= nums[i]) {
                result = max(i - left, result);
                left = i;
            }
        }
        result = max(size - left, result);
        return result;
    }
};

• Code - Python:

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        size = len(nums)
        left, result = 0, 0
        for i in range(1, size):
            if nums[i - 1] >= nums[i]:
                result = max(i - left, result)
                left = i
        return max(size - left, result)

• Code - C:

int findLengthOfLCIS(int* nums, int numsSize) {
    int left = 0;
    int result = 0;
    for (int i = 1; i < numsSize; i++) {
        if (nums[i - 1] >= nums[i]) {
            if (i - left > result) {
                result = i - left;
            }
            left = i;
        }
    }
    if (numsSize - left > result) {
        result = numsSize - left;
    }
    return result;
}

• Time Complexity: O(N), N is the length of input array

• Space Complexity: O(1)