Leetcode: Number of Longest Increasing Subsequence

Posted on June 18, 2019November 2, 2019 by braindenny

Number of Longest Increasing Subsequence

Similar Problems:

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

Example 1:

Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].

Example 2:

Input: [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.
Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution:

// https://code.dennyzhang.com/number-of-longest-increasing-subsequence
// Basic Ideas: dynamic programming
//    dp(i): ends with nums[i]
// Complexity: Time O(n*n), Space O(n)
type Node struct {
    max int
    count int
}
func findNumberOfLIS(nums []int) int {
    res, max := 0, 0
    l := make([]Node, len(nums))
    for i, num := range nums {
        l[i].max, l[i].count = 1, 1
        if i != 0 {
            for j:=i-1; j>=0; j-- {
                if nums[j] < num {
                    if l[j].max+1 > l[i].max {
                        l[i].max, l[i].count = l[j].max+1, l[j].count
                    } else {
                        if l[j].max+1 == l[i].max {
                            l[i].count += l[j].count
                        }
                    }
                }
            }
        }
        if l[i].max > max { 
            max, res = l[i].max, l[i].count
        } else {
            if l[i].max == max {
                res += l[i].count
            }
        }
    }
    return res
}

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