Formatted question description: https://leetcode.ca/all/673.html

673. Number of Longest Increasing Subsequence (Medium)

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.

Related Topics:
Dynamic Programming

Similar Questions:

Solution 1. DP

Consider the subproblem in range A[0..i] and the LIS must ends with A[i].

Let len[i] be the length of longest LIS ending with A[i].

Let cnt[i] be the count of longest LIS ending with A[i].

len[i] = 1 + max( len[j] | 0 <= j < i && A[j] < A[i] )
cnt[i] = sum( cnt[j] | 0 <= j < i && len[j] + 1 == len[i] )

The answer is

sum( cnt[i] | len[i] == maxLen )
    where maxLen = max( len[i] | 0 <= i < N )
// OJ: https://leetcode.com/problems/number-of-longest-increasing-subsequence/

// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int findNumberOfLIS(vector<int>& A) {
        if (A.empty()) return 0;
        int N = A.size(), ans = 0, maxLen = 0;
        vector<int> len(N, 1), cnt(N, 1);
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[j] >= A[i]) continue;
                if (len[i] < 1 + len[j]) len[i] = 1 + len[j], cnt[i] = cnt[j];
                else if (len[i] == 1 + len[j]) cnt[i] += cnt[j];
            }
            if (len[i] > maxLen) maxLen = len[i], ans = cnt[i];
            else if (len[i] == maxLen) ans += cnt[i];
        }
        return ans;
    }
};

Java

  • class Solution {
        public int findNumberOfLIS(int[] nums) {
            if (nums == null || nums.length == 0)
                return 0;
            if (nums.length == 1)
                return 1;
            int maxLength = 1;
            int length = nums.length;
            int[] dp = new int[length];
            dp[0] = 1;
            for (int i = 1; i < length; i++) {
                dp[i] = 1;
                for (int j = 0; j < i; j++) {
                    if (nums[j] < nums[i])
                        dp[i] = Math.max(dp[i], dp[j] + 1);
                }
                maxLength = Math.max(maxLength, dp[i]);
            }
            int[][] dp2d = new int[length][maxLength + 1];
            for (int i = 0; i < length; i++)
                dp2d[i][0] = 1;
            dp2d[0][1] = 1;
            for (int i = 1; i < length; i++) {
                dp2d[i][1] = 1;
                for (int j = 0; j < i; j++) {
                    if (nums[j] < nums[i]) {
                        for (int k = 1; k < maxLength; k++)
                            dp2d[i][k + 1] += dp2d[j][k];
                    }
                }
            }
            int count = 0;
            for (int i = 0; i < length; i++)
                count += dp2d[i][maxLength];
            return count;
        }
    }
    
  • // OJ: https://leetcode.com/problems/number-of-longest-increasing-subsequence/
    // Time: O(N^2)
    // Space: O(N)
    class Solution {
    public:
        int findNumberOfLIS(vector<int>& A) {
            if (A.empty()) return 0;
            int N = A.size(), ans = 0, maxLen = 0;
            vector<int> len(N, 1), cnt(N, 1);
            for (int i = 0; i < N; ++i) {
                for (int j = 0; j < i; ++j) {
                    if (A[j] >= A[i]) continue;
                    if (len[i] < 1 + len[j]) len[i] = 1 + len[j], cnt[i] = cnt[j];
                    else if (len[i] == 1 + len[j]) cnt[i] += cnt[j];
                }
                if (len[i] > maxLen) maxLen = len[i], ans = cnt[i];
                else if (len[i] == maxLen) ans += cnt[i];
            }
            return ans;
        }
    };
    
  • class Solution(object):
        def findNumberOfLIS(self, nums):
            """
            :type nums: List[int]
            :rtype: int
            """
            dp, longest = [[1, 1] for _ in range(len(nums))], 1
            for i in range(1, len(nums)):
                curr_longest, count = 1, 0
                for j in range(i):
                    if nums[j] < nums[i]:
                        curr_longest = max(curr_longest, dp[j][0] + 1)
                for j in range(i):
                    if dp[j][0] == curr_longest - 1 and nums[j] < nums[i]:
                        count += dp[j][1]
                dp[i] = [curr_longest, max(count, dp[i][1])]
                longest = max(longest, curr_longest)
            return sum([item[1] for item in dp if item[0] == longest])
    

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