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673. Number of Longest Increasing Subsequence (Medium)

Given an integer array nums, return the number of longest increasing subsequences.

Notice that the sequence has to be strictly increasing.

 

Example 1:

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

Example 2:

Input: nums = [2,2,2,2,2]
Output: 5
Explanation: The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.

 

Constraints:

  • 1 <= nums.length <= 2000
  • -106 <= nums[i] <= 106

Companies: Amazon, Google, Bloomberg

Related Topics:
Array, Dynamic Programming, Binary Indexed Tree, Segment Tree

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/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2. Top-down DP

// OJ: https://leetcode.com/problems/number-of-longest-increasing-subsequence
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int findNumberOfLIS(vector<int>& A) {
        int N = A.size(), maxLen = 0, ans = 0;
        vector<int> len(N), cnt(N);
        function<void(int)> dp = [&](int i) {
            if (i == N || len[i]) return;
            len[i] = 1;
            cnt[i] = 1;
            for (int j = 0; j < i; ++j) {
                if (A[j] >= A[i]) continue;
                dp(j);
                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];
            }
        };
        for (int i = 0; i < N; ++i) {
            dp(i);
            if (len[i] > maxLen) maxLen = len[i], ans = cnt[i];
            else if (len[i] == maxLen) ans += cnt[i];
        }
        return ans;
    }
};