LeetCode 673: Number of Longest Increasing Subsequence Solution

Master LeetCode problem 673 (Number of Longest Increasing Subsequence), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

673. Number of Longest Increasing Subsequence

Medium

Array Dynamic Programming Binary Indexed Tree Segment Tree

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will maintain two arrays lengths and counts. The lengths[i] will store the length of the longest increasing subsequence ending at index i, and counts[i] will store the number of such subsequences ending at index i.

We iterate through the array nums and for each element at index i, we iterate from 0 to i-1 to find the longest increasing subsequence ending at index i. If nums[i] > nums[j], we update lengths[i] and counts[i] based on the values of lengths[j] and counts[j]. If lengths[j] + 1 > lengths[i], we update both lengths[i] and counts[i]. If lengths[j] + 1 == lengths[i], we increment counts[i] by counts[j].

Finally, we find the maximum length of the longest increasing subsequence in the lengths array and sum up the corresponding counts in the counts array.

Time Complexity: O(n^2) where n is the length of the input array nums. Space Complexity: O(n) where n is the length of the input array nums.

Solution Code

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int n = nums.length;
        int[] lengths = new int[n];
        int[] counts = new int[n];
        Arrays.fill(lengths, 1);
        Arrays.fill(counts, 1);
        
        int maxLength = 1;
        int result = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    if (lengths[j] + 1 > lengths[i]) {
                        lengths[i] = lengths[j] + 1;
                        counts[i] = counts[j];
                    } else if (lengths[j] + 1 == lengths[i]) {
                        counts[i] += counts[j];
                    }
                }
            }
            maxLength = Math.max(maxLength, lengths[i]);
        }
        
        for (int i = 0; i < n; i++) {
            if (lengths[i] == maxLength) {
                result += counts[i];
            }
        }
        
        return result;
    }
}

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