What is the time complexity of Number of Longest Increasing Subsequence?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Number of Longest Increasing Subsequence?

This problem uses Array, Dynamic Programming, Binary Indexed Tree, Segment Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Longest Increasing Subsequence?

Insight 1: Dynamic Programming is suitable because the LIS problem exhibits optimal substructure. The LIS ending at index `i` can be built from the LIS ending at earlier indices. Insight 2: We need to track both the length of the LIS ending at each index and the number of such LISs. Two arrays, `lengths` and `counts`, are crucial.

Which companies ask LeetCode 673. Number of Longest Increasing Subsequence in interviews?

This problem is frequently asked at Commvault, Intuit.

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Number of Longest Increasing Subsequence - LeetCode 673 Solution

Q: How to solve Number of Longest Increasing Subsequence?

The provided code employs a dynamic programming approach. It iterates through the input array `nums` and, for each element `nums[i]`, it considers all preceding elements `nums[j]` (where `j < i`). If `nums[j] < nums[i]`, it means `nums[i]` can potentially extend an increasing subsequence ending at `nums[j]`. The `lengths` array stores the length of the LIS ending at each index, and the `counts` array stores the number of such LISs. The algorithm maintains the `max_len` and `result` variables to track the overall maximum LIS length and the number of LISs with that length. After processing all elements, the `result` will contain the desired count.

Number of Longest Increasing Subsequence - Complete Solution Guide

Number of Longest Increasing Subsequence is LeetCode problem 673, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 -10 6

Detailed Explanation

The problem asks us to find the number of Longest Increasing Subsequences (LIS) within a given integer array `nums`. An increasing subsequence is a sequence of numbers within the array that are strictly increasing (nums[i] < nums[j] for i < j). We need to return how many such subsequences have the maximum possible length. The input array can have any integers, including duplicates, and its length can range from 1 to 2000.

Solution Approach

The provided code employs a dynamic programming approach. It iterates through the input array `nums` and, for each element `nums[i]`, it considers all preceding elements `nums[j]` (where `j < i`). If `nums[j] < nums[i]`, it means `nums[i]` can potentially extend an increasing subsequence ending at `nums[j]`. The `lengths` array stores the length of the LIS ending at each index, and the `counts` array stores the number of such LISs. The algorithm maintains the `max_len` and `result` variables to track the overall maximum LIS length and the number of LISs with that length. After processing all elements, the `result` will contain the desired count.

Step-by-Step Algorithm

Step 1: Initialize two arrays, `lengths` and `counts`, each with the same size as the input array `nums`. Initialize all elements of `lengths` to 1 (as each element itself forms an LIS of length 1) and all elements of `counts` to 1 (as there's initially one LIS of length 1 ending at each index).
Step 2: Initialize `max_len` to 0 and `result` to 0. These variables will store the maximum LIS length found so far and the number of LISs with that maximum length.
Step 3: Iterate through the input array `nums` from index `i = 0` to `n-1`.
Step 4: For each `i`, iterate through the preceding elements `nums[j]` from index `j = 0` to `i-1`.
Step 5: If `nums[j] < nums[i]` (i.e., `nums[i]` can extend the subsequence ending at `nums[j]`):
a. If `lengths[j] + 1 > lengths[i]` (extending the subsequence at `j` results in a longer LIS at `i`): Update `lengths[i]` to `lengths[j] + 1` and `counts[i]` to `counts[j]` (because all LISs ending at `i` are now extensions of LISs ending at `j`).
b. Else if `lengths[j] + 1 == lengths[i]` (extending the subsequence at `j` results in an LIS of the same length at `i`): Add `counts[j]` to `counts[i]` (because we found additional LISs of the same length ending at `i`).
Step 6: After processing all `j` for a given `i`, update `max_len` and `result`:
a. If `lengths[i] > max_len`: Update `max_len` to `lengths[i]` and `result` to `counts[i]` (we found a new longest LIS).
b. Else if `lengths[i] == max_len`: Add `counts[i]` to `result` (we found additional LISs of the maximum length).
Step 7: After processing all elements in `nums`, return `result`.

Key Insights

Insight 1: Dynamic Programming is suitable because the LIS problem exhibits optimal substructure. The LIS ending at index `i` can be built from the LIS ending at earlier indices.
Insight 2: We need to track both the length of the LIS ending at each index and the number of such LISs. Two arrays, `lengths` and `counts`, are crucial.
Insight 3: When building an LIS, if extending a previous subsequence results in a longer subsequence at the current index, we update both the length and count. If it results in a subsequence of the same length, we add the count of the previous subsequence to the current count.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Binary Indexed Tree, Segment Tree.

Companies

Asked at: Commvault, Intuit.

Number of Longest Increasing Subsequence | LeetCode 673

Medium

Array Dynamic Programming Binary Indexed Tree Segment Tree

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }

        int[] lengths = new int[n];
        int[] counts = new int[n];

        for (int i = 0; i < n; i++) {
            lengths[i] = 1;
            counts[i] = 1;
        }

        int max_len = 0;
        int result = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    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];
                    }
                }
            }

            if (lengths[i] > max_len) {
                max_len = lengths[i];
                result = counts[i];
            } else if (lengths[i] == max_len) {
                result += counts[i];
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 673: Number of Longest Increasing Subsequence - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Binary Indexed Tree Segment Tree

Companies

CommvaultIntuit

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:673

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize two arrays, `lengths` and `counts`, each with the same size as the input array `nums`. Initialize all elements of `lengths` to 1 (as each element itself forms an LIS of length 1) and all elements of `counts` to 1 (as there's initially one LIS of length 1 ending at each index).
2
Step 2: Initialize `max_len` to 0 and `result` to 0. These variables will store the maximum LIS length found so far and the number of LISs with that maximum length.
3
Step 3: Iterate through the input array `nums` from index `i = 0` to `n-1`.
4
Step 4: For each `i`, iterate through the preceding elements `nums[j]` from index `j = 0` to `i-1`.
5
Step 5: If `nums[j] < nums[i]` (i.e., `nums[i]` can extend the subsequence ending at `nums[j]`):
6
a. If `lengths[j] + 1 > lengths[i]` (extending the subsequence at `j` results in a longer LIS at `i`): Update `lengths[i]` to `lengths[j] + 1` and `counts[i]` to `counts[j]` (because all LISs ending at `i` are now extensions of LISs ending at `j`).
7
b. Else if `lengths[j] + 1 == lengths[i]` (extending the subsequence at `j` results in an LIS of the same length at `i`): Add `counts[j]` to `counts[i]` (because we found additional LISs of the same length ending at `i`).
8
Step 6: After processing all `j` for a given `i`, update `max_len` and `result`:
9
a. If `lengths[i] > max_len`: Update `max_len` to `lengths[i]` and `result` to `counts[i]` (we found a new longest LIS).
10
b. Else if `lengths[i] == max_len`: Add `counts[i]` to `result` (we found additional LISs of the maximum length).
11
Step 7: After processing all elements in `nums`, return `result`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested for loops. The outer loop iterates `n` times, and the inner loop iterates up to `n-1` times in the worst case, resulting in O(n*(n-1)) = O(n^2) operations. The space complexity is due to the `lengths` and `counts` arrays, which both scale linearly with the size of the input array, hence O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Indexed Tree (Fenwick Tree)

This approach uses a Binary Indexed Tree to optimize the search for the best preceding subsequence to extend. It can reduce the time complexity compared to the pure DP solution.

Time: O(n log n)Space: O(n)

Trade-offs: More complex implementation, but improves the time complexity, making it suitable for larger input sizes.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the LIS problem exhibits optimal substructure. The LIS ending at index `i` can be built from the LIS ending at earlier indices.
2Insight 2: We need to track both the length of the LIS ending at each index and the number of such LISs. Two arrays, `lengths` and `counts`, are crucial.
3Insight 3: When building an LIS, if extending a previous subsequence results in a longer subsequence at the current index, we update both the length and count. If it results in a subsequence of the same length, we add the count of the previous subsequence to the current count.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array (`nums` is empty). The code explicitly handles this case by returning 0.
Edge case 2: Input array with all same elements (e.g., `[2, 2, 2, 2, 2]`). The algorithm correctly calculates the LIS length as 1 and the number of LISs as the number of elements in the array (which is 5 in this example).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing the `lengths` and `counts` arrays correctly. Each element should be considered an LIS of length 1 initially, so both arrays must be initialized to 1.

✗Common mistake 2: Failing to handle the case where extending a subsequence results in an LIS of the *same* length. In this scenario, the count of the current LIS must be incremented by the count of the extended subsequence.

✗Common mistake 3: Updating the overall `result` and `max_len` incorrectly. Make sure to check if a newly found LIS is longer than the current `max_len` and update both `max_len` and `result` accordingly. If it's the same length, simply add the current count to the result.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock market analysis - Identifying the longest period of increasing stock prices and determining how many such periods exist can provide valuable insights for traders.

🏦Real-world scenario 2: Bioinformatics - Analyzing DNA sequences to find the longest increasing sequence of specific nucleotides, which can be useful in gene prediction.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Longest Increasing Subsequence"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?