What is the time complexity of K Inverse Pairs Array?

The optimal solution has a time complexity of O(n*k).

What data structures are used to solve K Inverse Pairs Array?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving K Inverse Pairs Array?

Insight 1: Dynamic Programming can efficiently calculate the number of arrays with a specific number of inverse pairs. Insight 2: The number of inverse pairs can be built incrementally. Adding a new number `i` to an existing array can introduce 0 to `i-1` new inverse pairs.

Which companies ask LeetCode 629. K Inverse Pairs Array in interviews?

This problem is frequently asked at Works Applications.

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K Inverse Pairs Array - LeetCode 629 Solution

K Inverse Pairs Array - Complete Solution Guide

K Inverse Pairs Array is LeetCode problem 629, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

For an integer array nums , an inverse pair is a pair of integers [i, j] where 0 <= i < j < nums.length and nums[i] > nums[j] . Given two integers n and k, return the number of different arrays consisting of numbers from 1 to n such that there are exactly k inverse pairs . Since the answer can be huge, return it modulo 10 9 + 7 . Example 1: Input: n = 3, k = 0 Output: 1 Explanation: Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pairs. Example 2: Input: n = 3,

Detailed Explanation

The problem asks us to find the number of distinct arrays of length `n` containing numbers from 1 to `n` such that the number of inverse pairs in the array is exactly `k`. An inverse pair is defined as a pair of indices `(i, j)` where `0 <= i < j < n` and `nums[i] > nums[j]`. The result needs to be returned modulo `10^9 + 7` due to potentially large values.

Solution Approach

The solution utilizes dynamic programming. `dp[j]` represents the number of arrays of length `i` with `j` inverse pairs. We iteratively build up the solution from `i = 1` to `n`. In each iteration, we construct a new DP table `temp` based on the previous DP table `dp`. When adding the number `i` to an array of length `i-1`, the number `i` can be inserted at any position. Inserting it at the end adds 0 inverse pairs, inserting it before the last element adds 1 inverse pair, and so on, up to inserting it at the beginning adding `i-1` inverse pairs. The inner loop calculates `temp[j]` by considering all possible positions to insert `i` in arrays with up to `j` inverse pairs. The subtraction of `dp[j-i]` is to remove the arrays that have more than `j` inverse pairs if we placed `i` at the beginning.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `k+1` with all elements set to 0. Set `dp[0] = 1`, as there is only one way to have 0 inverse pairs (a sorted array).
Step 2: Iterate from `i = 1` to `n` (inclusive). This represents the size of the array we're building.
Step 3: In each iteration, create a temporary DP array `temp` of size `k+1`, initialized with all elements set to 0. Set `temp[0] = 1`.
Step 4: Iterate from `j = 1` to `k` (inclusive). This represents the number of inverse pairs we're aiming for.
Step 5: Calculate `temp[j]` by adding `temp[j-1]` and `dp[j]`. `temp[j-1]` represents the number of arrays of length `i` with `j-1` inverse pairs considering that we added number 'i' in a position that created one new inverse pair. `dp[j]` represents the number of arrays of length i-1 with 'j' inverse pairs. Adding these together and taking modulo helps to handle overflow cases.
Step 6: If `j >= i`, subtract `dp[j-i]` from `temp[j]` to account for invalid cases (where adding 'i' to an existing array requires at least `i` inverse pairs, and exceeding 'k'). This ensures the number of inverse pairs does not become negative after subtraction by adding MOD.
Step 7: After the inner loop completes, update `dp` with `temp`. This ensures that we're using the results of adding `i` numbers for the next iteration.
Step 8: Finally, return `dp[k]`, which contains the number of arrays of length `n` with exactly `k` inverse pairs.

Key Insights

Insight 1: Dynamic Programming can efficiently calculate the number of arrays with a specific number of inverse pairs.
Insight 2: The number of inverse pairs can be built incrementally. Adding a new number `i` to an existing array can introduce 0 to `i-1` new inverse pairs.
Insight 3: We can use a sliding window optimization by only keeping track of the last `i` values in the DP table. This significantly improves the space complexity.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(k)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Works Applications.

K Inverse Pairs Array | LeetCode 629

Hard

Dynamic Programming

Time: O(n*k)Space: O(k)

Solution Code

Available in:

class Solution {
    public int kInversePairs(int n, int k) {
        int MOD = 1000000007;

        int[] dp = new int[k + 1];
        dp[0] = 1;

        for (int i = 1; i <= n; i++) {
            int[] temp = new int[k + 1];
            temp[0] = 1;
            for (int j = 1; j <= k; j++) {
                long val = (long)(temp[j - 1] + dp[j]) % MOD;
                if (j >= i) {
                    val = (val - dp[j - i] + MOD) % MOD;
                }
                temp[j] = (int)val;
            }
            dp = temp;
        }

        return dp[k];
    }
}

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Problem Info

Description

Problem 629: K Inverse Pairs Array - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Works Applications

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:629

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

8 steps

1
Step 1: Initialize a DP array `dp` of size `k+1` with all elements set to 0. Set `dp[0] = 1`, as there is only one way to have 0 inverse pairs (a sorted array).
2
Step 2: Iterate from `i = 1` to `n` (inclusive). This represents the size of the array we're building.
3
Step 3: In each iteration, create a temporary DP array `temp` of size `k+1`, initialized with all elements set to 0. Set `temp[0] = 1`.
4
Step 4: Iterate from `j = 1` to `k` (inclusive). This represents the number of inverse pairs we're aiming for.
5
Step 5: Calculate `temp[j]` by adding `temp[j-1]` and `dp[j]`. `temp[j-1]` represents the number of arrays of length `i` with `j-1` inverse pairs considering that we added number 'i' in a position that created one new inverse pair. `dp[j]` represents the number of arrays of length i-1 with 'j' inverse pairs. Adding these together and taking modulo helps to handle overflow cases.
6
Step 6: If `j >= i`, subtract `dp[j-i]` from `temp[j]` to account for invalid cases (where adding 'i' to an existing array requires at least `i` inverse pairs, and exceeding 'k'). This ensures the number of inverse pairs does not become negative after subtraction by adding MOD.
7
Step 7: After the inner loop completes, update `dp` with `temp`. This ensures that we're using the results of adding `i` numbers for the next iteration.
8
Step 8: Finally, return `dp[k]`, which contains the number of arrays of length `n` with exactly `k` inverse pairs.

💡

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*k)

💾

Space Complexity

O(k)

📝

Why?

The outer loop runs `n` times, and the inner loop runs `k` times. The operations inside the inner loop are constant time. We only maintain two arrays of size `k+1`, so the space complexity is O(k).

📖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

Recursive approach with memoization

A recursive solution can be developed using memoization to avoid redundant calculations. The function would take n, k, and a memoization table as input. It explores all possibilities of adding numbers from 1 to n to form an array with k inverse pairs.

Time: O(n*k)Space: O(n*k)

Trade-offs: Although having the same time complexity, this approach could lead to a StackOverflow error if the recursion depth is too deep, and the constant factor in space complexity is higher as memoization table has size n*k.

✨

Key Insights

1Insight 1: Dynamic Programming can efficiently calculate the number of arrays with a specific number of inverse pairs.
2Insight 2: The number of inverse pairs can be built incrementally. Adding a new number `i` to an existing array can introduce 0 to `i-1` new inverse pairs.
3Insight 3: We can use a sliding window optimization by only keeping track of the last `i` values in the DP table. This significantly improves the space complexity.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: When k = 0, the only possible array is the sorted array [1, 2, ..., n], so the answer is 1. The code handles this correctly because dp[0] is initialized to 1.
Edge case 2: When n = 1, there can only be 0 inverse pairs. The code handles this correctly as the outer loop starts at i=1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to apply the modulo operator at each step, leading to integer overflow and incorrect results.

✗Common mistake 2: Incorrectly handling the subtraction in the DP update step (not adding MOD before subtracting can lead to negative values and incorrect results).

✗Common mistake 3: Using O(n*k) space complexity DP solution resulting in memory limit exceeded error especially in lower level languages like C.

✅

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: Ranking systems. Determining the number of possible ranking arrangements given a specific number of inversions (e.g., how many ways can a leaderboard be arranged if we know there are 'k' swaps needed to sort it).

🏦Real-world scenario 2: Data compression. The number of inversions in a sequence can be used to measure the 'unsortedness' of the sequence, which is related to the amount of information needed to represent it efficiently. This could inform the choice of compression algorithms.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: August 29, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "K Inverse Pairs Array"?

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

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