What is the time complexity of Count Increasing Quadruplets?

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

What is the space complexity of Count Increasing Quadruplets?

The space complexity is O(n^2).

What data structures are used to solve Count Increasing Quadruplets?

This problem uses Array, Dynamic Programming, Binary Indexed Tree, Enumeration, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Increasing Quadruplets?

Insight 1: The core idea is to iterate through all possible pairs (j, k) and for each pair, count how many i's exist to the left of j such that nums[i] < nums[k], and how many l's exist to the right of k such that nums[j] < nums[l]. Insight 2: Using prefix sums or similar data structures to precompute the counts of elements less than or greater than a certain value to the left and right respectively can significantly reduce the complexity.

Which companies ask LeetCode 2552. Count Increasing Quadruplets in interviews?

This problem is frequently asked at SAP.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Count Increasing Quadruplets - LeetCode 2552 Solution

Q: How to solve Count Increasing Quadruplets?

The provided solution uses dynamic programming to precompute the number of elements smaller than a certain value to the left of each index and the number of elements greater than a certain value to the right of each index. This is achieved by creating two 2D arrays, `less` and `greater`. Then, it iterates through all possible pairs (j, k) such that j < k. If nums[k] < nums[j], it finds the number of i's to the left of j such that nums[i] < nums[k] using the `less` array, and the number of l's to the right of k such that nums[j] < nums[l] using the `greater` array. The product of these two counts contributes to the final answer.

Q: Which companies ask LeetCode 2552. Count Increasing Quadruplets in interviews?

This problem is frequently asked at SAP.

Count Increasing Quadruplets - Complete Solution Guide

Count Increasing Quadruplets is LeetCode problem 2552, 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

Given a 0-indexed integer array nums of size n containing all numbers from 1 to n , return the number of increasing quadruplets . A quadruplet (i, j, k, l) is increasing if: 0 <= i < j < k < l < n , and nums[i] < nums[k] < nums[j] < nums[l] . Example 1: Input: nums = [1,3,2,4,5] Output: 2 Explanation: - When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l]. - When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l]. There are no other quadruplets, so we ret

Detailed Explanation

The problem requires us to find the number of increasing quadruplets (i, j, k, l) in a given 0-indexed integer array `nums` of size `n`, where all numbers from 1 to n are present exactly once. An increasing quadruplet satisfies the conditions: `0 <= i < j < k < l < n` and `nums[i] < nums[k] < nums[j] < nums[l]`. The task is to count how many such quadruplets exist in the array.

Solution Approach

The provided solution uses dynamic programming to precompute the number of elements smaller than a certain value to the left of each index and the number of elements greater than a certain value to the right of each index. This is achieved by creating two 2D arrays, `less` and `greater`. Then, it iterates through all possible pairs (j, k) such that j < k. If nums[k] < nums[j], it finds the number of i's to the left of j such that nums[i] < nums[k] using the `less` array, and the number of l's to the right of k such that nums[j] < nums[l] using the `greater` array. The product of these two counts contributes to the final answer.

Step-by-Step Algorithm

Step 1: Initialize two 2D arrays, `less` of size n x (n+1) and `greater` of size n x (n+1). `less[j][v]` will store the number of elements to the left of index j that are less than v. `greater[k][v]` will store the number of elements to the right of index k that are greater than v.
Step 2: Populate the `less` array. For each index j from 1 to n-1, iterate through possible values v from 1 to n. The value of `less[j][v]` is initially the same as `less[j-1][v]`. If nums[j-1] < v, increment `less[j][v]` by 1.
Step 3: Populate the `greater` array. For each index k from n-2 down to 0, iterate through possible values v from 1 to n. The value of `greater[k][v]` is initially the same as `greater[k+1][v]`. If nums[k+1] > v, increment `greater[k][v]` by 1.
Step 4: Iterate through all possible pairs (j, k) such that 0 <= j < k < n. For each pair, check if nums[k] < nums[j].
Step 5: If nums[k] < nums[j], find the number of i's to the left of j such that nums[i] < nums[k]. This is given by `less[j][nums[k]]`.
Step 6: If nums[k] < nums[j], find the number of l's to the right of k such that nums[j] < nums[l]. This is given by `greater[k][nums[j]]`.
Step 7: Multiply the two counts obtained in steps 5 and 6, and add the result to the `ans` variable.
Step 8: Return the `ans` variable, which represents the total number of increasing quadruplets.

Key Insights

Insight 1: The core idea is to iterate through all possible pairs (j, k) and for each pair, count how many i's exist to the left of j such that nums[i] < nums[k], and how many l's exist to the right of k such that nums[j] < nums[l].
Insight 2: Using prefix sums or similar data structures to precompute the counts of elements less than or greater than a certain value to the left and right respectively can significantly reduce the complexity.
Insight 3: Precomputing `less` and `greater` arrays allows the number of `i` and `l` to be calculated in constant time for given `j` and `k`.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming, Binary Indexed Tree, Enumeration, Prefix Sum.

Companies

Asked at: SAP.

Count Increasing Quadruplets | LeetCode 2552

Hard

Array Dynamic Programming Binary Indexed Tree Enumeration Prefix Sum

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

Solution Code

Available in:

class Solution {
    public long countQuadruplets(int[] nums) {
        int n = nums.length;
        
        int[][] less = new int[n][n + 1];
        for (int j = 1; j < n; j++) {
            for (int v = 1; v <= n; v++) {
                less[j][v] = less[j - 1][v];
                if (nums[j - 1] < v) {
                    less[j][v] += 1;
                }
            }
        }
        
        int[][] greater = new int[n][n + 1];
        for (int k = n - 2; k >= 0; k--) {
            for (int v = 1; v <= n; v++) {
                greater[k][v] = greater[k + 1][v];
                if (nums[k + 1] > v) {
                    greater[k][v] += 1;
                }
            }
        }
        
        long ans = 0;
        for (int j = 0; j < n; j++) {
            for (int k = j + 1; k < n; k++) {
                if (nums[k] < nums[j]) {
                    int count_i = less[j][nums[k]];
                    int count_l = greater[k][nums[j]];
                    ans += (long) count_i * count_l;
                }
            }
        }
        
        return ans;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2552: Count Increasing Quadruplets - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Binary Indexed Tree Enumeration Prefix Sum

Companies

SAP

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2552

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 two 2D arrays, `less` of size n x (n+1) and `greater` of size n x (n+1). `less[j][v]` will store the number of elements to the left of index j that are less than v. `greater[k][v]` will store the number of elements to the right of index k that are greater than v.
2
Step 2: Populate the `less` array. For each index j from 1 to n-1, iterate through possible values v from 1 to n. The value of `less[j][v]` is initially the same as `less[j-1][v]`. If nums[j-1] < v, increment `less[j][v]` by 1.
3
Step 3: Populate the `greater` array. For each index k from n-2 down to 0, iterate through possible values v from 1 to n. The value of `greater[k][v]` is initially the same as `greater[k+1][v]`. If nums[k+1] > v, increment `greater[k][v]` by 1.
4
Step 4: Iterate through all possible pairs (j, k) such that 0 <= j < k < n. For each pair, check if nums[k] < nums[j].
5
Step 5: If nums[k] < nums[j], find the number of i's to the left of j such that nums[i] < nums[k]. This is given by `less[j][nums[k]]`.
6
Step 6: If nums[k] < nums[j], find the number of l's to the right of k such that nums[j] < nums[l]. This is given by `greater[k][nums[j]]`.
7
Step 7: Multiply the two counts obtained in steps 5 and 6, and add the result to the `ans` variable.
8
Step 8: Return the `ans` variable, which represents the total number of increasing quadruplets.

💡

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^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The dominant factor in time complexity is the nested loops for j and k, resulting in O(n^2). The space complexity is determined by the size of the `less` and `greater` 2D arrays which scale quadratically with input size `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 (BIT) / Fenwick Tree

Instead of precomputing the `less` and `greater` arrays, a Binary Indexed Tree (BIT) can be used to query the number of elements less than or greater than a certain value in O(log n) time. This approach would iterate through j and k, and use BIT to count the possible i's and l's

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

Trade-offs: This approach trades off space for a slightly higher time complexity. It uses O(n) space for the BIT, but the time complexity increases due to the log n factor for BIT operations. Useful if memory is a concern.

Brute-force Approach

A naive brute-force approach involves iterating through all possible quadruplets (i, j, k, l) and checking if they satisfy the given conditions.

Time: O(n^4)Space: O(1)

Trade-offs: This approach has a much higher time complexity and is generally not feasible for larger input sizes. It's only useful for very small arrays due to simplicity.

✨

Key Insights

1Insight 1: The core idea is to iterate through all possible pairs (j, k) and for each pair, count how many i's exist to the left of j such that nums[i] < nums[k], and how many l's exist to the right of k such that nums[j] < nums[l].
2Insight 2: Using prefix sums or similar data structures to precompute the counts of elements less than or greater than a certain value to the left and right respectively can significantly reduce the complexity.
3Insight 3: Precomputing `less` and `greater` arrays allows the number of `i` and `l` to be calculated in constant time for given `j` and `k`.

📚

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 array or array with less than 4 elements. The solution implicitly handles this because the loops will not execute or not execute sufficiently to perform the comparisons.
Edge case 2: Array with already sorted elements (e.g., [1, 2, 3, 4]). The algorithm correctly identifies that there are no increasing quadruplets in such a case, and returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the precomputation of `less` and `greater` arrays, leading to incorrect counts of elements smaller or greater than a given value.

✗Common mistake 2: Not handling integer overflow. The product of `count_i` and `count_l` can be large, potentially exceeding the maximum value of an integer. Therefore, using `long` data type is essential.

✅

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: Data analysis where finding specific increasing patterns within a dataset can reveal valuable insights. For example, analyzing stock prices or sales data to identify trends.

🏦Real-world scenario 2: Ranking algorithms where identifying specific patterns is needed. Analyzing ranking of products to find relationships and how they influence the overall customer preference.

💡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

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

⚠️ 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: July 3, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Increasing Quadruplets"?

For this hard-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?