What is the time complexity of Count Number of Teams?

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

What data structures are used to solve Count Number of Teams?

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 Count Number of Teams?

Insight 1: The core idea is to iterate through each element of the array and consider it as the middle element (rating[j]) of a potential team. Insight 2: For each middle element, count the number of elements smaller and larger to its left and right.

Which companies ask LeetCode 1395. Count Number of Teams in interviews?

This problem is frequently asked at IBM.

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Count Number of Teams - LeetCode 1395 Solution

Count Number of Teams - Complete Solution Guide

Count Number of Teams is LeetCode problem 1395, 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

There are n soldiers standing in a line. Each soldier is assigned a unique rating value. You have to form a team of 3 soldiers amongst them under the following rules: Choose 3 soldiers with index ( i , j , k ) with rating ( rating[i] , rating[j] , rating[k] ). A team is valid if: ( rating[i] < rating[j] < rating[k] ) or ( rating[i] > rating[j] > rating[k] ) where ( 0 <= i < j < k < n ). Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams). Exampl

Detailed Explanation

The problem asks us to find the number of valid teams that can be formed from a given array of soldier ratings. A team consists of three soldiers at indices i, j, and k (where i < j < k) and their ratings must either be strictly increasing (rating[i] < rating[j] < rating[k]) or strictly decreasing (rating[i] > rating[j] > rating[k]). The goal is to count the number of such valid teams.

Solution Approach

The provided solution uses a brute-force approach. It iterates through each element of the `rating` array and considers it as the middle element of a potential team. For each such middle element, it iterates through the elements to its left and right, counting how many are smaller and larger than the middle element. Finally, it calculates the number of increasing and decreasing teams that can be formed with this middle element and adds it to the total count.

Step-by-Step Algorithm

Step 1: Initialize `total_teams` to 0. This variable will store the final count of valid teams.
Step 2: Iterate through the `rating` array using a loop with index `j` from 0 to `n-1`. Each element `rating[j]` is considered as the middle element of a potential team.
Step 3: Inside the outer loop, initialize `left_smaller`, `left_larger`, `right_smaller`, and `right_larger` to 0. These variables will store the count of elements smaller/larger than `rating[j]` to its left and right, respectively.
Step 4: Iterate through the elements to the left of `rating[j]` (from `i = 0` to `j - 1`). If `rating[i]` is smaller than `rating[j]`, increment `left_smaller`. Otherwise, increment `left_larger`.
Step 5: Iterate through the elements to the right of `rating[j]` (from `k = j + 1` to `n - 1`). If `rating[k]` is smaller than `rating[j]`, increment `right_smaller`. Otherwise, increment `right_larger`.
Step 6: Calculate the number of increasing teams with `rating[j]` as the middle element as `left_smaller * right_larger` and add it to `total_teams`.
Step 7: Calculate the number of decreasing teams with `rating[j]` as the middle element as `left_larger * right_smaller` and add it to `total_teams`.
Step 8: After the outer loop completes, return `total_teams`.

Key Insights

Insight 1: The core idea is to iterate through each element of the array and consider it as the middle element (rating[j]) of a potential team.
Insight 2: For each middle element, count the number of elements smaller and larger to its left and right.
Insight 3: The number of increasing teams with rating[j] as the middle element is the product of (number of elements smaller than rating[j] on the left) and (number of elements larger than rating[j] on the right). Similarly, the number of decreasing teams with rating[j] as the middle element is the product of (number of elements larger than rating[j] on the left) and (number of elements smaller than rating[j] on the right).
Insight 4: The problem constraints state that the rating values are unique, simplifying the comparisons. We don't have to deal with equal ratings.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

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

Companies

Asked at: IBM.

Count Number of Teams | LeetCode 1395

Medium

Array Dynamic Programming Binary Indexed Tree Segment Tree

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int numTeams(int[] rating) {
        int n = rating.length;
        if (n < 3) {
            return 0;
        }

        int total_teams = 0;
        for (int j = 0; j < n; j++) {
            int left_smaller = 0;
            int left_larger = 0;
            for (int i = 0; i < j; i++) {
                if (rating[i] < rating[j]) {
                    left_smaller += 1;
                } else {
                    left_larger += 1;
                }
            }
            
            int right_smaller = 0;
            int right_larger = 0;
            for (int k = j + 1; k < n; k++) {
                if (rating[k] < rating[j]) {
                    right_smaller += 1;
                } else {
                    right_larger += 1;
                }
            }

            // Count increasing teams (rating[i] < rating[j] < rating[k])
            total_teams += left_smaller * right_larger;
            
            // Count decreasing teams (rating[i] > rating[j] > rating[k])
            total_teams += left_larger * right_smaller;
        
        }
        return total_teams;
    }
}

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Problem Info

Description

Problem 1395: Count Number of Teams - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Binary Indexed Tree Segment Tree

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1395

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 `total_teams` to 0. This variable will store the final count of valid teams.
2
Step 2: Iterate through the `rating` array using a loop with index `j` from 0 to `n-1`. Each element `rating[j]` is considered as the middle element of a potential team.
3
Step 3: Inside the outer loop, initialize `left_smaller`, `left_larger`, `right_smaller`, and `right_larger` to 0. These variables will store the count of elements smaller/larger than `rating[j]` to its left and right, respectively.
4
Step 4: Iterate through the elements to the left of `rating[j]` (from `i = 0` to `j - 1`). If `rating[i]` is smaller than `rating[j]`, increment `left_smaller`. Otherwise, increment `left_larger`.
5
Step 5: Iterate through the elements to the right of `rating[j]` (from `k = j + 1` to `n - 1`). If `rating[k]` is smaller than `rating[j]`, increment `right_smaller`. Otherwise, increment `right_larger`.
6
Step 6: Calculate the number of increasing teams with `rating[j]` as the middle element as `left_smaller * right_larger` and add it to `total_teams`.
7
Step 7: Calculate the number of decreasing teams with `rating[j]` as the middle element as `left_larger * right_smaller` and add it to `total_teams`.
8
Step 8: After the outer loop completes, return `total_teams`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^2) because the solution uses nested loops. The outer loop iterates 'n' times, and for each iteration, the inner loops iterate a maximum of 'n' times (in the worst case). The space complexity is O(1) because the solution uses a constant amount of extra space, regardless of the input size. Only a few integer variables are used.

📖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) or Segment Tree

This approach involves using BIT or Segment Tree data structures to efficiently count the number of smaller and larger elements on the left and right sides of each element. We can create a BIT to store the counts of elements encountered so far and query it for each rating[j].

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

Trade-offs: This approach offers better time complexity than the brute-force approach, but requires more space and a more complex implementation. Use this when n is large.

✨

Key Insights

1Insight 1: The core idea is to iterate through each element of the array and consider it as the middle element (rating[j]) of a potential team.
2Insight 2: For each middle element, count the number of elements smaller and larger to its left and right.
3Insight 3: The number of increasing teams with rating[j] as the middle element is the product of (number of elements smaller than rating[j] on the left) and (number of elements larger than rating[j] on the right). Similarly, the number of decreasing teams with rating[j] as the middle element is the product of (number of elements larger than rating[j] on the left) and (number of elements smaller than rating[j] on the right).
4Insight 4: The problem constraints state that the rating values are unique, simplifying the comparisons. We don't have to deal with equal ratings.

📚

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: If the input array has less than 3 elements, no team can be formed. The code handles this by returning 0 at the beginning.
Edge case 2: An input array with all elements in increasing or decreasing order. The algorithm handles this correctly by counting all possible increasing or decreasing combinations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the condition i < j < k. It's crucial to iterate correctly to ensure the indices are in the right order when calculating the team count.

✗Common mistake 2: Incorrectly counting smaller/larger elements, leading to an incorrect team count.

✗Common mistake 3: Not handling the edge case of less than 3 ratings properly, resulting in incorrect output for small inputs.

✅

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: Analyzing stock prices. Finding patterns of increasing or decreasing stock prices over a period of time is similar to finding teams in this problem.

🏦Real-world scenario 2: Evaluating employee performance. If you have a sequence of employee performance ratings, you might want to identify individuals whose performance has consistently improved or declined compared to their peers.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 14, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 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 "Count Number of Teams"?

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.

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