What is the time complexity of Count Pairs Whose Sum is Less than Target?

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

What data structures are used to solve Count Pairs Whose Sum is Less than Target?

This problem uses Array, Two Pointers, Binary Search, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Pairs Whose Sum is Less than Target?

Insight 1: Brute-force approach is sufficient due to small input constraints. The problem's constraints (n <= 50) make a simple nested loop solution feasible without requiring optimization techniques that would be necessary for larger datasets. Insight 2: Nested loops provide a straightforward way to iterate through all possible pairs of indices (i, j) where i < j.

Which companies ask LeetCode 2824. Count Pairs Whose Sum is Less than Target in interviews?

This problem is frequently asked at Accenture, josh technology.

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Count Pairs Whose Sum is Less than Target - LeetCode 2824 Solution

Q: How to solve Count Pairs Whose Sum is Less than Target?

The provided code utilizes a brute-force approach. It iterates through all possible pairs of indices (i, j) using nested loops. For each pair, it checks if the sum of the corresponding elements is less than the `target`. If it is, the `count` is incremented. Finally, the function returns the total `count` of pairs that meet the condition.

Q: Which companies ask LeetCode 2824. Count Pairs Whose Sum is Less than Target in interviews?

This problem is frequently asked at Accenture, josh technology.

Count Pairs Whose Sum is Less than Target - Complete Solution Guide

Count Pairs Whose Sum is Less than Target is LeetCode problem 2824, a Easy 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 length n and an integer target , return the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target . Example 1: Input: nums = [-1,1,2,3,1], target = 2 Output: 3 Explanation: There are 3 pairs of indices that satisfy the conditions in the statement: - (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target - (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target - (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target Note that (0, 3) is not

Detailed Explanation

The problem asks you to count the number of pairs of indices (i, j) in a given integer array `nums` such that `i < j` and the sum of the elements at those indices (`nums[i] + nums[j]`) is strictly less than a given integer `target`. The array is 0-indexed, meaning the first element is at index 0. The problem emphasizes that `i` must be strictly less than `j`. The constraints limit the array size to a maximum of 50 elements, and the values within the array, as well as the `target`, are bounded between -50 and 50.

Solution Approach

The provided code utilizes a brute-force approach. It iterates through all possible pairs of indices (i, j) using nested loops. For each pair, it checks if the sum of the corresponding elements is less than the `target`. If it is, the `count` is incremented. Finally, the function returns the total `count` of pairs that meet the condition.

Step-by-Step Algorithm

Step 1: Initialize a `count` variable to 0. This variable will store the number of pairs that satisfy the condition.
Step 2: Iterate through the array using a nested loop. The outer loop iterates from `i = 0` to `n-1` (where `n` is the length of the array). The inner loop iterates from `j = i + 1` to `n-1`. This ensures that `i < j` for every pair considered.
Step 3: For each pair (i, j), calculate the sum `nums[i] + nums[j]`.
Step 4: Check if the sum is strictly less than the `target`. If it is (`nums[i] + nums[j] < target`), increment the `count`.
Step 5: After iterating through all pairs, return the final `count`.

Key Insights

Insight 1: Brute-force approach is sufficient due to small input constraints. The problem's constraints (n <= 50) make a simple nested loop solution feasible without requiring optimization techniques that would be necessary for larger datasets.
Insight 2: Nested loops provide a straightforward way to iterate through all possible pairs of indices (i, j) where i < j.
Insight 3: The condition `nums[i] + nums[j] < target` is crucial; it ensures that only pairs satisfying this inequality are counted. The strict inequality (<) is important and must not be mistakenly interpreted as less than or equal to (<=).

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Sorting.

Companies

Asked at: Accenture, josh technology.

Count Pairs Whose Sum is Less than Target | LeetCode 2824

Easy

Array Two Pointers Binary Search Sorting

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

Solution Code

Available in:

import java.util.List;

class Solution {
    public int countPairs(List<Integer> nums, int target) {
        int count = 0;
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums.get(i) + nums.get(j) < target) {
                    count++;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2824: Count Pairs Whose Sum is Less than Target - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

Accenturejosh technology

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2824

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

5 steps

1
Step 1: Initialize a `count` variable to 0. This variable will store the number of pairs that satisfy the condition.
2
Step 2: Iterate through the array using a nested loop. The outer loop iterates from `i = 0` to `n-1` (where `n` is the length of the array). The inner loop iterates from `j = i + 1` to `n-1`. This ensures that `i < j` for every pair considered.
3
Step 3: For each pair (i, j), calculate the sum `nums[i] + nums[j]`.
4
Step 4: Check if the sum is strictly less than the `target`. If it is (`nums[i] + nums[j] < target`), increment the `count`.
5
Step 5: After iterating through all pairs, return the final `count`.

💡

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 nested loops iterate through all possible pairs of indices in the array. The number of such pairs is proportional to n*(n-1)/2, which is O(n^2). The solution uses only a constant amount of extra space to store the `count` variable and loop counters; therefore, the space complexity is O(1).

📖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

Sorting and Two Pointers

Sort the array. Then, use two pointers (one at the beginning and one at the end). Move the pointers towards each other, counting pairs that satisfy the condition. This approach reduces the time complexity for larger arrays, but sorting adds overhead.

Time: O(n log n) (dominated by sorting)Space: O(1) or O(n) depending on sorting implementation

Trade-offs: This is preferable for larger arrays where O(n^2) becomes computationally expensive. The brute-force approach is more concise and suitable for small arrays as specified in the problem constraints.

✨

Key Insights

1Insight 1: Brute-force approach is sufficient due to small input constraints. The problem's constraints (n <= 50) make a simple nested loop solution feasible without requiring optimization techniques that would be necessary for larger datasets.
2Insight 2: Nested loops provide a straightforward way to iterate through all possible pairs of indices (i, j) where i < j.
3Insight 3: The condition `nums[i] + nums[j] < target` is crucial; it ensures that only pairs satisfying this inequality are counted. The strict inequality (<) is important and must not be mistakenly interpreted as less than or equal to (<=).

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: An empty array. The code handles this correctly; the outer loop won't execute, and the function will return 0.
Edge case 2: An array where all pairs sum to a value greater than or equal to the target. The code will correctly return 0 in this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly using `<=` instead of `<` in the condition. This leads to an incorrect count.

✗Common mistake 2: Not handling the indices correctly in the nested loops. Forgetting `j = i + 1` in the inner loop's initialization would lead to counting pairs (i, i) which is not allowed by the problem statement.

✅

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: Finding pairs of data points (e.g., stock prices, sensor readings) whose combined value falls below a certain threshold.

🏦Real-world scenario 2: Identifying pairs of items in a database that, when combined, meet a specific cost or weight constraint.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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 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

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

❓

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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 Pairs Whose Sum is Less than Target"?

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

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