What is the time complexity of Count the Number of Fair Pairs?

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

What data structures are used to solve Count the Number of Fair Pairs?

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

What are the key insights for solving Count the Number of Fair Pairs?

Insight 1: Sorting the array allows us to efficiently find pairs that sum up to a value within the specified range (lower, upper) using a two-pointer approach. Insight 2: Instead of directly counting fair pairs in the range [lower, upper], we can count pairs whose sum is less than or equal to `upper` and subtract the count of pairs whose sum is less than `lower`. This simplifies the counting process.

Which companies ask LeetCode 2563. Count the Number of Fair Pairs in interviews?

This problem is frequently asked at Qualcomm, Salesforce.

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Count the Number of Fair Pairs - LeetCode 2563 Solution

Q: How to solve Count the Number of Fair Pairs?

The solution involves sorting the input array and then using a two-pointer technique to count the number of pairs whose sum falls within the given range. The core idea is to count the number of pairs with a sum less than or equal to `upper` and then subtract the number of pairs with a sum less than `lower` (which is equivalent to less than or equal to `lower - 1`). This difference gives us the count of fair pairs.

Count the Number of Fair Pairs - Complete Solution Guide

Count the Number of Fair Pairs is LeetCode problem 2563, 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 a 0-indexed integer array nums of size n and two integers lower and upper , return the number of fair pairs . A pair (i, j) is fair if: 0 <= i < j < n , and lower <= nums[i] + nums[j] <= upper Example 1: Input: nums = [0,1,7,4,4,5], lower = 3, upper = 6 Output: 6 Explanation: There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5). Example 2: Input: nums = [1,7,9,2,5], lower = 11, upper = 11 Output: 1 Explanation: There is a single fair pair: (2,3). Constraints: 1 <= nums.leng

Detailed Explanation

The problem asks us to count the number of 'fair pairs' in a given array `nums`. A fair pair `(i, j)` is defined as a pair of indices where `0 <= i < j < n` (meaning `i` is strictly less than `j`, and both are valid indices in the array) and `lower <= nums[i] + nums[j] <= upper`. We are given the array `nums` and the integer bounds `lower` and `upper` as input, and we need to return the total count of such fair pairs.

Solution Approach

The solution involves sorting the input array and then using a two-pointer technique to count the number of pairs whose sum falls within the given range. The core idea is to count the number of pairs with a sum less than or equal to `upper` and then subtract the number of pairs with a sum less than `lower` (which is equivalent to less than or equal to `lower - 1`). This difference gives us the count of fair pairs.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order using a standard sorting algorithm (e.g., merge sort, quicksort).
Step 2: Define a helper function `count_less_equal(k)` that counts the number of pairs `(i, j)` such that `i < j` and `nums[i] + nums[j] <= k`. This function uses a two-pointer approach.
Step 3: In the `count_less_equal(k)` function, initialize two pointers, `left` to 0 and `right` to `len(nums) - 1`.
Step 4: While `left < right`, check if `nums[left] + nums[right] <= k`. If it is, then all pairs `(left, left+1)`, `(left, left+2)`, ..., `(left, right)` satisfy the condition, so increment the count by `right - left`. Then increment `left` to explore larger sums.
Step 5: If `nums[left] + nums[right] > k`, it means the current sum is too large. Decrement `right` to try smaller sums.
Step 6: Return the count from the `count_less_equal(k)` function.
Step 7: Calculate the result as `count_less_equal(upper) - count_less_equal(lower - 1)` and return it.

Key Insights

Insight 1: Sorting the array allows us to efficiently find pairs that sum up to a value within the specified range (lower, upper) using a two-pointer approach.
Insight 2: Instead of directly counting fair pairs in the range [lower, upper], we can count pairs whose sum is less than or equal to `upper` and subtract the count of pairs whose sum is less than `lower`. This simplifies the counting process.
Insight 3: It's crucial to consider that `i` must be strictly less than `j`. The two-pointer approach must maintain this invariant.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Qualcomm, Salesforce.

Count the Number of Fair Pairs | LeetCode 2563

Medium

Array Two Pointers Binary Search Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long countFairPairs(int[] nums, int lower, int upper) {
        Arrays.sort(nums);
        
        
        
        return count_less_equal(nums,upper) - count_less_equal(nums,lower - 1);
    }
    
    private long count_less_equal(int[] nums, int k) {
        long count = 0;
        int left = 0;
        int right = nums.length - 1;
        while (left < right) {
            if (nums[left] + nums[right] <= k) {
                count += (right - left);
                left++;
            } else {
                right--;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2563: Count the Number of Fair Pairs - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

QualcommSalesforce

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2563

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

7 steps

1
Step 1: Sort the input array `nums` in ascending order using a standard sorting algorithm (e.g., merge sort, quicksort).
2
Step 2: Define a helper function `count_less_equal(k)` that counts the number of pairs `(i, j)` such that `i < j` and `nums[i] + nums[j] <= k`. This function uses a two-pointer approach.
3
Step 3: In the `count_less_equal(k)` function, initialize two pointers, `left` to 0 and `right` to `len(nums) - 1`.
4
Step 4: While `left < right`, check if `nums[left] + nums[right] <= k`. If it is, then all pairs `(left, left+1)`, `(left, left+2)`, ..., `(left, right)` satisfy the condition, so increment the count by `right - left`. Then increment `left` to explore larger sums.
5
Step 5: If `nums[left] + nums[right] > k`, it means the current sum is too large. Decrement `right` to try smaller sums.
6
Step 6: Return the count from the `count_less_equal(k)` function.
7
Step 7: Calculate the result as `count_less_equal(upper) - count_less_equal(lower - 1)` and return it.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting the array takes O(n log n) time. The two-pointer approach in the `count_less_equal` function takes O(n) time. Since we call `count_less_equal` twice, the overall time complexity is dominated by the sorting step, making it O(n log n). The space complexity is O(1) because we are sorting the array in-place (or using a sorting algorithm with O(1) space) and using only a constant amount of extra space for variables.

📖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 Search

Instead of using two pointers after sorting, for each element `nums[i]`, we can use binary search to find the range of elements `nums[j]` (where j > i) such that `lower - nums[i] <= nums[j] <= upper - nums[i]`. The number of elements in this range contributes to the total count.

Time: O(n log n)Space: O(1)

Trade-offs: The binary search approach has the same time complexity as the two-pointer approach but might be slightly less efficient in practice due to the overhead of binary search for each element.

✨

Key Insights

1Insight 1: Sorting the array allows us to efficiently find pairs that sum up to a value within the specified range (lower, upper) using a two-pointer approach.
2Insight 2: Instead of directly counting fair pairs in the range [lower, upper], we can count pairs whose sum is less than or equal to `upper` and subtract the count of pairs whose sum is less than `lower`. This simplifies the counting process.
3Insight 3: It's crucial to consider that `i` must be strictly less than `j`. The two-pointer approach must maintain this invariant.

📚

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: Empty or very small input array (nums.length <= 1). The algorithm correctly handles this as the while loop `left < right` will not execute.
Edge case 2: All elements are the same. The two-pointer approach will still function correctly and count the pairs accordingly.
Edge case 3: `lower` and `upper` are equal. The solution correctly computes the number of pairs summing to the target value by subtracting the number of pairs less than lower-1 from number of pairs less than upper.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the array. Sorting is crucial for the two-pointer and binary search approaches to work correctly.

✗Common mistake 2: Incorrectly handling the condition `i < j`. The two-pointer approach must ensure that `left` is always less than `right` to avoid counting the same pair twice or including invalid pairs.

✗Common mistake 3: Integer overflow when calculating the sum `nums[left] + nums[right]`. Although the individual elements are constrained to -10^9 to 10^9, their sum could reach 2 * 10^9, which might exceed the maximum value of a 32-bit integer in some languages. Casting to `long long` avoids this issue, especially in C/C++ and Java.

✅

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 that fall within a certain range, such as identifying customers whose combined spending falls within a target revenue bracket.

🏦Real-world scenario 2: In signal processing, identifying frequency components that fall within a specific bandwidth by analyzing pairs of frequencies.

💡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: July 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

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.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 the Number of Fair Pairs"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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