What is the time complexity of Find K-th Smallest Pair Distance?

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

What data structures are used to solve Find K-th Smallest Pair Distance?

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

What are the key insights for solving Find K-th Smallest Pair Distance?

Insight 1: Sorting the input array allows us to use binary search to efficiently find the k-th smallest distance. Insight 2: The problem can be solved by using binary search on the possible range of distances. The lower bound is 0, and the upper bound is the difference between the largest and smallest numbers in the array.

Which companies ask LeetCode 719. Find K-th Smallest Pair Distance in interviews?

This problem is frequently asked at Flipkart, Pinterest.

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Find K-th Smallest Pair Distance - LeetCode 719 Solution

Q: How to solve Find K-th Smallest Pair Distance?

The solution uses a combination of sorting and binary search. First, the input array `nums` is sorted. Then, a binary search is performed on the range of possible distances to find the k-th smallest distance. For each `mid` value during the binary search, we count the number of pairs with a distance less than or equal to `mid`. If the count is less than `k`, it means that the k-th smallest distance is greater than `mid`, so we search in the upper half. Otherwise, it means the k-th smallest distance is less than or equal to `mid`, so we search in the lower half. The `count_pairs` function efficiently calculates the number of pairs using a two-pointer approach.

Find K-th Smallest Pair Distance - Complete Solution Guide

Find K-th Smallest Pair Distance is LeetCode problem 719, 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

The distance of a pair of integers a and b is defined as the absolute difference between a and b . Given an integer array nums and an integer k , return the k th smallest distance among all the pairs nums[i] and nums[j] where 0 <= i < j < nums.length . Example 1: Input: nums = [1,3,1], k = 1 Output: 0 Explanation: Here are all the pairs: (1,3) -> 2 (1,1) -> 0 (3,1) -> 2 Then the 1 st smallest distance pair is (1,1), and its distance is 0. Example 2: Input: nums = [1,1,1], k = 2 Output: 0 Example

Detailed Explanation

The problem asks us to find the k-th smallest distance among all possible pairs of numbers in a given integer array `nums`. The distance between a pair of numbers `a` and `b` is defined as the absolute difference `|a - b|`. The input consists of an integer array `nums` and an integer `k`. The output is the k-th smallest distance. We need to consider all possible pairs `(nums[i], nums[j])` where `0 <= i < j < nums.length`.

Solution Approach

The solution uses a combination of sorting and binary search. First, the input array `nums` is sorted. Then, a binary search is performed on the range of possible distances to find the k-th smallest distance. For each `mid` value during the binary search, we count the number of pairs with a distance less than or equal to `mid`. If the count is less than `k`, it means that the k-th smallest distance is greater than `mid`, so we search in the upper half. Otherwise, it means the k-th smallest distance is less than or equal to `mid`, so we search in the lower half. The `count_pairs` function efficiently calculates the number of pairs using a two-pointer approach.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize `low` to 0 and `high` to `nums[-1] - nums[0]`. These represent the lower and upper bounds of the possible distances.
Step 3: Perform binary search between `low` and `high` until `low` is equal to `high`.
Step 4: In each iteration of the binary search, calculate the `mid` value as `low + (high - low) // 2`.
Step 5: Use the `count_pairs` function to count the number of pairs with a distance less than or equal to `mid`.
Step 6: If the `count` is less than `k`, it means that the k-th smallest distance is greater than `mid`, so update `low` to `mid + 1`.
Step 7: Otherwise, it means that the k-th smallest distance is less than or equal to `mid`, so update `high` to `mid`.
Step 8: When the loop terminates, `low` will be equal to `high`, and this value will be the k-th smallest distance. Return `low`.

Key Insights

Insight 1: Sorting the input array allows us to use binary search to efficiently find the k-th smallest distance.
Insight 2: The problem can be solved by using binary search on the possible range of distances. The lower bound is 0, and the upper bound is the difference between the largest and smallest numbers in the array.
Insight 3: For a given distance `mid`, we need to efficiently count the number of pairs with a distance less than or equal to `mid`. This can be done in O(n) time using two pointers after sorting.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Flipkart, Pinterest.

Find K-th Smallest Pair Distance | LeetCode 719

Hard

Array Two Pointers Binary Search Sorting

Time: O(nlogn)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int smallestDistancePair(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;

        // Inner function to count pairs
        class Helper {
            public int countPairs(int dist) {
                int count = 0;
                int right = 0;
                for (int left = 0; left < n; left++) {
                    while (right < n && nums[right] - nums[left] <= dist) {
                        right++;
                    }
                    count += right - left - 1;
                }
                return count;
            }
        }
        Helper helper = new Helper();

        int low = 0;
        int high = nums[n - 1] - nums[0];

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (helper.countPairs(mid) < k) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }

        return low;
    }
}

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Problem Info

Description

Problem 719: Find K-th Smallest Pair Distance - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sorting

Companies

FlipkartPinterest

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:719

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: Sort the input array `nums` in ascending order.
2
Step 2: Initialize `low` to 0 and `high` to `nums[-1] - nums[0]`. These represent the lower and upper bounds of the possible distances.
3
Step 3: Perform binary search between `low` and `high` until `low` is equal to `high`.
4
Step 4: In each iteration of the binary search, calculate the `mid` value as `low + (high - low) // 2`.
5
Step 5: Use the `count_pairs` function to count the number of pairs with a distance less than or equal to `mid`.
6
Step 6: If the `count` is less than `k`, it means that the k-th smallest distance is greater than `mid`, so update `low` to `mid + 1`.
7
Step 7: Otherwise, it means that the k-th smallest distance is less than or equal to `mid`, so update `high` to `mid`.
8
Step 8: When the loop terminates, `low` will be equal to `high`, and this value will be the k-th smallest distance. Return `low`.

💡

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(nlogn)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting the array takes O(n log n) time. The binary search iterates log(m) times, where m is the range of distances. Inside the binary search, the `count_pairs` function uses a two-pointer approach to count the number of pairs with a distance less than or equal to the current `mid` value. This takes O(n) time. Therefore, the overall time complexity is dominated by sorting and the binary search combined, resulting in O(n log n + n log m). The space complexity is O(1) because the algorithm only uses a few extra 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

Heap-based approach

Instead of binary search, we could use a min-heap to keep track of the smallest distances. We would initially add all pair distances to the heap and then extract the minimum k times. This is less efficient in this case.

Time: O(n^2 + k log n^2) which is O(n^2 + k log n) since the number of pairs is n(n-1)/2Space: O(n^2) in the worst case to store all pairs distances.

Trade-offs: This approach is generally less efficient than the binary search approach due to the quadratic time complexity for generating all pairs and the logarithmic cost for heap operations. It might be useful if the number of pairs to consider were significantly smaller (e.g., if `k` was very small and we didn't need *all* pairs).

✨

Key Insights

1Insight 1: Sorting the input array allows us to use binary search to efficiently find the k-th smallest distance.
2Insight 2: The problem can be solved by using binary search on the possible range of distances. The lower bound is 0, and the upper bound is the difference between the largest and smallest numbers in the array.
3Insight 3: For a given distance `mid`, we need to efficiently count the number of pairs with a distance less than or equal to `mid`. This can be done in O(n) time using two pointers after sorting.

📚

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: When all elements in `nums` are the same, the distance between all pairs is 0. If k is within the possible pair count, the solution correctly returns 0.
Edge case 2: When k is equal to 1, the solution returns the smallest distance between any two numbers. This is correctly handled by the binary search.
Edge case 3: When k is equal to n*(n-1)/2, the solution returns the largest distance between any two numbers when considering only the valid pairs

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the input array before applying binary search. Sorting is crucial for the two-pointer approach within `count_pairs` to work correctly.

✗Common mistake 2: Incorrectly calculating the `mid` value in the binary search. Using `low + (high - low) // 2` prevents potential integer overflow issues.

✗Common mistake 3: Not properly handling the edge cases where all numbers in the array are the same.

✅

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 the k-th closest data point in a dataset. For example, in recommendation systems, you might want to find the k most similar users based on their profile data.

🏦Real-world scenario 2: In manufacturing, determining the k-th smallest deviation in production measurements to identify potential quality control issues.

💡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: November 27, 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

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.

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 "Find K-th Smallest Pair Distance"?

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

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