Find Dupes: An O(NLogN) Solution

-
Here is the problem (#423 in my solved list): https://leetcode.com/problems/find-the-duplicate-number/

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Example 1:
Input: [1,3,4,2,2]
Output: 2
Example 2:
Input: [3,1,3,4,2]
Output: 3
Note:
  1. You must not modify the array (assume the array is read only).
  2. You must use only constant, O(1) extra space.
  3. Your runtime complexity should be less than O(n2).
  4. There is only one duplicate number in the array, but it could be repeated more than once.

The key here are the two statements (conditions) in yellow above: no extra space, and some time complexity better than O(N^2). The first option that came to my mind then would be an O(NLogN): sort the array, and do a linear scan checking which two adjacent elements are the same, and when you find it, return. Code is below, cheers, ACC.

public class Solution
{
    public int FindDuplicate(int[] nums)
    {
        Array.Sort(nums);

        for (int i = 0; i < nums.Length - 1; i++)
        {
            if (nums[i] == nums[i + 1]) return nums[i];
        }
        return -1;
    }
}

Comments

  1. TarasSeptember 2, 2019 at 7:13 PM

    Great solution! It's possible to get a linear complexity though by looking at this problem from a different angle - imagine that each array element is a pointer to the next "node" forming a linked list. Now what does it mean to have 2 (or more) nodes in linked list that point to the same next node? Correct, a cycle. So we can use a famous Tortoise and Hare algorithm to find a cycle and its starting point (duplicate):

    class Solution {
    public:
    int findDuplicate(vector& nums) {
    int slow = nums.front();
    int fast = nums.front();
    while (true) {
    slow = nums[slow];
    fast = nums[nums[fast]];
    if (slow == fast) break;
    }
    slow = nums.front();
    while (slow != fast) {
    slow = nums[slow];
    fast = nums[fast];
    }
    return slow;
    }
    };

    ReplyDelete

Post a Comment

Popular posts from this blog

Advent of Code - Day 6, 2024: BFS and FSM

-
Image
To solve this Advent of Code (parts 1 and 2), I used Breadth-First Search (BFS) to keep moving the guard as well as Finite State Machine (FSM) to control the directions. This solves part 1 easily. Part 2 can be solved using the same technique, takes a little longer since you need to try every empty cell, but it eventually does the trick (takes a min or so). Code is down below, cheers, ACC. Day 6 - Advent of Code 2024 using System.IO; using System.Collections; using System; using System.Text; using System.Text.RegularExpressions; using System.ComponentModel.DataAnnotations; Process2(); void Process() { string fileName = "input.txt"; FileInfo fileInfo = new FileInfo(fileName); StreamReader sr = fileInfo.OpenText(); List map = new List (); int startRow = 0; int startCol = 0; while (!sr.EndOfStream) { string line = sr.ReadLine().Trim(); map.Add(new StringBuilder(line)); int indexStart = line.IndexOf('^'); ...
Read more

Golang vs. C#: performance characteristics (simple case study)

-
Image
This is an interesting example to study the performance characteristics of Golang compared to C# for a very particular case (not generalized though).  The same code was written in C# and Golang (literally translated from one to the other). Below you'll be able to see the source code for both. The problem is a simple post-order DFS in a tree-like graph. The code in C# times out (TLE = Time Limited Exceeded). The code in Golang not only works but beats 100% of the other Golang submissions. Take a look the image below: Some potential reasons for the performance gains from Go in this particular example are: 1. Language and Runtime Differences:    Go is compiled directly to machine code, while C# typically runs on a virtual machine (CLR).    Go has a simpler runtime with less overhead compared to the .NET runtime.    Go's garbage collector is designed for low latency, which can lead to better performance in some scenarios. 2. Data Structures:    G...
Read more

Advent of Code - Day 7, 2024: Backtracking and Eval

-
Image
Problem today requires backtracking and eval (evaluating a mathematical expression in the format of a string). Glad that it used a left-2-right evaluation, makes for an easy parsing and not overly complicated evaluation function. Code is down below for parts 1 and 2. Cheers, ACC. Day 7 - Advent of Code 2024 using System.IO; using System.Collections; using System; using System.Text; using System.Text.RegularExpressions; using System.ComponentModel.DataAnnotations; Process2(); void Process() { string fileName = "input.txt"; long retVal = 0; FileInfo fileInfo = new FileInfo(fileName); StreamReader sr = fileInfo.OpenText(); while (!sr.EndOfStream) { string line = sr.ReadLine().Trim(); string[] parts1 = line.Split(':'); long target = Int64.Parse(parts1[0]); string[] numbers = parts1[1].Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries); if (TryAll1(numbers, target, "", 0)) r...
Read more