[Array] Check If N and Its Double Exist - Easy

To check if a number ( N ) and its double ( 2N ) are in an array, we can use smart methods. We can use data structures like hash sets. The main way is to go through the array and store elements in a hash set. This helps us find out if ( N ) and ( 2N ) are there very quickly. This way is much faster than a simple method.

In this article, we will look at different ways to solve the problem of checking if a number and its double exist in an array. First, we will understand the problem. Then, we will see the best solutions using HashSet in Java, Python, and C++. After that, we will talk about simple methods in these languages. Lastly, we will look at the time complexity of all methods and answer common questions.

  • [Array] Determine If N and Its Double Exists in an Array - Easy
  • Understanding the Problem Statement
  • Optimal Solution Using HashSet in Java
  • Brute Force Approach in Java
  • Optimal Solution Using HashSet in Python
  • Brute Force Approach in Python
  • Optimal Solution Using HashSet in C++
  • Brute Force Approach in C++
  • Time Complexity Analysis of All Approaches
  • Frequently Asked Questions

If you want to learn more about similar topics, you can check these articles: Array Two Sum - Easy, Array Contains Duplicate - Easy, and Array Maximum Subarray - Easy.

Understanding the Problem Statement

We need to find out if there is an integer ( n ) in a given array. This integer must have its double, ( 2n ), also in the same array.

Input

  • We have an integer array arr with size ( m ) where ( 0 m ).
  • The integers in this array can be from (-10^5) to (10^5).

Output

  • We will return true if we find an integer ( n ) in the array such that ( 2n ) is also in the array.
  • We return false if we do not find such an integer.

Examples

  1. Input: arr = [10, 2, 5, 3]
    Output: true
    Explanation: Here, ( n = 5 ) and ( 2n = 10 ) are both in the array.

  2. Input: arr = [7, 1, 14, 11]
    Output: true
    Explanation: In this case, ( n = 7 ) and ( 2n = 14 ) are both in the array.

  3. Input: arr = [3, 1, 7, 11]
    Output: false
    Explanation: There is no integer in this array that has its double also in it.

Constraints

  • We should handle special cases like empty arrays or arrays with negative numbers. We want our solution to work well for all cases.

This problem is about finding ways to search for numbers in an array. We can use methods like HashSets or simple loops to solve it.

Optimal Solution Using HashSet in Java

We want to find out if a number ( n ) and its double ( 2n ) are in an array. We can do this quickly by using a HashSet in Java. The HashSet gives us fast lookups. This means we can check for numbers in almost constant time ( O(1) ).

Implementation Steps:

  1. We go through the array and save each number in a HashSet.
  2. For each number ( n ), we check if ( 2n ) is in the HashSet.
  3. If we find it, we return true. If we finish checking all numbers and do not find it, we return false.

Java Code Example:

import java.util.HashSet;

public class DoubleExist {
    public static boolean checkIfExist(int[] arr) {
        HashSet<Integer> set = new HashSet<>();
        for (int num : arr) {
            // Check if double of the number exists in the set
            if (set.contains(num * 2) || (num % 2 == 0 && set.contains(num / 2))) {
                return true;
            }
            set.add(num);
        }
        return false;
    }

    public static void main(String[] args) {
        int[] array = {10, 2, 5, 3};
        System.out.println(checkIfExist(array)); // Output: true
    }
}

Key Properties:

  • Time Complexity: ( O(n) ). Here ( n ) is the number of elements in the array.
  • Space Complexity: ( O(n) ). This is for saving numbers in the HashSet.
  • We also think about edge cases where ( n ) can be zero or negative.

This method works well. It uses the HashSet to make our search for ( 2n ) faster. If you want to learn more about array techniques, you can check Array Two Sum.

Brute Force Approach in Java

The brute force way to check if an integer ( N ) and its double ( 2N ) are in an array is to go through the array and look for both values. This method takes a lot of time. It has a time complexity of ( O(n^2) ) because of the loops inside loops. So, it is not very efficient for big datasets.

Here is an example of how we can use this method in Java:

public class CheckDoubleExistence {
    public static boolean checkIfExist(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < arr.length; j++) {
                if (i != j && arr[i] == 2 * arr[j]) {
                    return true;
                }
            }
        }
        return false;
    }

    public static void main(String[] args) {
        int[] array = {10, 2, 5, 3};
        System.out.println(checkIfExist(array)); // Output: true
    }
}

Explanation of the Code:

  • We use two loops to go through the array.
  • The first loop picks an element ( arr[i] ). The second loop checks if there is another element ( arr[j] ) such that ( arr[i] ) equals ( 2 arr[j] ).
  • We make sure i is not equal to j so we do not compare an element with itself.

This brute force method is simple but not good for large arrays. It checks every possible pair. For larger datasets, we should think about a better way like using HashSet which works in linear time ( O(n) ). For better solutions, check the section on Optimal Solution Using HashSet in Java.

Optimal Solution Using HashSet in Python

We want to find out if a number n and its double 2n are in an array using Python. We can use a HashSet for quick checks. This way, we can look for both n and 2n in one go.

Implementation

Here is a simple Python code that shows this good solution using a set:

def checkIfExist(arr):
    seen = set()
    
    for num in arr:
        if num * 2 in seen or (num % 2 == 0 and num // 2 in seen):
            return True
        seen.add(num)
    
    return False

# Example usage
arr = [10, 2, 5, 3]
result = checkIfExist(arr)
print(result)  # Output: True

Explanation

  • Data Structure: We use a set to keep track of numbers we have seen.
  • Loop: We go through each number in the array:
    • First, we check if double the number is in the set.
    • Then, we check if half the number is in the set (only for even numbers).
  • Time Complexity: O(n). Here, n is how many elements are in the array. Each check and add in a set takes O(1) on average.

This method is quick and simple. It gives us a clear way to find if n and 2n are in the array. For more related topics, we can look at Array Two Sum or Array Contains Duplicate.

Brute Force Approach in Python

To check if an integer ( n ) and its double ( 2n ) are in an array, we can use a simple brute force approach in Python. We can do this with nested loops. This means we will check every pair of elements in the array. This method works but it has a time complexity of ( O(n^2) ).

Here is a simple code for the brute force approach:

def checkIfExist(arr):
    n = len(arr)
    for i in range(n):
        for j in range(n):
            if i != j and arr[i] == 2 * arr[j]:
                return True
    return False

# Example usage:
arr = [10, 2, 5, 3]
print(checkIfExist(arr))  # Output: True

Explanation:

  • We go through each element ( arr[i] ).
  • For each ( arr[i] ), we check the array again with ( arr[j] ).
  • We see if ( arr[i] ) is equal to ( 2 arr[j] ). We make sure that ( i ) is not the same as ( j ) to avoid easy cases.
  • If we find such a pair, we return True. If not, we return False after checking all pairs.

This method is easy to understand but it is not the best choice for big arrays. It can be slow because of its quadratic time complexity. For bigger datasets, we can think about using a HashSet to make it faster. We can talk about that in the better solutions parts.

Optimal Solution Using HashSet in C++

To find out if a number ( n ) and its double ( 2n ) are in an array using C++, we can use a HashSet or unordered_set in C++. This way, we can check if these values are there in constant time. This gives us a total time complexity of linear.

Implementation Steps:

  1. Create an unordered_set to keep the elements from the array.
  2. Go through the array and add each number to the set.
  3. For each number, check if its double or half is in the set.

C++ Code Example:

#include <iostream>
#include <unordered_set>
#include <vector>

bool checkIfDoubleExists(const std::vector<int>& arr) {
    std::unordered_set<int> numSet;

    for (int num : arr) {
        numSet.insert(num);
    }

    for (int num : arr) {
        if (numSet.count(num * 2) > 0 || (num % 2 == 0 && numSet.count(num / 2) > 0)) {
            return true;
        }
    }

    return false;
}

int main() {
    std::vector<int> arr = {10, 2, 5, 3};
    if (checkIfDoubleExists(arr)) {
        std::cout << "N and its double exist in the array." << std::endl;
    } else {
        std::cout << "N and its double do not exist in the array." << std::endl;
    }
    return 0;
}

Explanation of the Code:

  • We create unordered_set<int> numSet to store unique numbers from the input array.
  • The first loop fills the set with numbers from the array.
  • The second loop checks if each number has its double or half in the set.
  • The method count looks for the numbers in constant average time.

Key Properties:

  • Time Complexity: ( O(n) ) where ( n ) is the number of elements in the array.
  • Space Complexity: ( O(n) ) for keeping elements in the set.

This way, we can check for the presence of ( n ) and its double in the array using C++. For more algorithms related to arrays, check Array Two Sum or Array Contains Duplicate.

Brute Force Approach in C++

In the brute force method, we check if a number ( N ) and its double ( 2N ) are in an array. We do this by going through each item in the array. We check for both values. This method takes time ( O(n^2) ) because we compare each item with every other item.

Here is how we can write the brute force solution in C++:

#include <iostream>
#include <vector>
using namespace std;

bool checkIfExists(vector<int>& arr) {
    int n = arr.size();
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (i != j && arr[i] == 2 * arr[j]) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    vector<int> arr = {10, 2, 5, 3};
    bool result = checkIfExists(arr);
    cout << (result ? "True" : "False") << endl; // Output: True
    return 0;
}

Explanation:

  • The first loop goes through each element ( arr[i] ).
  • The second loop checks every other element to see if it is half of ( arr[i] ).
  • If we find both ( N ) and ( 2N ), we return true. If not, we return false after checking all pairs.

This method is simple but not good for large arrays. For better speed, we can use a hash set. This can give us a better time of linear complexity. We will talk about this in the optimal solution section.

If you want to read more about different array problems, look at Array Two Sum or Array Contains Duplicate.

Time Complexity Analysis of All Approaches

When we look at time complexity of different ways to check if N and its double are in an array, we can see both the best and brute force methods.

Optimal Solution Using HashSet

  • Time Complexity: O(N)
    • We process each element in the array one time when we add to the HashSet. Then we check for N/2 and 2*N again.
    • The total complexity is linear with the size of the input array.

Brute Force Approach

  • Time Complexity: O(N^2)
    • This way uses nested loops to check all possible pairs of elements in the array.
    • For each element, we may need to look through the whole array to find its double. This gives us a quadratic complexity.

Optimal Solution Using HashSet in Python

  • Time Complexity: O(N)
    • Like the Java version, we use a HashSet to keep the elements and check for N/2 and 2*N.
    • This also gives us linear time complexity.

Brute Force Approach in Python

  • Time Complexity: O(N^2)
    • This method also uses nested loops. It checks each element with every other element.
    • The result is still a quadratic time complexity.

Optimal Solution Using HashSet in C++

  • Time Complexity: O(N)
    • For C++, we use unordered_set to get linear time complexity.
    • We insert and check elements in a good way.

Brute Force Approach in C++

  • Time Complexity: O(N^2)
    • Just like in other languages, the brute force method checks each element with all others in a nested way.
    • The complexity stays quadratic.

Summary of Time Complexities

Approach Time Complexity
Optimal (HashSet) O(N)
Brute Force O(N^2)

Understanding the time complexity of these approaches is very important for making performance better, especially when we work with large data sets. For more about related array problems, we can check articles like Array Two Sum and Array Contains Duplicate.

Frequently Asked Questions

1. What does it mean to check if N and its double exist in an array?

When we check if N and its double exist in an array, we want to find out if a number N is in the array. We also check if the number 2 * N is also in that array. This problem often comes up in algorithm tasks. We can solve it using different methods, like using a HashSet for better performance.

2. How can I implement the optimal solution using HashSet in Java?

To use the best solution with HashSet in Java, we can go through the array and keep each number in the HashSet. While we add each number, we check if it is half of any number we already added or if it is double any of those numbers. This way, we keep it fast with O(n) time, which is good for big data sets.

import java.util.HashSet;

public boolean checkIfExist(int[] arr) {
    HashSet<Integer> set = new HashSet<>();
    for (int num : arr) {
        if (set.contains(num * 2) || (num % 2 == 0 && set.contains(num / 2))) {
            return true;
        }
        set.add(num);
    }
    return false;
}

3. What is the brute force approach for checking N and its double in Python?

The brute force way to check if N and its double are in an array uses two loops. Each number gets compared with all other numbers in the array. This method has a time complexity of O(n^2) and is not as good as using sets. But it can be easy to understand for small arrays.

def checkIfExist(arr):
    for i in range(len(arr)):
        for j in range(len(arr)):
            if i != j and arr[i] == 2 * arr[j]:
                return True
    return False

4. How can I analyze the time complexity of different approaches?

To look at the time complexity of different ways to check if N and its double are in an array, we should think about both the good and brute force methods. The best solution with HashSet takes O(n) time because we go through the array once and check in constant time. The brute force method takes O(n^2) time because of the two loops. This shows how hash methods can be more efficient.

5. Are there any similar problems I can practice on?

Yes, there are many similar problems we can practice to get better at arrays and algorithms. For example, we can try problems like Array Two Sum or Array Contains Duplicate. These will help us improve our skills with arrays and hash tables.