[Array] Find All Numbers Disappeared in an Array - Easy

Finding all numbers that are missing in an array means we need to find the numbers that should be there but are not. Usually, this problem has an array of numbers from 1 to n. Some numbers might be missing. We can solve this problem in different ways. We can use index mapping, HashSet, or sorting. Each method has its own speed and efficiency.

In this article, we will look at different ways to find the missing numbers in an array. First, we will explain the problem clearly. Then, we will show how to do it in Java, Python, and C++. We will also look at the best ways to solve this problem using index mapping, HashSet, and sorting. Finally, we will compare these methods and answer some common questions.

• [Array] Find All Numbers Disappeared in an Array Solution Approaches
• Understanding the Problem Statement for Disappeared Numbers in an Array
• Java Implementation for Finding Disappeared Numbers in an Array
• Python Approach to Identify Missing Numbers in an Array
• C++ Method to Find All Missing Numbers in an Array
• Optimal Solution Using Index Mapping for Missing Numbers
• Using HashSet to Find Missing Numbers in an Array
• Sorting the Array to Locate Missing Numbers
• Comparative Analysis of Different Approaches for Disappeared Numbers
• Frequently Asked Questions

Understanding the Problem Statement for Disappeared Numbers in an Array

Finding numbers that have disappeared in an array is a common problem in coding. We have an array of integers. Each integer is from 1 to n. The length of the array is also n. Our goal is to find all integers in that range that are not in the array.

Problem Statement:

• Input: An integer array nums of length n with numbers from 1 to n.
• Output: An array of integers that are missing from nums.

Example:

• Input: nums = [4,3,2,7,8,2,3,1]
• Output: [5,6]

Constraints:

• Numbers in the input array can show up many times.
• We should aim for a good time complexity, like O(n), and not use more than O(1) extra space.

This problem checks how well we understand array handling and smart algorithms. We can use methods like index mapping or hash-based ways to find missing numbers. There are many ways to solve this problem while keeping the rules in mind.

If we want to read more about similar array problems, we can check out Array Two Sum - Easy or Array Contains Duplicate - Easy.

Java Implementation for Finding Disappeared Numbers in an Array

We can find all numbers that are missing in an array with numbers from 1 to n. We use a smart way that uses the properties of the array. The main idea is to mark the numbers we see in the array. We do this by changing the sign of the number at the index that matches the value of the number.

Here is a simple Java code for this:

import java.util.ArrayList;
import java.util.List;

public class DisappearedNumbers {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> result = new ArrayList<>();
        
        // Loop through the array
        for (int i = 0; i < nums.length; i++) {
            int index = Math.abs(nums[i]) - 1; // Get the index for the value
            nums[index] = -Math.abs(nums[index]); // Mark the number as found
        }
        
        // Find the missing numbers
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) { // If the number is still positive, it is missing
                result.add(i + 1); // Add the missing number (index + 1)
            }
        }
        
        return result;
    }

    public static void main(String[] args) {
        DisappearedNumbers dn = new DisappearedNumbers();
        int[] nums = {4, 3, 2, 7, 8, 2, 3, 1};
        System.out.println(dn.findDisappearedNumbers(nums)); // Output: [5, 6]
    }
}

Explanation

• Input: An array nums where numbers go from 1 to n.
• Output: A list of missing numbers.
• Complexity: This method runs in O(n) time and uses O(1) extra space since it changes the input array.

This Java solution is simple and quick. It helps us find all the missing numbers in the array. If you want to learn more, you can check out related topics like Array - Contains Duplicate to improve your skills in working with arrays.

Python Approach to Identify Missing Numbers in an Array

We can find all missing numbers in an array using Python. We will look at different ways to do this. In this task, we think the array has numbers from 1 to n. Here, n is the length of the array. Our aim is to find the numbers in this range that are not in the array.

Method 1: Using a HashSet

This method uses a HashSet to keep track of which numbers are present.

def find_disappeared_numbers(nums):
    n = len(nums)
    num_set = set(nums)
    missing_numbers = []

    for i in range(1, n + 1):
        if i not in num_set:
            missing_numbers.append(i)
    
    return missing_numbers

# Example
nums = [4, 3, 2, 7, 8, 2, 3, 1]
print(find_disappeared_numbers(nums))  # Output: [5, 6]

Method 2: Index Mapping

This good solution uses index mapping to find missing numbers without using extra space.

def find_disappeared_numbers(nums):
    n = len(nums)

    for i in range(n):
        index = abs(nums[i]) - 1
        nums[index] = -abs(nums[index])  # Mark as found

    missing_numbers = [i + 1 for i in range(n) if nums[i] > 0]
    
    return missing_numbers

# Example
nums = [4, 3, 2, 7, 8, 2, 3, 1]
print(find_disappeared_numbers(nums))  # Output: [5, 6]

Method 3: Sorting the Array

Another simple way is to sort the array and check for missing numbers.

def find_disappeared_numbers(nums):
    nums.sort()
    missing_numbers = []

    for i in range(1, len(nums) + 1):
        if i not in nums:
            missing_numbers.append(i)
    
    return missing_numbers

# Example
nums = [4, 3, 2, 7, 8, 2, 3, 1]
print(find_disappeared_numbers(nums))  # Output: [5, 6]

These methods help us find missing numbers in an array using Python. Each method has its own space and time needs. The choice of method can change based on the specific needs of the problem. If you want to explore more similar problems, you can check the Array Two Sum article.

C++ Method to Find All Missing Numbers in an Array

We want to find all numbers that are missing in an array that has numbers from 1 to n. We can do this with a C++ method that uses the properties of array indices. The main idea is to mark the indices that relate to the numbers in the array.

C++ Implementation

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

vector<int> findDisappearedNumbers(vector<int>& nums) {
    int n = nums.size();
    
    // Mark the numbers that are present by negating the values at the corresponding indices
    for (int i = 0; i < n; i++) {
        int index = abs(nums[i]) - 1;
        if (nums[index] > 0) {
            nums[index] = -nums[index];
        }
    }

    // Collect all the indices that are not marked (positive values)
    vector<int> missingNumbers;
    for (int i = 0; i < n; i++) {
        if (nums[i] > 0) {
            missingNumbers.push_back(i + 1);
        }
    }
    
    return missingNumbers;
}

int main() {
    vector<int> nums = {4, 3, 2, 7, 8, 2, 3, 1};
    vector<int> missing = findDisappearedNumbers(nums);
    
    // Output the missing numbers
    cout << "Missing numbers: ";
    for (int num : missing) {
        cout << num << " ";
    }
    return 0;
}

Explanation of the Code

The function findDisappearedNumbers takes a vector of integers. It goes through each number. It uses the number to find the index (value - 1) and marks it by negating the number at that index. After marking, it checks for indices that stay positive. This shows that those numbers are missing. Finally, it returns a vector that has all the missing numbers.

Key Points

This method runs in O(n) time and uses O(1) extra space. This is good for large datasets. It changes the input array, which is okay if we don’t need the original values after we finish. This way helps us find missing numbers easily by using index mapping.

For more about array manipulations, you can check topics like Array Two Sum or Array Contains Duplicate.

Optimal Solution Using Index Mapping for Missing Numbers

We can use the index mapping method to find missing numbers in an array. This method works well because the numbers in the array should be from 1 to n. Here, n is the length of the array. By putting each number in the right index, we can find the missing numbers quickly and without using much space.

Approach:

1. We go through the array. For each number, we place it at the index that matches its value (number - 1).
2. After rearranging the array, we check again. We look for indices that do not have the expected number. Those indices show us the missing numbers.

Time Complexity:

• O(n), where n is the length of the array.

Space Complexity:

• O(1), since we do not use any extra space except for the output list.

Java Implementation:

import java.util.ArrayList;
import java.util.List;

public class FindDisappearedNumbers {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            int index = Math.abs(nums[i]) - 1;
            nums[index] = -Math.abs(nums[index]);
        }
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) {
                result.add(i + 1);
            }
        }
        return result;
    }
}

Python Implementation:

def find_disappeared_numbers(nums):
    result = []
    for num in nums:
        index = abs(num) - 1
        nums[index] = -abs(nums[index])
    for i in range(len(nums)):
        if nums[i] > 0:
            result.append(i + 1)
    return result

C++ Implementation:

#include <vector>
#include <cmath>

std::vector<int> findDisappearedNumbers(std::vector<int>& nums) {
    std::vector<int> result;
    for (int i = 0; i < nums.size(); i++) {
        int index = abs(nums[i]) - 1;
        nums[index] = -abs(nums[index]);
    }
    for (int i = 0; i < nums.size(); i++) {
        if (nums[i] > 0) {
            result.push_back(i + 1);
        }
    }
    return result;
}

Example:

If we have an input array like [4,3,2,7,8,2,3,1], the output will be [5, 6]. These numbers are missing in the range from 1 to 8.

This method using index mapping is very efficient. It helps us find missing numbers in the array with less effort. For more on array problems, you can check out Array Two Sum and Array Contains Duplicate.

Using HashSet to Find Missing Numbers in an Array

We can use a HashSet to find missing numbers in an array. This method is simple and works well. First, we store all the numbers in the HashSet. Then, we check which numbers from the expected range are missing. This way, we can do it in average time of O(n) because HashSet operations take constant time.

Implementation Steps:

1. Create a HashSet to hold the numbers from the input array.
2. Go through the array and add each number to the HashSet.
3. Look for missing numbers from 1 to n (where n is the size of the array).
4. Gather the missing numbers and return them.

Example Code:

Here are examples in Java, Python, and C++.

Java Implementation:

import java.util.*;

public class MissingNumbers {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        Set<Integer> numSet = new HashSet<>();
        for (int num : nums) {
            numSet.add(num);
        }
        
        List<Integer> missingNumbers = new ArrayList<>();
        for (int i = 1; i <= nums.length; i++) {
            if (!numSet.contains(i)) {
                missingNumbers.add(i);
            }
        }
        
        return missingNumbers;
    }
}

Python Implementation:

def find_disappeared_numbers(nums):
    num_set = set(nums)
    missing_numbers = []
    for i in range(1, len(nums) + 1):
        if i not in num_set:
            missing_numbers.append(i)
    return missing_numbers

C++ Implementation:

#include <vector>
#include <unordered_set>

class Solution {
public:
    std::vector<int> findDisappearedNumbers(std::vector<int>& nums) {
        std::unordered_set<int> numSet(nums.begin(), nums.end());
        std::vector<int> missingNumbers;
        
        for (int i = 1; i <= nums.size(); i++) {
            if (numSet.find(i) == numSet.end()) {
                missingNumbers.push_back(i);
            }
        }
        
        return missingNumbers;
    }
};

Key Points:

• Efficiency: This method is fast with time complexity of O(n) and space complexity of O(n).
• Versatility: We can use this logic in different programming languages with small changes.
• Use Case: The HashSet method works well when the array is big and we need good performance.

This way of using HashSet is a clear solution to find all missing numbers in an array. We like to use it in many cases. For more related array problems, we can check out Array Two Sum and Array Contains Duplicate.

[Array] Find All Numbers Disappeared in an Array - Easy

[Array] Find All Numbers Disappeared in an Array Solution Approaches

We can find all numbers that are missing in an array that goes from 1 to n. We have different ways to solve this. Each way has good points based on what we need.

Understanding the Problem Statement for Disappeared Numbers in an Array

The job is to find all numbers between 1 and n that are not in the given array. The input is an array of size n. The numbers in the array are also between 1 and n.

Java Implementation for Finding Disappeared Numbers in an Array

In Java, we can use a simple way with a boolean array to check which numbers we see.

import java.util.ArrayList;
import java.util.List;

public class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        boolean[] seen = new boolean[nums.length + 1];
        List<Integer> result = new ArrayList<>();
        
        for (int num : nums) {
            seen[num] = true;
        }
        
        for (int i = 1; i < seen.length; i++) {
            if (!seen[i]) {
                result.add(i);
            }
        }
        
        return result;
    }
}

Python Approach to Identify Missing Numbers in an Array

In Python, we can use a set to keep track of the numbers we see. Then we can find the missing ones.

def findDisappearedNumbers(nums):
    n = len(nums)
    num_set = set(nums)
    return [i for i in range(1, n + 1) if i not in num_set]

C++ Method to Find All Missing Numbers in an Array

In C++, we can use a similar way as in Java. We can use a vector to keep track of the numbers we see.

#include <vector>

class Solution {
public:
    std::vector<int> findDisappearedNumbers(std::vector<int>& nums) {
        std::vector<bool> seen(nums.size() + 1, false);
        std::vector<int> result;

        for (int num : nums) {
            seen[num] = true;
        }

        for (int i = 1; i < seen.size(); i++) {
            if (!seen[i]) {
                result.push_back(i);
            }
        }

        return result;
    }
};

Optimal Solution Using Index Mapping for Missing Numbers

A good solution is to use the input array itself to mark which numbers are present.

def findDisappearedNumbers(nums):
    for i in range(len(nums)):
        index = abs(nums[i]) - 1
        nums[index] = -abs(nums[index])
    
    return [i + 1 for i in range(len(nums)) if nums[i] > 0]

Using HashSet to Find Missing Numbers in an Array

Using a HashSet helps us find missing numbers quickly with average O(1) time for lookups.

def findDisappearedNumbers(nums):
    num_set = set(nums)
    return [i for i in range(1, len(nums) + 1) if i not in num_set]

Sorting the Array to Locate Missing Numbers

Sorting the array helps us also find missing numbers. After sorting, we can check each number with what we expect.

def findDisappearedNumbers(nums):
    nums.sort()
    missing_numbers = []
    for i in range(1, len(nums) + 1):
        if i not in nums:
            missing_numbers.append(i)
    return missing_numbers

Comparative Analysis of Different Approaches for Disappeared Numbers

• Boolean Array: Easy to use but takes more space
• Set: Good for lookups but needs extra space too
• Index Mapping: Best for space and time. It changes the input array
• Sorting: Simple but takes O(n log n) time

Frequently Asked Questions

1. What is the time complexity of these approaches?
   • The time complexity can go from O(n) for the best solutions like index mapping and hash sets to O(n log n) for sorting.
2. Can I change the input array?
   • If we can change it, index mapping is the best way.
3. Is there a way to solve this problem without extra space?
   • Yes, we can use index mapping to track numbers in the input array.

For more reading about similar array problems, check articles like Array: Two Sum and Array: Contains Duplicate.

Comparative Analysis of Different Approaches for Disappeared Numbers

When we try to find all numbers that are missing in an array, we can use different methods. Each method has its good points and some downsides. Here, we look at the most common ways to find these missing numbers.

1. Index Mapping

• Time Complexity: O(n)
• Space Complexity: O(1)
• In this method, we change the original array. We use the indices to show which numbers are present. Each number from [1, n] goes to its right index (number - 1). After we process the array, the unmarked indices show the missing numbers.

public List<Integer> findDisappearedNumbers(int[] nums) {
    List<Integer> missing = new ArrayList<>();
    for (int i = 0; i < nums.length; i++) {
        int index = Math.abs(nums[i]) - 1;
        nums[index] = -Math.abs(nums[index]);
    }
    for (int i = 1; i <= nums.length; i++) {
        if (nums[i - 1] > 0) {
            missing.add(i);
        }
    }
    return missing;
}

2. Using HashSet

• Time Complexity: O(n)
• Space Complexity: O(n)
• This method uses a HashSet to keep the numbers from the input array. We go through the range [1, n] to check which numbers are missing easily.

def findDisappearedNumbers(nums):
    num_set = set(nums)
    return [i for i in range(1, len(nums) + 1) if i not in num_set]

3. Sorting the Array

• Time Complexity: O(n log n)
• Space Complexity: O(1) (in-place sorting)
• Here, we first sort the array. Then we check through it to find any missing numbers. Sorting takes more time, but it can help with smaller datasets.

#include <vector>
#include <algorithm>

std::vector<int> findDisappearedNumbers(std::vector<int>& nums) {
    std::sort(nums.begin(), nums.end());
    std::vector<int> missing;
    for (int i = 1; i <= nums.size(); i++) {
        if (i != nums[i - 1]) {
            missing.push_back(i);
        }
    }
    return missing;
}

4. Mathematical Approach

• Time Complexity: O(n)
• Space Complexity: O(1)
• We use the formula for the sum of the first n natural numbers, ( ). This helps us find the expected sum. We compare it with the actual sum of the numbers in the array. This shows us the missing numbers.

def findDisappearedNumbers(nums):
    n = len(nums)
    expected_sum = n * (n + 1) // 2
    actual_sum = sum(nums)
    missing_number = expected_sum - actual_sum
    return [missing_number]

5. Comparative Overview

• Performance: Index Mapping is best for both time and space. Sorting is slower but simple to use.
• Modifications: Index mapping changes the original array. HashSet and sorting keep the input data unchanged.
• Complexity: Think about the problem’s limits when picking a method. For big datasets, we want to be efficient. Index mapping or the math method is better for that.

In short, each method has its good sides based on what we need. This includes time efficiency, space usage, and keeping the original array the same. For more info on related array problems, check Array: Two Sum or Array: Contains Duplicate.

Frequently Asked Questions

1. What is the problem statement for finding all numbers disappeared in an array?

The problem is to find which numbers are missing in an array. The numbers we look for are from 1 to n. The array is of size n, and we expect each number to be in that range. But some numbers may be gone. We can solve this problem in different ways. We can use index mapping, HashSet, or sorting methods.

2. How can I implement a Java solution for finding disappeared numbers in an array?

In Java, we can find missing numbers by using a simple loop. We go through the array and mark which numbers are present. By changing the indices based on the values, we can find which numbers are missing. This way runs in O(n) time and needs O(1) extra space.

3. What is the Python approach to identify missing numbers in an array?

In Python, we can find missing numbers by using lists and sets. First, we create a set of all expected numbers. Then we compare it with the set of numbers that are in the array. This way, we can easily see which numbers are missing. This method is easy to understand and works well for this problem.

4. How does the C++ method work for finding all missing numbers in an array?

The C++ method for finding missing numbers usually means going through the array. We use the indices to track which numbers are there. We can change the array to mark existing numbers. After that, we can go through it again to collect the missing numbers. This method works well for time and space.

5. What are the advantages of using a HashSet to find missing numbers in an array?

Using a HashSet helps us find missing numbers quickly. It allows fast lookups and stores the elements we find in the array. This way, we can easily check which numbers are not there from the expected range. The HashSet method is simple and effective. It has a time complexity of O(n) and space complexity of O(n).

For more reading on related array problems and solutions, check these articles: Array Two Sum, Array Best Time to Buy and Sell Stock, and Array Contains Duplicate.