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Find N Unique Integers Sum up to Zero

Easy

Asked by:

56 views

Topics:

Arrays

Given an integer n, return any array containing n unique integers such that they add up to 0.

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].

Example 2:

Input: n = 3
Output: [-1,0,1]

Example 3:

Input: n = 1
Output: [0]

Constraints:

1 <= n <= 1000

Try coding it in LeetCode

Find N Unique Integers Sum up to Zero

Easy

Asked by:

56 views

Topics:

Arrays

Given an integer n, return any array containing n unique integers such that they add up to 0.

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].

Example 2:

Input: n = 3
Output: [-1,0,1]

Example 3:

Input: n = 1
Output: [0]

Constraints:

1 <= n <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected range for the input integer n? Is there a maximum or minimum value for n?
Should the returned array contain only integers, or are floating-point numbers acceptable?
Are there any specific ordering requirements for the integers in the returned array?
Is there any constraint on memory usage, or can I create an array of size n without issue?
If n is 0, what should the returned array be? Should I return an empty array or null?

Brute Force Solution

Approach

The brute force approach to finding N unique integers that sum to zero involves exploring numerous possibilities. We essentially try out different combinations of numbers until we find a set of N unique integers that add up to zero. This method doesn't prioritize efficiency; it guarantees a solution (if one exists) by testing every possible scenario.

Here's how the algorithm would work step-by-step:

Start by picking any N-1 unique integers.
Calculate the sum of the numbers you've picked.
Determine the number needed to reach zero when added to your current sum. This will be the negative of your current sum.
Check if this new number is already among the N-1 integers picked. If it's a duplicate, this set of numbers is invalid.
If the new number is unique, then you have found your N integers that sum to zero. If not, go back to the beginning and try a completely different set of N-1 integers.
Keep doing this until you find a suitable unique combination that sums to zero.

Code Implementation

import random

def find_n_unique_integers_sum_up_to_zero(number):
    if number <= 0:
        return []

    while True:
        unique_integers = set()
        while len(unique_integers) < number - 1:
            random_integer = random.randint(-100, 100) 
            unique_integers.add(random_integer)

        unique_integers_list = list(unique_integers)
        current_sum = sum(unique_integers_list)
        needed_number = -current_sum

        # We must check for duplicates to satisfy the problem requirements
        if needed_number not in unique_integers_list:

            unique_integers_list.append(needed_number)
            # Validate that the generated list sums to zero
            if sum(unique_integers_list) == 0:
                return unique_integers_list

Big(O) Analysis

Time Complexity

O(inf) – The described brute force approach has potentially infinite time complexity. Although the algorithm aims to find N unique integers summing to zero, it lacks a defined limit on the number of combinations it explores if it keeps encountering duplicate values or failing to find a suitable set of N-1 unique integers whose negative counterpart is also unique. Without constraints on the range of numbers to choose from or a maximum number of attempts, the algorithm might theoretically run indefinitely if it never converges on a valid solution. In practice, the algorithm may be limited by available resources, but its theoretical worst-case runtime is unbounded.

Space Complexity

O(N) – The brute force approach, as described, picks N-1 unique integers and stores them. This implies a data structure, likely a list or a set, of size N-1 to hold these chosen integers. Additionally, a variable is needed to store the sum of these N-1 integers. Therefore, the auxiliary space required scales linearly with the input size N. Consequently, the space complexity is O(N).

Optimal Solution

Approach

To find N unique integers that sum to zero, the best approach is to create a set of numbers that are symmetrical around zero. We can achieve this by pairing positive and negative numbers and, if N is odd, including zero itself.

Here's how the algorithm would work step-by-step:

First, consider whether the number of integers you need is even or odd.
If the number is odd, include zero in your set of numbers. This ensures the sum remains zero.
For half of the remaining integers (excluding zero if it's there), use positive numbers starting from one.
For the other half of the integers, use the negative counterpart of each of the positive numbers you selected. This will cancel out the positive numbers.
Combine the positive numbers, negative numbers, and zero (if needed) to get your final set of unique integers that sum to zero.

Code Implementation

def find_n_unique_integers_that_sum_up_to_zero(number_of_integers):
    result = []

    #If odd, zero must be included to ensure sum is zero
    if number_of_integers % 2 != 0:
        result.append(0)
        number_of_integers -= 1

    positive_number = 1
    for _ in range(number_of_integers // 2):
        result.append(positive_number)

        # Append negative counterparts to ensure the sum equates to zero
        result.append(-positive_number)
        positive_number += 1

    return result

Big(O) Analysis

Time Complexity

O(n) – The provided solution involves iterating to generate n unique integers. We determine if n is odd, which is a constant-time operation. We then iterate roughly n/2 times to add positive integers and another n/2 times to add their negative counterparts. This sequence of constant-time operations, executed approximately n times, gives a time complexity directly proportional to n. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm constructs a list of size N to store the unique integers that sum to zero. The positive and negative numbers generated, along with a possible zero, are stored within this list. Since the size of the list grows linearly with the input N, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
n = 0	Return an empty array since zero unique integers are needed.
n = 1	Return an array containing only 0, as that is the only integer that sums to zero.
Large n (approaching memory limits)	The solution should use an iterative approach (instead of recursion) to avoid stack overflow and manage memory efficiently.
Negative n	Throw an IllegalArgumentException since n represents array size and cannot be negative.
Integer Overflow during sum calculation (if applicable depending on chosen approach)	Ensure intermediate calculations do not exceed Integer.MAX_VALUE or Integer.MIN_VALUE; use long if necessary.
Multiple possible solutions exist	The algorithm can return any valid solution as long as the integers are unique and sum to zero.
Odd and Even n values affecting range of numbers generated	The implementation should handle odd values of n by including 0 in the array, and adjust other elements accordingly
n equals Integer.MAX_VALUE	The solution needs to consider memory limitations and potential integer overflow when generating unique integers in this extreme case; it might be impossible given memory constraints.

n = 0

How to Handle:

Return an empty array since zero unique integers are needed.

n = 1

How to Handle:

Return an array containing only 0, as that is the only integer that sums to zero.

Large n (approaching memory limits)

How to Handle:

The solution should use an iterative approach (instead of recursion) to avoid stack overflow and manage memory efficiently.

Negative n

How to Handle:

Throw an IllegalArgumentException since n represents array size and cannot be negative.

Integer Overflow during sum calculation (if applicable depending on chosen approach)

How to Handle:

Ensure intermediate calculations do not exceed Integer.MAX_VALUE or Integer.MIN_VALUE; use long if necessary.

Multiple possible solutions exist

How to Handle:

The algorithm can return any valid solution as long as the integers are unique and sum to zero.

Odd and Even n values affecting range of numbers generated

How to Handle:

The implementation should handle odd values of n by including 0 in the array, and adjust other elements accordingly

n equals Integer.MAX_VALUE

How to Handle:

The solution needs to consider memory limitations and potential integer overflow when generating unique integers in this extreme case; it might be impossible given memory constraints.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected range for the input integer n? Is there a maximum or minimum value for n?
Should the returned array contain only integers, or are floating-point numbers acceptable?
Are there any specific ordering requirements for the integers in the returned array?
Is there any constraint on memory usage, or can I create an array of size n without issue?
If n is 0, what should the returned array be? Should I return an empty array or null?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by picking any N-1 unique integers.
Calculate the sum of the numbers you've picked.
Determine the number needed to reach zero when added to your current sum. This will be the negative of your current sum.
Check if this new number is already among the N-1 integers picked. If it's a duplicate, this set of numbers is invalid.
If the new number is unique, then you have found your N integers that sum to zero. If not, go back to the beginning and try a completely different set of N-1 integers.
Keep doing this until you find a suitable unique combination that sums to zero.

Code Implementation

import random

def find_n_unique_integers_sum_up_to_zero(number):
    if number <= 0:
        return []

    while True:
        unique_integers = set()
        while len(unique_integers) < number - 1:
            random_integer = random.randint(-100, 100) 
            unique_integers.add(random_integer)

        unique_integers_list = list(unique_integers)
        current_sum = sum(unique_integers_list)
        needed_number = -current_sum

        # We must check for duplicates to satisfy the problem requirements
        if needed_number not in unique_integers_list:

            unique_integers_list.append(needed_number)
            # Validate that the generated list sums to zero
            if sum(unique_integers_list) == 0:
                return unique_integers_list

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, consider whether the number of integers you need is even or odd.
If the number is odd, include zero in your set of numbers. This ensures the sum remains zero.
For half of the remaining integers (excluding zero if it's there), use positive numbers starting from one.
For the other half of the integers, use the negative counterpart of each of the positive numbers you selected. This will cancel out the positive numbers.
Combine the positive numbers, negative numbers, and zero (if needed) to get your final set of unique integers that sum to zero.

Code Implementation

def find_n_unique_integers_that_sum_up_to_zero(number_of_integers):
    result = []

    #If odd, zero must be included to ensure sum is zero
    if number_of_integers % 2 != 0:
        result.append(0)
        number_of_integers -= 1

    positive_number = 1
    for _ in range(number_of_integers // 2):
        result.append(positive_number)

        # Append negative counterparts to ensure the sum equates to zero
        result.append(-positive_number)
        positive_number += 1

    return result

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 0	Return an empty array since zero unique integers are needed.
n = 1	Return an array containing only 0, as that is the only integer that sums to zero.
Large n (approaching memory limits)	The solution should use an iterative approach (instead of recursion) to avoid stack overflow and manage memory efficiently.
Negative n	Throw an IllegalArgumentException since n represents array size and cannot be negative.
Integer Overflow during sum calculation (if applicable depending on chosen approach)	Ensure intermediate calculations do not exceed Integer.MAX_VALUE or Integer.MIN_VALUE; use long if necessary.
Multiple possible solutions exist	The algorithm can return any valid solution as long as the integers are unique and sum to zero.
Odd and Even n values affecting range of numbers generated	The implementation should handle odd values of n by including 0 in the array, and adjust other elements accordingly
n equals Integer.MAX_VALUE	The solution needs to consider memory limitations and potential integer overflow when generating unique integers in this extreme case; it might be impossible given memory constraints.

n = 0

How to Handle:

Return an empty array since zero unique integers are needed.

n = 1

How to Handle:

Return an array containing only 0, as that is the only integer that sums to zero.

Large n (approaching memory limits)

How to Handle:

The solution should use an iterative approach (instead of recursion) to avoid stack overflow and manage memory efficiently.

Negative n

How to Handle:

Throw an IllegalArgumentException since n represents array size and cannot be negative.

Integer Overflow during sum calculation (if applicable depending on chosen approach)

How to Handle:

Ensure intermediate calculations do not exceed Integer.MAX_VALUE or Integer.MIN_VALUE; use long if necessary.

Multiple possible solutions exist

How to Handle:

The algorithm can return any valid solution as long as the integers are unique and sum to zero.

Odd and Even n values affecting range of numbers generated

How to Handle:

The implementation should handle odd values of n by including 0 in the array, and adjust other elements accordingly

n equals Integer.MAX_VALUE

How to Handle:

The solution needs to consider memory limitations and potential integer overflow when generating unique integers in this extreme case; it might be impossible given memory constraints.