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Consecutive Numbers Sum

Hard

Asked by:

17 views

Topics:

Math

Given an integer n, return the number of ways you can write n as the sum of consecutive positive integers.

Example 1:

Input: n = 5
Output: 2
Explanation: 5 = 2 + 3

Example 2:

Input: n = 9
Output: 3
Explanation: 9 = 4 + 5 = 2 + 3 + 4

Example 3:

Input: n = 15
Output: 4
Explanation: 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

Constraints:

1 <= n <= 10⁹

Consecutive Numbers Sum

Hard

Asked by:

17 views

Topics:

Math

Given an integer n, return the number of ways you can write n as the sum of consecutive positive integers.

Example 1:

Input: n = 5
Output: 2
Explanation: 5 = 2 + 3

Example 2:

Input: n = 9
Output: 3
Explanation: 9 = 4 + 5 = 2 + 3 + 4

Example 3:

Input: n = 15
Output: 4
Explanation: 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

Constraints:

1 <= n <= 10⁹

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 upper bound for the input integer n?
Is n guaranteed to be a positive integer, or should I handle potential edge cases like zero or negative input?
If n cannot be expressed as a sum of consecutive positive integers, what should I return?
Can you provide a few more examples, especially for smaller values of n, to ensure I understand the problem correctly?
Are we looking for distinct sums, or can overlapping sequences contribute to the count?

Brute Force Solution

Approach

The brute force strategy involves testing all possible combinations. For this problem, we'll try every possible starting number and count how many consecutive numbers, beginning at that start, sum to our target number. We repeat until our starting number is too large to contribute to the sum.

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

Begin by selecting a number as the starting point.
Then, keep adding consecutive numbers after the starting number, and calculate the sum.
If the sum is equal to the target number, then we found a valid set of consecutive numbers.
If the sum is greater than the target number, then there's no need to continue with that starting number; move to the next.
Keep repeating this process with different starting numbers.
Stop when the starting number is more than half of the target number, because at that point any series of consecutive numbers will exceed the target sum.
Finally, count how many valid sets of consecutive numbers we found that sum to the target number.

Code Implementation

def consecutive_numbers_sum(target_number):
    count = 0
    max_start_number = target_number // 2

    for start_number in range(1, max_start_number + 1):
        current_sum = 0
        current_number = start_number

        # Keep adding consecutive numbers to calculate the sum
        while current_sum < target_number:
            current_sum += current_number

            if current_sum == target_number:
                count += 1
                break

            current_number += 1

    return count

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The outer loop iterates from 1 up to n/2, where n is the input number. The inner loop adds consecutive numbers until the sum equals or exceeds n. In the worst case, the inner loop iterates a few times for each value in the outer loop. Therefore, the total number of iterations of the inner loop across all iterations of the outer loop is bounded by a constant multiple of n. Thus, the time complexity is O(n).

Space Complexity

O(1) – The provided plain English explanation outlines an iterative approach that primarily uses a starting number and a running sum to check for consecutive number sequences. It doesn't describe the creation of any auxiliary data structures like arrays, lists, or hash maps to store intermediate results. Therefore, the space complexity is determined by the few constant-size variables used for the starting number and the running sum, which are independent of the input number N. Consequently, the auxiliary space complexity is O(1), indicating constant space.

Optimal Solution

Approach

The problem asks us to find how many ways a number can be expressed as the sum of consecutive numbers. Instead of checking every possible sequence, we can use math to find a smarter way to determine potential solutions. We focus on the length of possible sequences to avoid unnecessary calculations.

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

Consider that a number can be represented by a sequence of 'k' consecutive integers. We want to find the possible values of 'k'.
Recognize that the sum of 'k' consecutive integers can be expressed with a formula related to 'k' and the starting integer of the sequence.
Rearrange the formula to express the starting integer in terms of the target number and 'k'. This helps us figure out if a valid sequence exists for a specific 'k'.
Notice that the starting integer must be a positive whole number. So, for each possible 'k', check if the formula gives us a positive whole number starting integer.
Iterate through possible values of 'k' starting from 1. Stop when 'k' becomes too large, as further increasing 'k' won't give a valid solution.
For each 'k', if we find a positive whole number starting integer, increment our solution count. This tells us we've found a valid way to represent the target number as a sum of 'k' consecutive integers.
The final solution count is the number of ways the number can be expressed as a sum of consecutive numbers.

Code Implementation

def consecutive_numbers_sum(number):
    solution_count = 0
    for sequence_length in range(1, number + 1): 
        # If sequence is too long, exit loop.
        if (number - (sequence_length * (sequence_length - 1) // 2)) <= 0:
            break

        starting_integer = (number - (sequence_length * (sequence_length - 1) // 2)) / sequence_length

        # Check if it is an integer.
        if starting_integer == int(starting_integer):
            # Starting integer must be positive.
            if starting_integer > 0:
                solution_count += 1

    return solution_count

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(sqrt(n)) – The algorithm iterates through possible lengths 'k' of consecutive number sequences. The crucial observation is that the loop condition depends on 'k' being less than sqrt(2*n), derived from the formula relating the starting integer, k, and the target number n. Therefore, the loop executes approximately sqrt(n) times. Each iteration involves constant-time arithmetic operations. Thus, the overall time complexity is O(sqrt(n)).

Space Complexity

O(1) – The algorithm iterates through possible lengths 'k' of consecutive integer sequences and calculates a potential starting integer for each 'k'. It uses a constant number of variables like 'count' (for the number of solutions) and 'k' (for the length of the sequence). The space used does not depend on the input number N, as we are only performing arithmetic calculations and storing a few integer variables. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
n = 1	The only possible sum is 1 itself, so the function should return 1.
n = a large prime number	The only possible sum is the number itself, so the function should return 1; ensure the algorithm scales well for large n.
n = a power of 2	The only possible sum is the number itself because all other sums will involve odd divisors, so the function should return 1.
Integer overflow during calculations	Use appropriate data types (long) or check for potential overflows to avoid incorrect results for large n.
n is a small number like 2 or 3	Check the algorithm handles small numbers correctly; 2 = 2, 3 = 3, 1+2; so should return 1 and 2 respectively.
n has many factors	The algorithm should efficiently count the number of odd factors to determine the valid consecutive sums.
The consecutive sequence starts from 1	The algorithm should correctly identify and count sequences like 1 + 2 + 3 + ... = n.
The consecutive sequence only contains 1 number.	The algorithm should include the sequence consisting of only n when counting.

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 upper bound for the input integer n?
Is n guaranteed to be a positive integer, or should I handle potential edge cases like zero or negative input?
If n cannot be expressed as a sum of consecutive positive integers, what should I return?
Can you provide a few more examples, especially for smaller values of n, to ensure I understand the problem correctly?
Are we looking for distinct sums, or can overlapping sequences contribute to the count?

Brute Force Solution

Approach

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

Begin by selecting a number as the starting point.
Then, keep adding consecutive numbers after the starting number, and calculate the sum.
If the sum is equal to the target number, then we found a valid set of consecutive numbers.
If the sum is greater than the target number, then there's no need to continue with that starting number; move to the next.
Keep repeating this process with different starting numbers.
Stop when the starting number is more than half of the target number, because at that point any series of consecutive numbers will exceed the target sum.
Finally, count how many valid sets of consecutive numbers we found that sum to the target number.

Code Implementation

def consecutive_numbers_sum(target_number):
    count = 0
    max_start_number = target_number // 2

    for start_number in range(1, max_start_number + 1):
        current_sum = 0
        current_number = start_number

        # Keep adding consecutive numbers to calculate the sum
        while current_sum < target_number:
            current_sum += current_number

            if current_sum == target_number:
                count += 1
                break

            current_number += 1

    return count

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Consider that a number can be represented by a sequence of 'k' consecutive integers. We want to find the possible values of 'k'.
Recognize that the sum of 'k' consecutive integers can be expressed with a formula related to 'k' and the starting integer of the sequence.
Rearrange the formula to express the starting integer in terms of the target number and 'k'. This helps us figure out if a valid sequence exists for a specific 'k'.
Notice that the starting integer must be a positive whole number. So, for each possible 'k', check if the formula gives us a positive whole number starting integer.
Iterate through possible values of 'k' starting from 1. Stop when 'k' becomes too large, as further increasing 'k' won't give a valid solution.
For each 'k', if we find a positive whole number starting integer, increment our solution count. This tells us we've found a valid way to represent the target number as a sum of 'k' consecutive integers.
The final solution count is the number of ways the number can be expressed as a sum of consecutive numbers.

Code Implementation

def consecutive_numbers_sum(number):
    solution_count = 0
    for sequence_length in range(1, number + 1): 
        # If sequence is too long, exit loop.
        if (number - (sequence_length * (sequence_length - 1) // 2)) <= 0:
            break

        starting_integer = (number - (sequence_length * (sequence_length - 1) // 2)) / sequence_length

        # Check if it is an integer.
        if starting_integer == int(starting_integer):
            # Starting integer must be positive.
            if starting_integer > 0:
                solution_count += 1

    return solution_count

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 1	The only possible sum is 1 itself, so the function should return 1.
n = a large prime number	The only possible sum is the number itself, so the function should return 1; ensure the algorithm scales well for large n.
n = a power of 2	The only possible sum is the number itself because all other sums will involve odd divisors, so the function should return 1.
Integer overflow during calculations	Use appropriate data types (long) or check for potential overflows to avoid incorrect results for large n.
n is a small number like 2 or 3	Check the algorithm handles small numbers correctly; 2 = 2, 3 = 3, 1+2; so should return 1 and 2 respectively.
n has many factors	The algorithm should efficiently count the number of odd factors to determine the valid consecutive sums.
The consecutive sequence starts from 1	The algorithm should correctly identify and count sequences like 1 + 2 + 3 + ... = n.
The consecutive sequence only contains 1 number.	The algorithm should include the sequence consisting of only n when counting.