Consecutive Numbers Sum
“Consecutive Numbers Sum” is about finding a sequence of numbers and involves the concept of mathematical formulas, sequence, and loop. Here are 10 problems to prepare for it:
412. Fizz Buzz: This problem requires you to loop through numbers from 1 to n and perform certain checks, helping you understand basic loops and conditions.
204. Count Primes: This problem introduces you to mathematical problems and prime numbers.
509. Fibonacci Number: It’s a basic problem that involves the concept of sequences and recursion.
118. Pascal’s Triangle: This problem will help you understand how sequences can be generated from preceding numbers.
119. Pascal’s Triangle II: This problem further explores the concept of sequences.
268. Missing Number: This problem requires you to find a missing number in a sequence, introducing you to the idea of dealing with incomplete sequences.
1365. How Many Numbers Are Smaller Than the Current Number: This problem requires looping through an array and performing a comparison, which helps in understanding loops and conditions.
448. Find All Numbers Disappeared in an Array: This problem involves finding missing numbers in an array, which strengthens the concept of dealing with sequences.
283. Move Zeroes: This problem is a simple manipulation of an array which helps you understand inplace modifications in an array.
977. Squares of a Sorted Array: This problem helps you understand the sorting of an array in a different order.
These problems are selected to progressively introduce you to the concept of sequences and how to work with them.


Problem Classification
This problem belongs to the domain of “Number Theory”. The problem involves dealing with numbers and the interesting properties they have when they are represented as a sum of consecutive integers. The problem also involves understanding patterns and sequences, which is common in these domains.
Problem Components:
 Input: An integer
n
(1 <= n <= 109)  Output: The number of ways
n
can be expressed as the sum of consecutive positive integers.  Function: A function to calculate the number of ways to write
n
as the sum of consecutive positive integers.
This problem is a “Counting” problem, where the task is to count the number of ways that a certain condition can be satisfied. It can also be seen as a “Decomposition” problem, as we are decomposing the input number n
into sums of consecutive numbers.
This problem requires us to understand and leverage the properties of numbers, especially the property of how numbers can be broken down into sums of consecutive integers. It might involve understanding patterns and using possible optimizations to count the ways efficiently within the given constraints.
Clarification Questions
What are the clarification questions we can ask about this problem?
Identifying Problem Isomorphism
“Consecutive Numbers Sum” can be approximately mapped to “Arithmetic Slices II  Subsequence”.
Both problems revolve around the concept of sequences with specific properties. In “Consecutive Numbers Sum”, we need to find how many ways we can represent a given number as the sum of consecutive integers. On the other hand, “Arithmetic Slices II  Subsequence” requires finding the total number of subsequences of at least length 3 that are arithmetic progressions.
The common theme here is the handling of sequences and mathematical progressions. However, there’s a substantial difference in the nature of these problems. The first one deals with finding representations of a number as a sum of consecutive numbers, whereas the second one is about finding arithmetic sequences within a given array.
“Arithmetic Slices II  Subsequence” is more complex, as it requires handling a larger amount of data (the entire array) and finding subsequences, while “Consecutive Numbers Sum” deals with one number at a time and doesn’t require examining subsequences.
Therefore, this mapping is approximate, based on the general theme of sequences, but the problems are not identical.
Constraints
There are indeed several characteristics or conditions that can be exploited in this problem to find an efficient solution:
Positive Integers: The problem is strictly dealing with positive integers. This allows us to avoid considering negative numbers or fractional numbers, simplifying our task.
Consecutive Integers: The problem asks for sums of consecutive integers, which form an arithmetic progression. This allows us to use the formula for the sum of an arithmetic progression to quickly compute the sum and possibly reverseengineer the formula to check if a given number can be expressed as the sum of a certain number of consecutive integers.
Limitation of n: The upper limit of
n
is 10^9, which can be exploited in optimizing the solution. Given that we need to expressn
as the sum of consecutive numbers, the number of terms in the series will be considerably smaller thann
. This can drastically reduce the search space for possible solutions.Patterns in Possible Sums: For a given
n
, the possible number of ways it can be written as a sum of consecutive positive integers can have patterns. Noticing these patterns can help us find an efficient solution. For example, observe that an odd number can always be represented as the sum of two consecutive numbers, or a numbern
can be represented as the sum ofn
consecutive numbers starting from 1 ifn
is a triangular number.Factoring: The number of ways a number
n
can be represented as the sum of consecutive numbers is related to the number of its divisors. For example, if a number has an odd divisord
, it means there is a sequence ofd
consecutive numbers that sum ton
. This can be used to count the ways more efficiently.
By understanding and exploiting these properties and patterns, we can develop an efficient algorithm to solve this problem.
Thought Process
This problem is a mathematical problem in the domain of number theory. The cue in the problem statement is the need to represent the given number as the sum of “consecutive positive integers”. This immediately suggests the use of arithmetic series, since the sum of an arithmetic series with an odd number of terms can be calculated as n = number of terms * middle term
. The middle term can be n/number of terms
, which indicates that n
must be divisible by number of terms
.
Here’s a stepbystep thought process to solve the problem:
Understanding the Problem: A critical step is understanding that the problem is essentially asking us to find different arithmetic series where the sum equals
n
. Each such series represents one way to writen
as the sum of consecutive positive integers.Identifying the Formula: The sum
S
of an arithmetic series is given byS = n/2 * (a + l)
wheren
is the number of terms,a
is the first term andl
is the last term. If the series consists ofn
terms, andn
is odd, the first and the last terms are(n1)/2
terms away from the middle termm
, i.e.,S = m * n
.Factoring n: If
n
can be written as the productm * n
, then it can be represented as the sum ofn
consecutive integers, withm
as the middle term. But since all terms have to be positive,m
must be at least(n1)/2
. This meansn
must be a divisor ofn
.Counting Divisors: The number of ways
n
can be written as the sum of consecutive positive integers is equal to the number of divisors ofn
. In the case wheren
is even, the divisor must also be even, as the sum of an even number of terms is always even.Coding the Solution: We can implement this solution by writing a function that calculates the divisors of
n
and counts them.
Here is the Python3 code:


 Understanding the Effect of Changing Parameters: If
n
increases, the number of divisors would generally increase, resulting in more ways to expressn
as the sum of consecutive positive integers. However, ifn
is a power of 2, there would be only one way to expressn
, regardless of how largen
is. This is because a power of 2 has only one odd divisor, which is 1.
This problem is a good example of how a seemingly complicated question can be simplified with the help of number theory and careful analysis. It also shows how mathematical formulas can be translated into efficient code.
Solution Approach and Analysis
Approaching this problem involves a good understanding of arithmetic series and factoring. The key insight is that if n
can be expressed as the sum of k
consecutive numbers, then n
is divisible by k
.
Identify the Series: The sum of
k
consecutive numbers can be represented ask*(midPoint)
, wheremidPoint
is the middle number in the series. Whenk
is odd, the midpoint is an integer. Whenk
is even, the midpoint is a half integer.Divisibility Condition: We want
n = k*midPoint
, which implies thatn
must be divisible byk
. Whenk
is odd,midPoint
can be an integer, and son
is divisible byk
. Whenk
is even,midPoint
is a half integer. In order forn
to be an integer,n
must be divisible by2*k
. So, for every odd divisor ofn
, there exists an oddk
such thatn
can be expressed as the sum ofk
consecutive integers. For every even divisor ofn/2
, there exists an evenk
such thatn
can be expressed as the sum ofk
consecutive integers.Counting Divisors: The problem boils down to counting the number of odd divisors of
n
and the number of even divisors ofn/2
. This is straightforward: factorizen
andn/2
, count the number of factors, and add them up.Edge Cases: Remember to exclude the case where
n
itself is the sum of consecutive numbers, as this would require the series to start from zero, which is not a positive integer.
Let’s illustrate this with an example:
Suppose n = 15
, we need to find out the number of odd divisors of 15 and even divisors of 15/2 = 7.5. However, 7.5 is not an integer, so it does not have any divisors. The divisors of 15 are 1, 3, 5, and 15. Among them, 1, 3, and 5 are valid because they represent sums starting from a positive integer (15 = 7+8, 15 = 4+5+6, 15 = 2+3+4+5+6), but 15 is not because it would represent the sum 15 + 0, and 0 is not a positive integer. So there are 3 ways to express 15 as the sum of consecutive positive integers.
If n
changes, the count of divisors would change and thus the number of ways to express n
as the sum of consecutive integers would change. For example, if n = 9
, the number of ways would be 3: 9 = 2+3+4, 9 = 4+5, and 9 = 9.
Language Agnostic Coding Drills
Dissection of the code and identification of distinct concepts:
Variable Initialization: This is a fundamental concept that involves setting up and initializing variables. In this case,
divisors
andi
are initialized.Loops: This code contains several loops, including a while loop that runs until
n
is no longer divisible by 2, and another that runs whilei * i <= n
. Loops are a central concept in any programming language as they allow for repetitive tasks.Conditionals: There are multiple condition checks in the code. For example, the code checks if
n % 2 == 0
and ifn % i == 0
. This is a key concept as conditionals allow us to execute different parts of the code based on certain conditions.Operators: The code uses arithmetic and assignment operators, including
+=
,*=
, and/=
. Understanding how these operators work is fundamental to programming.Functions and Method Definitions: The code is encapsulated in a function, which is a key concept as functions allow us to structure and reuse code.
List of concepts/drills in order of increasing difficulty:
Variable Initialization: This is the easiest concept to understand and implement. You just need to assign initial values to your variables.
Operators: These are also relatively easy to understand, but they can become more complex when combined with other operations.
Conditionals: While not inherently complex, conditionals can become challenging when used in combination with other concepts such as loops and operators.
Loops: The understanding and correct implementation of loops can be challenging, especially when the termination condition is complex.
Functions and Method Definitions: This is the most complex concept, as it involves understanding not only how to define functions but also how and when to use them.
Problemsolving approach leading to the final solution:
First, we initialize our variables. In this problem, we have
divisors
andi
.Next, we perform a while loop operation to remove factors of 2 from
n
. This is important because it simplifies the problem by reducingn
to a state where we only need to handle odd divisors.Now we begin our main loop. We perform a while loop operation that runs while
i * i <= n
, checking ifn
is divisible byi
. If it is, we dividen
byi
and increase the count. We then multiplydivisors
bycount + 1
. This effectively applies the formula for the number of divisors ofn
.If after the loop
n
is greater than 1, we multiplydivisors
by 2. This takes care of the scenario wheren
is a prime number.Finally, we return
divisors
, which is our answer.
The strategy here is to reduce the given problem into a simpler one by dividing out the factors of 2. We then find the number of odd divisors of n
by continuously dividing n
by each odd number, starting from 3, until n
becomes 1 or a prime number. The count of these divisors represents the total number of ways n
can be expressed as the sum of consecutive positive integers.
Targeted Drills in Python
Pythonbased coding drills for identified concepts:
a. Variable Initialization:
1 2 3
# This drill is about initializing variables x = 10 print(x)
b. Operators:
1 2 3 4 5
# This drill is about using operators x = 10 y = x + 2 # addition operator z = x * 2 # multiplication operator print(y, z)
c. Conditionals:
1 2 3 4 5 6
# This drill is about using conditionals x = 10 if x % 2 == 0: # checking if x is even print("Even") else: print("Odd")
d. Loops:
1 2 3
# This drill is about using loops for i in range(10): # print numbers from 0 to 9 print(i)
e. Functions and Method Definitions:
1 2 3 4 5
# This drill is about defining and using functions def greet(name): print(f"Hello, {name}!") greet("Python Learner")
Problemspecific concept:
a. While Loops with Conditions:
1 2 3 4 5
# This drill demonstrates using while loops with conditions n = 20 while n % 2 == 0: # Dividing out factors of 2 n /= 2 print(n)
This drill is essential for our problem as it simplifies the given problem by removing the factor of 2 from
n
, thereby reducing it to a state where we only need to handle odd divisors.Assembling the drills into a comprehensive solution:
To solve the problem, we first use the variable initialization drill to set up our variables,
divisors
andi
. Then, we use the problemspecific while loop with conditions to remove factors of 2 fromn
. After that, we set up a while loop where we keep dividingn
by each odd number starting from 3 untiln
becomes 1 or a prime number. Inside this loop, we use the operators and conditionals to count the number of timesn
can be divided byi
and to updaten
anddivisors
accordingly. Finally, we check ifn
is greater than 1 (i.e., if it’s a prime number), and if so, we multiplydivisors
by 2. All these pieces together build up to our final solution, which we return using thereturn
statement inside our function definition.
Q&A
The solution to this problem is not challenging; the hard part is knowing when one can stop checking.
For a first attempt, a brute force of sorts was to iterate from 2 through N on smaller numbers and sqrt(N) on the larger ones, for each iteration summing i numbers, with the middle number roughly N/i. This solution barely passed, at about 7 seconds.
It took some googling to discover a mathematical approach, bringing the time complexity down to O(N).
It took reading the solution to learn the formula of k = sqrt((2N + (1/4))  1/2); allowing the O(sqrt(N)) complexity.
In an interview, there is no way I would have gotten this in O(N) let alone anything better.
Thus, to the question outlined in the subject; in what roles would someone expect this sort of question? I have interviewed at many companies for general SWE and have never come across a question that relied on knowledge of any mathematical formulae.
Was it just luck? Or are these question types specific to roles involving, for example, ML or similar?
While this problem does have a mathematical flavor to it, it’s not necessarily exclusive to roles involving Machine Learning or related fields. This kind of problem could come up in interviews for general Software Engineering roles, especially at companies known for challenging algorithmic interview questions like Google, Facebook, and Amazon. The interview process for these companies often tests the candidate’s ability to apply mathematical reasoning and number theory in creative ways, even if the daytoday role doesn’t strictly require it.
That said, it’s not so much about knowing specific mathematical formulas but rather about demonstrating a capacity for mathematical thinking, problemsolving skills, and the ability to apply known techniques in novel ways. This problem, for example, can be approached from a brute force perspective and then optimized through recognizing patterns and applying mathematical insights, which is a valuable skill in many software engineering contexts.
Ultimately, the breadth of interview questions can be quite vast, and while certain topics may come up more frequently in certain roles (like graph theory for networking roles, or probability for roles related to data), it’s important for candidates to have a wellrounded foundation in computer science and problemsolving.
However, in specialized roles such as Machine Learning, Data Science, or Quantitative Analysis, you might find more interview questions that require knowledge of specific mathematical formulae or concepts. These fields heavily rely on statistics, linear algebra, calculus, and other areas of mathematics, so having a strong understanding of these concepts is essential. But even in these cases, problemsolving and application are often valued more than rote memorization of formulas.
Let’s start with a brute force approach.
Brute Force Approach:
The brute force approach to this problem involves attempting all possible combinations of consecutive integers and counting those that sum to n
. We can start by trying to express n
as the sum of two consecutive numbers, then three, and so on, up to n
.
This can be done by writing a loop that starts from 1
and goes up to n
. For each iteration i
, we try to find sequences of i
consecutive integers that sum to n
. We start each sequence from 1
and keep adding the next consecutive integer until we either reach n
(in which case we increment our count of successful sequences) or surpass n
(in which case we move on to the next i
).
This brute force solution has a time complexity of O(n^2)
, which can be slow if n
is a large number. So we need to find a better approach.
Optimized Approach:
A more optimized solution can make use of mathematical insights.
The key observation here is that if n
can be expressed as a sum of k
consecutive integers, then n = k * m + (k  1) * k / 2
, where m
is the start of the sequence.
Rearranging this equation gives us m = (n  (k  1) * k / 2) / k
. From this, we can see that for a given k
, n
can be expressed as a sum of k
consecutive integers if and only if (n  (k  1) * k / 2)
is divisible by k
.
So instead of trying all possible combinations of consecutive numbers, we can just iterate over possible values of k
and check if (n  (k  1) * k / 2)
is divisible by k
. If it is, then we know n
can be expressed as a sum of k
consecutive numbers, and we increment our count.
This optimized solution significantly reduces the time complexity from O(n^2)
to O(sqrt(n))
, as we only need to iterate from 1
to sqrt(n)
. This is because k
can at most be sqrt(n)
for the value of m
to be a positive integer.
By using this optimized approach, we can solve the problem much more efficiently.
I wanted to know how is this question not the same as subarray sum equals to k? I think I am missing something here?
Why cant we just create a map/set of runningSum and check everytime if runningSum n is present in the set or not?
That’s a good question! Indeed, both problems involve dealing with sums and sequences, but they differ in their specifics.
The problem “Subarray Sum Equals K” is defined as: Given an array of integers and an integer K, you need to find the total number of continuous subarrays whose sum equals to K. In this problem, you’re dealing with an already given array, and you’re looking for all subarrays that sum up to a target value, K.
The key difference between this problem and the current problem is that the “Subarray Sum Equals K” problem deals with a predefined array that may contain any integers, not necessarily consecutive. Additionally, the subarray can start and end at any position within the array.
On the other hand, the problem “Number of Ways to Write N as a Sum of Consecutive Positive Integers” is about finding sequences of consecutive positive integers that sum up to a given number, N. Here, you’re not given any array to start with, and the integers must be consecutive. Moreover, these sequences can start from any positive integer, not just from the first element of an array.
As a result, the map/set strategy for “Subarray Sum Equals K” problem doesn’t directly apply here, because we’re not given an array, and we need to deal with sequences of consecutive integers instead. In other words, the two problems are asking for different things, and the approach used for one does not directly translate to the other.
I am getting a TLE on a testcase for n=53965645, but when I add it to my testcases and run it, it passes the testcase… What should I do?? int consecutiveNumbersSum(int n) { long int sum = 0; int ctr = 1; if(n==1  n==2) return 1; for(int i=1; i<n/2+1; i++) { for(int j=i; j<=n/2+2; j++) { if(sum<n) { sum += j; } else { if(sum==n) ctr++; sum = 0; break; } } } return ctr; }
The issue you’re encountering is due to the time complexity of your solution. Your current approach has a time complexity of roughly O(n^2), which can be too slow for larger inputs like n = 53965645
.
Your code currently iterates over every number up to n/2+1
, and for each of those numbers, it starts another loop to add numbers until the sum is equal to or greater than n
. This process is quite timeconsuming, especially for large inputs.
A more efficient solution to this problem exists with a time complexity of O(sqrt(n)).
Here is an optimized approach:
 Initiate a variable
count
to 0. This variable will keep track of the total number of ways.  Iterate
i
from 1 while(i * (i + 1)) / 2 <= n
. This part uses the sum formula of an arithmetic progression, ensuring that the sum does not exceedn
. In each iteration, check if
n  (i * (i + 1)) / 2
is divisible byi
. If it is, incrementcount
.
 In each iteration, check if
 At the end of the loop,
count
will be the answer.
This approach works because it efficiently checks for all possible lengths of sequences (i.e., i
) that can sum up to n
.
Please try to implement this approach in your code, and it should solve the TLE issue you’re currently facing.
Remember, in coding problems, often brute force solutions can lead to time limit exceeded errors. It’s crucial to try and find patterns or mathematical properties that can help you optimize your solution. In this case, the key insight is understanding how arithmetic progressions work and using that to limit the search space effectively.