Arranging Coins

This problem is about finding the largest integer k for which the sum of the first k integers (which is given by the formula k*(k+1)/2) is less than or equal to n.

The sum of the first k integers can be represented by the equation k*(k+1)/2. If we set this equal to n and solve for k, we would end up with a quadratic equation. However, a more straightforward method to solve this problem is to use a simple iteration.

Here’s a Python solution:

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class Solution:
    def arrangeCoins(self, n: int) -> int:
        k = 0
        while n > 0:
            k += 1
            n -= k
        return k - 1 if n < 0 else k

This code initializes k to 0 and starts a loop that continues until n is no longer positive. On each iteration, k is incremented by 1, and n is decreased by the current value of k. If at any point n becomes negative, it means the last row is incomplete, so k - 1 is returned. If n is exactly zero at the end of the loop, it means the last row is complete, so k is returned.

Abstract Representation of the Problem

The problem is about understanding the construction of an abstract staircase made up of units (which could be anything, but in our problem are represented by coins). The staircase is built following a specific rule: each successive level of the staircase has one unit more than the previous level.

Our task is to find out the maximum height of the staircase we can build given a certain amount of units. In the process, we may find that there aren’t enough units to complete a level - in which case, we disregard the incomplete level.

Abstracting further, this problem is essentially about summing a series of increasing integers starting from 1 (i.e., 1, 2, 3, 4, …). The goal is to find the maximum integer ‘k’ such that the sum from 1 to ‘k’ is less than or equal to the given total number of units. If the sum becomes more than the total number of units at any ‘k’, we should disregard that level.

Thus, the problem can be seen as a quest to find the largest ‘k’ for which the sum of first ‘k’ integers is less than or equal to the total number of units available.

Q&A

1+2+3+…+x = n -> (1+x)x/2 = n -> x^2+x = 2n -> x^2+x+1/4 = 2n +1/4 -> (x+1/2)^2 = 2n +1/4 -> (x+0.5) = sqrt(2n+0.25) -> x = -0.5 + sqrt(2n+0.25)

You are correct. You just rearranged the sum formula of the first ’n’ natural numbers to solve for ‘x’. The standard sum formula is n*(n+1)/2, and you’ve rearranged it to solve for ’n’ (or ‘x’ in your formula). The equation you derived is an implementation of the quadratic formula used to find the roots of a quadratic equation.

However, remember to consider the constraints of our problem. We’re counting levels in the staircase (which have to be whole numbers) and the number of units/coins is also a whole number. Therefore, it is necessary to take the floor of the value of ‘x’ you calculate, because the staircase levels must be a complete row, and we can’t have a fraction of a row.

In Python, you would implement it as:

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import math

class Solution:
    def arrangeCoins(self, n: int) -> int:
        return int(math.sqrt(2 * n + 0.25) - 0.5)

This function uses the formula you provided to calculate ‘x’, but it also wraps the result in int() to ensure we get a whole number of rows. The math.sqrt function is used to take the square root.

staircases are going to be 1 + 2 + 3 + 4 ….. + m-1 + m = (m* (m + 1))/2 you have to match m * (m+1) // 2 to highest possible m value such that the expression is less than n sqrt(n) on a 32 bit integer is log(sqrt(INT_MAX)) which is 15

For simplicity, lets consider ours is 4 instead of 15 and n value is 45

so we have to find a 4 bit integer x such that (x * x+1)/2 is less than 47. lets keep adding bits to position from msb to see how much far we can go

lets start, adding at 4th pos : 1000 (8) : 8 * 9 / 2 = 36 < 47 : add bit to 4th position adding at 3rd pos : 1100 (12) : 12 * 13 / 2 = 78 > 47 : dont add bit to 3rd position (remains 1000) adding at 2nd pos : 1010 (10) : 10 * 11/ 2 = 55 > 47 : dont add bit to 2th position(remains 1000) adding at 1st pos : 1001 (9) : 9 * 10 / 2 : 45 < 47 : add bit to 1st pos (ans = 1001)

final ans = 9

Based on your explanation, you want to find the largest number m for which the sum of integers from 1 to m is less than or equal to n. You are using a binary search-like strategy to find m.

Your code starts by setting m to the highest power of 2 and then checks if the sum of integers from 1 to m is less than or equal to n. If it is, you keep the bit set at that position. If it’s not, you unset the bit at that position and continue to the next bit.

Here’s how you might implement this in Python:

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class Solution:
    def arrangeCoins(self, n: int) -> int:
        left, right = 0, n
        while left <= right:
            mid = (left + right) // 2
            curr = mid * (mid + 1) // 2
            if curr == n:
                return mid
            if n < curr:
                right = mid - 1
            else:
                left = mid + 1
        return right

The binary search algorithm works by dividing the problem space into two halves on each iteration. It starts by checking the middle value. If the middle value is the target value, it returns the index. If the target value is less than the middle value, it discards the right half and continues with the left half. If the target value is more than the middle value, it discards the left half and continues with the right half. The algorithm continues until the target value is found or until there are no more values to check.

The time complexity of this algorithm is O(log n) and the space complexity is O(1).

The problem “Arranging Coins” asks you to determine the largest number of complete staircases that can be built from a given number of coins. Here, the coins form an arithmetic sequence where each step has one more coin than the previous step.

A simpler problem would be “Minimum Moves to Equal Array Elements”. This problem is conceptually similar, but instead of building staircases, you’re making moves to make all the elements of an array equal.

An approximately isomorphic problem is “Container With Most Water”. The reasoning here is a bit indirect, but both problems deal with optimizing some kind of allocation (coins into staircases or the area of water container), given a constraint (total number of coins or dimensions of container).

A more complex problem would be “Candy”. This problem adds the extra complexity of having to allocate resources (candies) according to a set of rules (children with a higher rating get more candies), which is similar to the staircase rule in the “Arranging Coins” problem.

Here is the ordered list of problems from simplest to most complex:

“Minimum Moves to Equal Array Elements” - Make moves to make all the elements of an array equal.
“Arranging Coins” - Determine the largest number of complete staircases that can be built from a given number of coins.
“Container With Most Water” - Find an optimal allocation given a constraint.
“Candy” - Allocate resources according to a set of rules.

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def arrangeCoins(self, n: int) -> int:
    return int((-1 + (1 + 8*n) ** 0.5) // 2)

Problem Classification

This problem can be classified into the following categories:

Mathematical problem: The problem requires calculation of rows which can be fully formed with a given number of coins. This includes understanding the arithmetic progression of sum of numbers.
Optimization problem: Given a limited resource (coins), the problem is to optimize the construction of complete rows.
Geometric problem: The problem involves creating a specific geometric structure, a staircase, with a given number of elements (coins).
Iterative/Sequential problem: The solution involves processing each row of the staircase sequentially until the resources (coins) are exhausted.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

Visual Model of the Problem

The “Arranging Coins” problem is about forming a complete stair-case with a given number of coins. Each step of the staircase contains an increasing amount of coins and the task is to determine the maximum number of complete steps that can be formed.

To visualize this problem, you can think of physically laying out the coins. Let’s say you have 10 coins. You start forming the staircase:

The first step has 1 coin.
The second step needs 2 coins.
The third step needs 3 coins.
The fourth step would need 4 coins, but you only have 3 left.

Therefore, the maximum number of complete steps you can form is 3.

This problem can be represented visually in the form of a staircase:

*
* *
* * *
* * * ?

Each * represents a coin and the ? indicates that you don’t have enough coins to complete the fourth step. This visualization helps you see how the staircase is built up and how the number of coins affects the number of complete steps you can form.

Language Agnostic Coding Drills

Understanding basic arithmetic sequences: Before tackling the problem, one needs to understand that the number of coins required for each row forms an arithmetic sequence (1, 2, 3, 4, 5, …, k). The sum of an arithmetic sequence can be found using the formula: sum = n/2 * (a + l) where n is the number of terms, a is the first term, and l is the last term.
Understanding the Quadratic Formula: The quadratic formula is a solution to the quadratic equation ax^2 + bx + c = 0. The formula is x = [ -b ± sqrt(b^2 - 4ac) ] / 2a. In this problem, the quadratic formula is used in a slightly rearranged form to find the number of complete rows (which is the same as finding the largest integer ‘x’ for which the sum of the first ‘x’ natural numbers is less than or equal to ’n’).
Understanding floating-point arithmetic: Since the result of a square root operation could be a floating-point number, we need to handle the conversion of this floating-point number back to an integer.

Step by step problem-solving approach:

Step 1: Understand that the problem can be modeled as finding the largest integer ‘k’ such that the sum of the first ‘k’ natural numbers is less than or equal to ’n’.
Step 2: Realize that the sum of the first ‘k’ natural numbers forms an arithmetic series and can be computed using the formula k*(k+1)/2.
Step 3: The problem then reduces to solving the quadratic inequality k*(k+1)/2 <= n. Rearranging terms gives us a quadratic equation in standard form.
Step 4: Use the quadratic formula to solve for ‘k’. Remember to round down to the nearest integer because ‘k’ represents the number of complete rows and it has to be an integer.
Step 5: Return the calculated ‘k’ as the solution.

This approach hinges on the understanding of arithmetic series and quadratic equations, and the ability to manipulate and apply the related formulas.

Targeted Drills in Python

Drill 1 - Understanding basic arithmetic sequences

Practice calculating the sum of an arithmetic sequence.

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def sum_arithmetic_sequence(n: int) -> int:
    return n * (n + 1) // 2

print(sum_arithmetic_sequence(5))  # Outputs: 15

Drill 2 - Understanding the Quadratic Formula

Practice applying the quadratic formula. Here, we solve the equation x^2 + 3x - 4 = 0.

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import math

def solve_quadratic(a: int, b: int, c: int) -> tuple:
    discriminant = b**2 - 4*a*c
    sqrt_discriminant = math.sqrt(abs(discriminant))
    sol1 = (-b - sqrt_discriminant) / (2*a)
    sol2 = (-b + sqrt_discriminant) / (2*a)
    return (sol1, sol2)

print(solve_quadratic(1, 3, -4))  # Outputs: (-4.0, 1.0)

Drill 3 - Understanding floating-point arithmetic

Practice converting floating-point numbers to integers.

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def float_to_int(f: float) -> int:
    return int(f)

print(float_to_int(3.7))  # Outputs: 3

Problem-specific Drill - Solving for the number of complete rows in a staircase
Putting it all together, the coding drill specific to this problem involves utilizing our understanding of arithmetic sequences, the quadratic formula, and floating-point arithmetic to solve the given problem.
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def arrange_coins(n: int) -> int: return int((-1 + (1 + 8*n) ** 0.5) // 2) print(arrange_coins(8)) # Outputs: 3

These drills cover the individual components of the final solution and practicing them would help in understanding and solving the given problem.