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Find the Pivot Integer

Easy

Google

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Topics:

ArraysBinary Search

Given a positive integer n, find the pivot integer x such that:

The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively.

Return the pivot integer x. If no such integer exists, return -1. It is guaranteed that there will be at most one pivot index for the given input.

Example 1:

Input: n = 8
Output: 6
Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.

Example 2:

Input: n = 1
Output: 1
Explanation: 1 is the pivot integer since: 1 = 1.

Example 3:

Input: n = 4
Output: -1
Explanation: It can be proved that no such integer exist.

Constraints:

1 <= n <= 1000

Try coding it in LeetCode

Find the Pivot Integer

Easy

Google

3 views

Topics:

ArraysBinary Search

Given a positive integer n, find the pivot integer x such that:

The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively.

Return the pivot integer x. If no such integer exists, return -1. It is guaranteed that there will be at most one pivot index for the given input.

Example 1:

Input: n = 8
Output: 6
Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.

Example 2:

Input: n = 1
Output: 1
Explanation: 1 is the pivot integer since: 1 = 1.

Example 3:

Input: n = 4
Output: -1
Explanation: It can be proved that no such integer exist.

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 range of possible values for the input integer 'n'? Can 'n' be negative?
If no such pivot integer exists, what should I return?
Is 'n' guaranteed to be a positive integer, or should I handle the case where 'n' is zero or negative?
Could you clarify the expected behavior when 'n' is a very large number and calculating the sums might lead to integer overflow?
Are we looking for the smallest such pivot integer if multiple pivot integers exist?

Brute Force Solution

Approach

We're trying to find a special number. The brute force method means we'll try every single number within the given range to see if it's the right one.

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

Start with the first possible number in the range.
Calculate the sum of all numbers from the beginning of the range up to this number.
Calculate the sum of all numbers from this number to the end of the range.
Compare the two sums. If they are equal, then we've found our special number and we're done.
If the sums are not equal, move on to the next number in the range and repeat the process.
Keep doing this until you either find the special number or you've tried every number in the range.

Code Implementation

def find_pivot_integer_brute_force(number_range): 
    for pivot_integer_candidate in range(1, number_range + 1):

        # Calculate the sum from 1 to the candidate
        first_sum = sum(range(1, pivot_integer_candidate + 1))

        # Calculate the sum from the candidate to n
        second_sum = sum(range(pivot_integer_candidate, number_range + 1))

        # Check if both sums are equal
        if first_sum == second_sum:
            return pivot_integer_candidate

    # No pivot integer found in the range
    return -1

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each number from 1 to n (where n is the input). For each number, it calculates the sum of numbers from 1 to that number, and then calculates the sum of numbers from that number to n. Both sum calculations require iterating through a portion of the range 1 to n. Therefore, for each of the n numbers, a summation operation that could take up to n steps is performed. This results in approximately n * n operations, thus the time complexity is O(n²).

Space Complexity

O(1) – The provided algorithm calculates sums iteratively without storing any data structures dependent on the input range. It uses a few variables to store the current number and the two sums being compared, but the number of these variables is constant regardless of the input range N. Therefore, the space complexity is constant.

Optimal Solution

Approach

The goal is to find a number where the sum of numbers before it equals the sum of numbers after it. Instead of checking every number individually, we use a bit of math to quickly narrow down the possibilities.

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

First, calculate the sum of all numbers from one up to the maximum number we're considering.
Then, think about what it means for a number to be the "pivot": the total sum is split into the sum before the number, the number itself, and the sum after the number.
Realize that the sum before the pivot and the sum after the pivot must be equal. That means if you add the number to the sum before the number, it equals half of the total sum plus half of the number.
Rearrange this to solve for the number. You'll get a formula showing how the number depends on the total sum.
Calculate the pivot using the formula. If the result is a whole number that falls within the given range, you've found the pivot.
If the result is not a whole number or isn't in the range, then there's no pivot number that satisfies the condition.

Code Implementation

def find_pivot_integer(maximum_number):
    total_sum = (maximum_number * (maximum_number + 1)) // 2

    # Applying the formula derived from the problem constraints.
    pivot_integer = (total_sum // (maximum_number + 1))

    # Check if pivot is an integer and within the range.
    if pivot_integer * (maximum_number + 1) == total_sum and \
pivot_integer >= 1 and pivot_integer <= maximum_number:

        return int(pivot_integer)
    else:
        # No pivot integer exists within the given range.
        return -1

Big(O) Analysis

Time Complexity

O(1) – The algorithm calculates the total sum from 1 to n, which takes constant time. It then uses a direct formula derived from the problem's constraints to compute the potential pivot integer. The remaining steps involve basic arithmetic operations and range checks, all of which take constant time. Therefore, the time complexity does not scale with the input size n, resulting in O(1) complexity.

Space Complexity

O(1) – The algorithm calculates the total sum and then iteratively computes a potential pivot integer using a formula. It only uses a few variables to store intermediate calculations like the total sum and the potential pivot. These variables consume a fixed amount of memory irrespective of the input number, n. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Input 'n' is zero	Return -1 since the range 1 to n will be empty, meaning there is no pivot integer.
Input 'n' is a negative number	Return -1, as the problem specifies a range from 1 to n (inclusive).
Input 'n' is 1	Return 1, as the sum from 1 to 1 equals the sum from 1 to 1.
Input 'n' is a large number that could lead to integer overflow when calculating sums	Use long data type or appropriate modular arithmetic to prevent integer overflow during sum calculations.
No pivot integer exists within the range 1 to n	Return -1 after iterating through all possible pivot integers and not finding a match.
Input 'n' is very large causing time limit exceed	Optimize the sum calculation to O(1) by using the formula n*(n+1)/2, and then iterating only to calculate the left sum.
n is close to the maximum integer value	Be careful to use proper datatypes (e.g., long) to store intermediate sums that could overflow.
Multiple pivot integers exist (this is impossible based on the problem constraints but should be considered).	The problem constraints guarantee at most one solution, so returning the first valid pivot found is sufficient (though an assertion could be used to confirm only one exists).

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 range of possible values for the input integer 'n'? Can 'n' be negative?
If no such pivot integer exists, what should I return?
Is 'n' guaranteed to be a positive integer, or should I handle the case where 'n' is zero or negative?
Could you clarify the expected behavior when 'n' is a very large number and calculating the sums might lead to integer overflow?
Are we looking for the smallest such pivot integer if multiple pivot integers exist?

Brute Force Solution

Approach

We're trying to find a special number. The brute force method means we'll try every single number within the given range to see if it's the right one.

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

Start with the first possible number in the range.
Calculate the sum of all numbers from the beginning of the range up to this number.
Calculate the sum of all numbers from this number to the end of the range.
Compare the two sums. If they are equal, then we've found our special number and we're done.
If the sums are not equal, move on to the next number in the range and repeat the process.
Keep doing this until you either find the special number or you've tried every number in the range.

Code Implementation

def find_pivot_integer_brute_force(number_range): 
    for pivot_integer_candidate in range(1, number_range + 1):

        # Calculate the sum from 1 to the candidate
        first_sum = sum(range(1, pivot_integer_candidate + 1))

        # Calculate the sum from the candidate to n
        second_sum = sum(range(pivot_integer_candidate, number_range + 1))

        # Check if both sums are equal
        if first_sum == second_sum:
            return pivot_integer_candidate

    # No pivot integer found in the range
    return -1

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, calculate the sum of all numbers from one up to the maximum number we're considering.
Then, think about what it means for a number to be the "pivot": the total sum is split into the sum before the number, the number itself, and the sum after the number.
Realize that the sum before the pivot and the sum after the pivot must be equal. That means if you add the number to the sum before the number, it equals half of the total sum plus half of the number.
Rearrange this to solve for the number. You'll get a formula showing how the number depends on the total sum.
Calculate the pivot using the formula. If the result is a whole number that falls within the given range, you've found the pivot.
If the result is not a whole number or isn't in the range, then there's no pivot number that satisfies the condition.

Code Implementation

def find_pivot_integer(maximum_number):
    total_sum = (maximum_number * (maximum_number + 1)) // 2

    # Applying the formula derived from the problem constraints.
    pivot_integer = (total_sum // (maximum_number + 1))

    # Check if pivot is an integer and within the range.
    if pivot_integer * (maximum_number + 1) == total_sum and \
pivot_integer >= 1 and pivot_integer <= maximum_number:

        return int(pivot_integer)
    else:
        # No pivot integer exists within the given range.
        return -1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input 'n' is zero	Return -1 since the range 1 to n will be empty, meaning there is no pivot integer.
Input 'n' is a negative number	Return -1, as the problem specifies a range from 1 to n (inclusive).
Input 'n' is 1	Return 1, as the sum from 1 to 1 equals the sum from 1 to 1.
Input 'n' is a large number that could lead to integer overflow when calculating sums	Use long data type or appropriate modular arithmetic to prevent integer overflow during sum calculations.
No pivot integer exists within the range 1 to n	Return -1 after iterating through all possible pivot integers and not finding a match.
Input 'n' is very large causing time limit exceed	Optimize the sum calculation to O(1) by using the formula n*(n+1)/2, and then iterating only to calculate the left sum.
n is close to the maximum integer value	Be careful to use proper datatypes (e.g., long) to store intermediate sums that could overflow.
Multiple pivot integers exist (this is impossible based on the problem constraints but should be considered).	The problem constraints guarantee at most one solution, so returning the first valid pivot found is sufficient (though an assertion could be used to confirm only one exists).