How hard is LeetCode 2485. Find the Pivot Integer?

LeetCode 2485. Find the Pivot Integer is rated Easy on LeetCode.

What is the time complexity of LeetCode 2485. Find the Pivot Integer?

The Python solution on this page runs in O(n), where n is the given positive integer n.

What is the space complexity of LeetCode 2485. Find the Pivot Integer?

The Python solution on this page uses O(1) auxiliary space.

What topics does LeetCode 2485. Find the Pivot Integer cover?

LeetCode 2485. Find the Pivot Integer is tagged Math and Prefix Sum on LeetCode.

EasyNo. 2485

Find the Pivot Integer — LeetCode 2485 Python Solution

Pattern: Prefix Sum
Topics: Math, Prefix Sum
Reading time: 2 min
Source: leetcode.com

The problem

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.

Example

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

Python solution

PythonSolution

class Solution:
    def pivotInteger(self, n: int) -> int:
        for x in range(1, n + 1):
            if (1 + x) * x == (x + n) * (n - x + 1):
                return x
        return -1

Complexity

Time: O(n)where n is the given positive integer n
Space · auxiliary: O(1)

Pattern: Prefix Sum

Precompute running totals once so any range query becomes a single subtraction. LeetCode 2485. Find the Pivot Integer is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Prefix Sum.

The prefix sum guide has the Python template for the pattern and the 157 LeetCode problems that use it.

Related problems
No.	Problem	Difficulty
1352	Product of the Last K Numbers	Medium
1588	Sum of All Odd Length Subarrays	Easy
1685	Sum of Absolute Differences in a Sorted Array	Medium
2483	Minimum Penalty for a Shop	Medium
0238	Product of Array Except Self	Medium
0303	Range Sum Query - Immutable	Easy

Frequently asked questions

How hard is LeetCode 2485. Find the Pivot Integer?: LeetCode 2485. Find the Pivot Integer is rated Easy on LeetCode.
What is the time complexity of LeetCode 2485. Find the Pivot Integer?: The Python solution on this page runs in O(n), where n is the given positive integer n.
What is the space complexity of LeetCode 2485. Find the Pivot Integer?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2485. Find the Pivot Integer cover?: LeetCode 2485. Find the Pivot Integer is tagged Math and Prefix Sum on LeetCode.

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

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Transcribing...

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

The problem

Example

Python solution

Complexity

Pattern: Prefix Sum

Related problems

Frequently asked questions

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Approach

Approach

Approach