How hard is LeetCode 2495. Number of Subarrays Having Even Product?

LeetCode 2495. Number of Subarrays Having Even Product is rated Medium on LeetCode.

What is the time complexity of LeetCode 2495. Number of Subarrays Having Even Product?

The Python solution on this page runs in O(n), where n is the length of the array `nums`.

What is the space complexity of LeetCode 2495. Number of Subarrays Having Even Product?

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

What topics does LeetCode 2495. Number of Subarrays Having Even Product cover?

LeetCode 2495. Number of Subarrays Having Even Product is tagged Array, Math and Dynamic Programming on LeetCode.

Is LeetCode 2495. Number of Subarrays Having Even Product a premium problem?

Yes. LeetCode 2495. Number of Subarrays Having Even Product is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 2495LeetCode Premium

Number of Subarrays Having Even Product — LeetCode 2495 Python Solution

Pattern: Dynamic Programming
Topics: Array, Math, Dynamic Programming
Reading time: 2 min
Source: leetcode.com

The problem

Given a 0-indexed integer array nums, return the number of subarrays of nums having an even product.

Example

Input: nums = [9,6,7,13]
Output: 6
Explanation: There are 6 subarrays with an even product:

Python solution

PythonSolution

class Solution:
    def evenProduct(self, nums: List[int]) -> int:
        ans, last = 0, -1
        for i, v in enumerate(nums):
            if v % 2 == 0:
                last = i
            ans += last + 1
        return ans

Complexity

Time: O(n)where n is the length of the array `nums`
Space · auxiliary: O(1)

Pattern: Dynamic Programming

Define a state, write the transition, and stop recomputing the same subproblem. LeetCode 2495. Number of Subarrays Having Even Product is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Dynamic Programming.

The dynamic programming guide has the Python template for the pattern and the 481 LeetCode problems that use it.

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Frequently asked questions

How hard is LeetCode 2495. Number of Subarrays Having Even Product?: LeetCode 2495. Number of Subarrays Having Even Product is rated Medium on LeetCode.
What is the time complexity of LeetCode 2495. Number of Subarrays Having Even Product?: The Python solution on this page runs in O(n), where n is the length of the array `nums`.
What is the space complexity of LeetCode 2495. Number of Subarrays Having Even Product?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2495. Number of Subarrays Having Even Product cover?: LeetCode 2495. Number of Subarrays Having Even Product is tagged Array, Math and Dynamic Programming on LeetCode.
Is LeetCode 2495. Number of Subarrays Having Even Product a premium problem?: Yes. LeetCode 2495. Number of Subarrays Having Even Product is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

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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.

Type your message...

GPT-6.1 Terra

Transcribing...

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...

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: Dynamic Programming

Related problems

Frequently asked questions

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Approach

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