How hard is LeetCode 1583. Count Unhappy Friends?

LeetCode 1583. Count Unhappy Friends is rated Medium on LeetCode.

What is the time complexity of LeetCode 1583. Count Unhappy Friends?

The Python solution on this page runs in O(n^2).

What is the space complexity of LeetCode 1583. Count Unhappy Friends?

The Python solution on this page uses O(n^2) auxiliary space.

What topics does LeetCode 1583. Count Unhappy Friends cover?

LeetCode 1583. Count Unhappy Friends is tagged Array and Simulation on LeetCode.

MediumNo. 1583

Count Unhappy Friends — LeetCode 1583 Python Solution

Topics: Array, Simulation
Reading time: 4 min
Source: leetcode.com

The problem

You are given a list of preferences for n friends, where n is always even. For each person i, preferences[i] contains a list of friends sorted in the order of preference.

Example

Input: n = 4, preferences = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], pairs = [[0, 1], [2, 3]]
Output: 2
Explanation: Friend 1 is unhappy because:

Python solution

PythonSolution

class Solution:
    def unhappyFriends(
        self, n: int, preferences: List[List[int]], pairs: List[List[int]]
    ) -> int:
        d = [{x: j for j, x in enumerate(p)} for p in preferences]
        p = {}
        for x, y in pairs:
            p[x] = y
            p[y] = x
        ans = 0
        for x in range(n):
            y = p[x]
            for i in range(d[x][y]):
                u = preferences[x][i]
                v = p[u]
                if d[u][x] < d[u][v]:
                    ans += 1
                    break
        return ans

Complexity

Time: O(n2)
Space · auxiliary: O(n2)

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Frequently asked questions

How hard is LeetCode 1583. Count Unhappy Friends?: LeetCode 1583. Count Unhappy Friends is rated Medium on LeetCode.
What is the time complexity of LeetCode 1583. Count Unhappy Friends?: The Python solution on this page runs in O(n²).
What is the space complexity of LeetCode 1583. Count Unhappy Friends?: The Python solution on this page uses O(n²) auxiliary space.
What topics does LeetCode 1583. Count Unhappy Friends cover?: LeetCode 1583. Count Unhappy Friends is tagged Array and Simulation 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

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