Please analyze this context and help me answer the question or solve the problem.
"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."
→ 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:
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 -> indexApproach
At each index r, the window's left edge is:
Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.
Please analyze this context and help me answer the question or solve the problem.
"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."
→ 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:
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 -> indexApproach
At each index r, the window's left edge is:
Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.
Please analyze this context and help me answer the question or solve the problem.
"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."
→ 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:
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 -> indexApproach
At each index r, the window's left edge is:
Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.
Interview Coder Review: Features, Workflow, Alternatives
Interview Coder review covering its public features, desktop workflow, coding support, trade-offs, and practical alternatives for technical interviews.
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