65. Valid Number
Asked at LinkedIn
Difficulty
Hard
Time
O(n)
Space
O(1)
Pattern
String
Companies
1
Problem
You are given a string s. Decide whether it is a valid number in the usual programming-language sense: an optional sign, then either an integer or a decimal, and then optionally an exponent part.
A valid number has this shape: an optional + or -, then a mantissa, then optionally e or E followed by an exponent. The mantissa is either plain digits (an integer) or digits with exactly one dot (a decimal). The exponent is an optional sign followed by digits — no dot is allowed there.
Some rules that trip people up: a dot alone (".") is not a number, and a dot must have at least one digit somewhere on one of its sides ("4." and ".9" are fine). The e part must contain at least one digit, so "1e" and "e3" fail. A sign is only allowed at the very start or right after the e/E, so "--6" and "1e+2+3" fail. Any letter other than e/E makes the string invalid.
The shape of a valid number
Constraints
- ●
1 <= s.length <= 20 - ●
s contains only English letters, digits, '+', '-', and '.'
Examples
s = "0"trues = "e"falses = "."falseAsked At
1 companies| Company | Difficulty | |
|---|---|---|
| Hard | View all LinkedIn questions → |
How to Think About It
Write down the grammar first. A valid number is: optional sign, then digits with at most one dot and at least one digit, then optionally e/E followed by an optional sign and at least one digit (no dot).
Key insight: a single left-to-right scan with three flags handles every case — seenDigit, seenDot, and seenExp. Each character is only legal if the flags say it is allowed at this point.
Rules per character: a digit sets seenDigit. A sign is only legal at index 0 or right after an e. A dot is illegal after a previous dot or after an exponent. An e is illegal if you already saw one or if no digit came before it — and after it you reset seenDigit = false so the exponent must have its own digits.
Walkthrough for "-90E3": - at index 0 OK; 9,0 set seenDigit; E allowed (digit seen, no exp yet) -> seenExp = true, seenDigit = false; 3 sets seenDigit. End: seenDigit is true -> valid.
Walkthrough for "1e": after e, seenDigit was reset and nothing follows -> invalid.
Edge cases worth testing aloud: ".", "+.", ".e1", "4e+", "+-5", "95a54e53", "0089" (valid), "-.9" (valid).
Optimal Approach
Step 1: Initialize seenDigit = seenDot = seenExp = false.
Step 2: For each index i and char ch:
If digit: seenDigit = true.
If + or -: valid only if i == 0 or s[i-1] is e/E.
If .: invalid if seenDot or seenExp; otherwise seenDot = true.
If e/E: invalid if seenExp or not seenDigit; otherwise seenExp = true and seenDigit = false.
Anything else: invalid.
Step 3: Return seenDigit.
The final seenDigit check covers both "no digits at all" and "exponent with no digits".
Time: O(n). Space: O(1).
One scan with three flags
What Trips People Up in Real Interviews
Using float(s) or a language parser. Those accept inputs like "inf", "nan", or "1_000" that are invalid here — and the interviewer wants the logic, not a library call.
Forgetting to reset seenDigit after the exponent. Without it, "1e" is accepted because a digit was seen before the e.
Allowing a dot inside the exponent. "99e2.5" must be rejected — dots are only legal before the e.
Diving straight into code. On this problem the grammar is the solution; state it in one sentence, then implement it with flags. It is much easier to defend than 20 nested ifs.
Accepting a sign in the middle, like "6+1". A sign is only legal at position 0 or right after e/E.
Solution Code
O(n) timedef isNumber(s):
seen_digit = seen_dot = seen_exp = False
for i, ch in enumerate(s):
if ch.isdigit():
seen_digit = True
elif ch in '+-':
if i > 0 and s[i - 1] not in 'eE':
return False
elif ch == '.':
if seen_dot or seen_exp:
return False
seen_dot = True
elif ch in 'eE':
if seen_exp or not seen_digit:
return False
seen_exp = True
seen_digit = False
else:
return False
return seen_digitFrequently Asked Questions
What is the Valid Number problem?
Valid Number asks whether a string represents a legal decimal or scientific-notation number, such as `"2"`, `"-0.1"`, `"4."`, `".5"`, or `"-90E3"`, while rejecting things like `"e3"`, `"99e2.5"`, `"--6"`, or `"."`. It is rated Hard not because of algorithms but because of edge cases — the interviewer wants to see you define the grammar precisely and encode it without a pile of special cases.
How do you solve Valid Number?
The optimal approach runs in O(n) time and O(1) space. Scroll up to "Optimal Approach" for the step-by-step walkthrough, diagrams, and code in Python, C++ and Java.
What companies ask Valid Number?
Valid Number is asked at LinkedIn. It is a hard difficulty problem.
What are common mistakes on Valid Number?
- Using `float(s)` or a language parser. Those accept inputs like `"inf"`, `"nan"`, or `"1_000"` that are invalid here — and the interviewer wants the logic, not a library call.
- Forgetting to reset `seenDigit` after the exponent. Without it, `"1e"` is accepted because a digit was seen before the `e`.
- Allowing a dot inside the exponent. `"99e2.5"` must be rejected — dots are only legal before the `e`.
- Diving straight into code. On this problem the grammar is the solution; state it in one sentence, then implement it with flags. It is much easier to defend than 20 nested ifs.
- Accepting a sign in the middle, like `"6+1"`. A sign is only legal at position 0 or right after `e`/`E`.