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Isomorphic String

Easy

The mapping must work both ways

Problem

Return whether two strings are isomorphic: characters of one can be consistently mapped to characters of the other (one-to-one).

In simple words

Each letter must map to exactly one partner both ways — track two mappings.

The idea

Walk both strings in step, recording the mapping from each character of one to the corresponding character of the other. Keeping two maps is essential: a single map would allow two different characters to map onto the same target, which is not isomorphic.

The trick

  • Two maps, one in each direction — one map alone accepts 'badc' -> 'baba'.
  • Different lengths are an immediate no.
  • O(n) time, O(alphabet) space.

This one walks through the worked example rather than tracing the algorithm frame by frame — a full walkthrough is still to be drawn. The code and the idea below are the real solution.

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Step 1 of 2. Here's the example — egg, add Values: 0.

1/2
Optimal
timeO(n)spaceO(1)
1map s->t and t->s2for i: if mappings conflict: return false; else set them3return true

Input

array
[0]

Output

answer
—

Check yourself

3 quick questions about this walkthrough. A wrong answer costs nothing.

Examples

Example 1

Input:
s = "egg", t = "add"
Output:
true
Explanation:
e->a, g->d is a clean one-to-one mapping.

Example 2

Input:
s = "foo", t = "bar"
Output:
false
Explanation:
o would need to map to two letters → false.

Example 3

Input:
s = "paper", t = "title"
Output:
true
Explanation:
Every letter maps consistently.

Practice this problem:LeetCode(opens in a new tab)Search GeeksforGeeks(opens in a new tab)

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