Isomorphic status codes

Scenario

A build system runs a pipeline twice, in two environments. Each step ends with a one-letter status code, so a run is a string such as egg.

You want to know whether the two runs follow the same pattern. Steps that share a status in one run must share a status in the other. Steps with different statuses must differ in both. The letters themselves may differ.

Your task

Write are_isomorphic(a, b).

a and b are the status strings of the two runs. They have the same length.
The runs follow the same pattern when you can replace each letter of a with one letter so that you get b. Every occurrence of a letter must become the same letter, and two different letters of a must become different letters.
Return true when the runs follow the same pattern, and false otherwise.

Examples

Example 1 (traced)#

text2 lines

Input:  a = "egg", b = "add"
Output: true

The trace reads both strings together and keeps two maps: from a letter of a to a letter of b, and back.

i	x (from a)	y (from b)	a to b	b to a	Consistent?
0	e	a	e: a	a: e	yes
1	g	d	e: a, g: d	a: e, d: g	yes
2	g	d	same	same	yes: g still maps to d

No conflict appears, so the output is true.

Example 2#

text2 lines

Input:  a = "foo", b = "bar"
Output: false

The letter f would map to b, then the letter o to a and again to r. The two os map to different letters, so the output is false.

Example 3#

text2 lines

Input:  a = "ab", b = "aa"
Output: false

The letters a and b would both map to a. Two different letters cannot share a target, so the output is false.

Constraints

1 ≤ a.length ≤ 100,000, and b has the same length
Both strings use lowercase letters a to z.

What does n ≤ 10^5 tell you?

Comparing every pair of positions costs about n × n / 2 steps, which is 5 × 10^9. Two maps cost one step per position. Each map holds at most 26 letters, so the space is constant.

Scenario

A build system runs a pipeline twice, in two environments. Each step ends with a one-letter status code, so a run is a string such as egg.

Your task

Write are_isomorphic(a, b).

a and b are the status strings of the two runs. They have the same length.
The runs follow the same pattern when you can replace each letter of a with one letter so that you get b. Every occurrence of a letter must become the same letter, and two different letters of a must become different letters.
Return true when the runs follow the same pattern, and false otherwise.

Examples

Example 1 (traced)#

text2 lines

Input:  a = "egg", b = "add"
Output: true

The trace reads both strings together and keeps two maps: from a letter of a to a letter of b, and back.

i	x (from a)	y (from b)	a to b	b to a	Consistent?
0	e	a	e: a	a: e	yes
1	g	d	e: a, g: d	a: e, d: g	yes
2	g	d	same	same	yes: g still maps to d

No conflict appears, so the output is true.

Example 2#

text2 lines

Input:  a = "foo", b = "bar"
Output: false

The letter f would map to b, then the letter o to a and again to r. The two os map to different letters, so the output is false.

Example 3#

text2 lines

Input:  a = "ab", b = "aa"
Output: false

The letters a and b would both map to a. Two different letters cannot share a target, so the output is false.

Constraints

1 ≤ a.length ≤ 100,000, and b has the same length
Both strings use lowercase letters a to z.

What does n ≤ 10^5 tell you?

Comparing every pair of positions costs about n × n / 2 steps, which is 5 × 10^9. Two maps cost one step per position. Each map holds at most 26 letters, so the space is constant.