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Leetcode #1153: String Transforms Into Another String

In this guide, we solve Leetcode #1153 String Transforms Into Another String in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given two strings str1 and str2 of the same length, determine whether you can transform str1 into str2 by doing zero or more conversions. In one conversion you can convert all occurrences of one character in str1 to any other lowercase English character.

Quick Facts

  • Difficulty: Hard
  • Premium: Yes
  • Tags: Graph, Hash Table, String

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

Input: str1 = "aabcc", str2 = "ccdee" Output: true Explanation: Convert 'c' to 'e' then 'b' to 'd' then 'a' to 'c'. Note that the order of conversions matter.

Python Solution

class Solution: def canConvert(self, str1: str, str2: str) -> bool: if str1 == str2: return True if len(set(str2)) == 26: return False d = {} for a, b in zip(str1, str2): if a not in d: d[a] = b elif d[a] != b: return False return True

Complexity

The time complexity is O(n)O(n)O(n), and the space complexity is O(C)O(C)O(C). The space complexity is O(C)O(C)O(C).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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