String Transforms Into Another String

String Transforms Into Another String - Problem

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.

Return true if and only if you can transform str1 into str2.

Input & Output

Example 1 — Simple Mapping

$ Input: str1 = "aab", str2 = "xxy"

› Output: true

💡 Note: We can convert 'a' to 'x' and 'b' to 'y': aab → xab → xxy. Each character in str1 maps consistently to one character in str2.

Example 2 — Conflicting Mapping

$ Input: str1 = "leetcode", str2 = "codeleet"

› Output: false

💡 Note: Character 'e' would need to map to both 'e' and 't', which is impossible. We cannot have one character map to multiple different characters.

Example 3 — All Characters Used

$ Input: str1 = "abcdefghijklmnopqrstuvwxyz", str2 = "bcdefghijklmnopqrstuvwxyza"

› Output: false

💡 Note: This creates a cycle where 'a'→'b'→'c'→...→'z'→'a', but all 26 characters are used in str2, so we have no temporary character to break the cycle.

Constraints

1 ≤ str1.length ≤ 10⁴
str2.length == str1.length
str1 and str2 consist of lowercase English letters only

Visualization

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Asked in

G Google 35 f Facebook 28

The key insight is that each character in str1 must map to exactly one character in str2, and cycles can only be resolved if there's an unused character as temporary storage. Best approach is the Hash Map Validation method. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force Simulation

⏱️ Time: O(26^26) Space: O(n)

Generate all possible character mappings and simulate the conversion process for each combination. This explores every potential transformation path.

Hash Map Validation

⏱️ Time: O(n) Space: O(1)

Create a mapping from str1 characters to str2 characters and verify two key rules: each character maps to exactly one target, and handle cycles properly.

Brute Force Simulation — Algorithm Steps

Generate all possible character mappings
For each mapping, simulate the conversion
Check if result matches str2

Visualization

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Step-by-Step Walkthrough

Find Mismatch

Identify first character that differs between strings

Try Mapping

Attempt to map source character to target character

Recurse

Continue with updated string until match found or impossible

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char* str1, char* str2) {
    int len1 = strlen(str1);
    int len2 = strlen(str2);
    if (len1 != len2) return false;
    if (strcmp(str1, str2) == 0) return true;
    
    // Build mapping from str1 to str2
    char mapping[26];
    bool hasMapping[26];
    for (int i = 0; i < 26; i++) {
        mapping[i] = 0;
        hasMapping[i] = false;
    }
    
    for (int i = 0; i < len1; i++) {
        char c1 = str1[i];
        char c2 = str2[i];
        int idx = c1 - 'a';
        if (hasMapping[idx]) {
            if (mapping[idx] != c2) {
                return false; // Conflicting mapping
            }
        } else {
            mapping[idx] = c2;
            hasMapping[idx] = true;
        }
    }
    
    // Check if transformation is possible
    // If all 26 characters are used in str2, cycles cannot be resolved
    bool uniqueCharsInStr2[26] = {false};
    int uniqueCount = 0;
    for (int i = 0; i < len2; i++) {
        int idx = str2[i] - 'a';
        if (!uniqueCharsInStr2[idx]) {
            uniqueCharsInStr2[idx] = true;
            uniqueCount++;
        }
    }
    
    if (uniqueCount == 26) {
        // Check if there are any cycles
        for (int i = 0; i < 26; i++) {
            if (hasMapping[i]) {
                char c1 = 'a' + i;
                char c2 = mapping[i];
                if (c1 != c2) { // This is a transformation needed
                    return false;
                }
            }
        }
    }
    
    return true;
}

int main() {
    char str1[10001], str2[10001];
    fgets(str1, sizeof(str1), stdin);
    str1[strcspn(str1, "\n")] = 0;
    fgets(str2, sizeof(str2), stdin);
    str2[strcspn(str2, "\n")] = 0;
    
    bool result = solution(str1, str2);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(26^26)

Up to 26 characters, each can map to any of 26 characters

✓ Linear Growth

Space Complexity

O(n)

Space for string copies during simulation

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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