README_EN.md

1153. String Transforms Into Another String

中文文档

Description

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.

 

Example 1:

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.

Example 2:

Input: str1 = "leetcode", str2 = "codeleet"
Output: false
Explanation: There is no way to transform str1 to str2.

 

Constraints:

• 1 <= str1.length == str2.length <= 104
• str1 and str2 contain only lowercase English letters.

Solutions

Python3

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

Java

class Solution {
    public boolean canConvert(String str1, String str2) {
        if (str1.equals(str2)) {
            return true;
        }
        int m = 0;
        int[] cnt = new int[26];
        int n = str1.length();
        for (int i = 0; i < n; ++i) {
            if (++cnt[str2.charAt(i) - 'a'] == 1) {
                ++m;
            }
        }
        if (m == 26) {
            return false;
        }
        int[] d = new int[26];
        for (int i = 0; i < n; ++i) {
            int a = str1.charAt(i) - 'a';
            int b = str2.charAt(i) - 'a';
            if (d[a] == 0) {
                d[a] = b + 1;
            } else if (d[a] != b + 1) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    bool canConvert(string str1, string str2) {
        if (str1 == str2) {
            return true;
        }
        int cnt[26]{};
        int m = 0;
        for (char& c : str2) {
            if (++cnt[c - 'a'] == 1) {
                ++m;
            }
        }
        if (m == 26) {
            return false;
        }
        int d[26]{};
        for (int i = 0; i < str1.size(); ++i) {
            int a = str1[i] - 'a';
            int b = str2[i] - 'a';
            if (d[a] == 0) {
                d[a] = b + 1;
            } else if (d[a] != b + 1) {
                return false;
            }
        }
        return true;
    }
};

Go

func canConvert(str1 string, str2 string) bool {
	if str1 == str2 {
		return true
	}
	s := map[rune]bool{}
	for _, c := range str2 {
		s[c] = true
		if len(s) == 26 {
			return false
		}
	}
	d := [26]int{}
	for i, c := range str1 {
		a, b := int(c-'a'), int(str2[i]-'a')
		if d[a] == 0 {
			d[a] = b + 1
		} else if d[a] != b+1 {
			return false
		}
	}
	return true
}

TypeScript

function canConvert(str1: string, str2: string): boolean {
    if (str1 === str2) {
        return true;
    }
    if (new Set(str2).size === 26) {
        return false;
    }
    const d: Map<string, string> = new Map();
    for (const [i, c] of str1.split('').entries()) {
        if (!d.has(c)) {
            d.set(c, str2[i]);
        } else if (d.get(c) !== str2[i]) {
            return false;
        }
    }
    return true;
}

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