Using Hash Set to Determine if a String is the Permutation of Another String

Given two strings s1 and s2, write an algorithm to determine if s1 is one permutation of s2.

For example:
s1 = “abc”, s2 = “bca”
output: true

s1 = “abc”, s2 = “bad”
output: false

Algorithm to Determine if a String is the Permutation of Another String

The fastest way to determine this is to use hash sets. If both strings (s1 and s2) are of different sizes, the result has to be false. Otherwise, we can convert both into two hash sets and simply compare the equality of both hash sets.

In C++, you can directly compare the the equality of both hash sets, either set (based on Red-Black Tree) or unordered_set (based on Hash Map):

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class Solution {
public:
    bool CheckPermutation(string s1, string s2) {
        if (s1.size() != s2.size()) return false;
        unordered_set<char> a(begin(s1), end(s1));
        unordered_set<char> b(begin(s2), end(s2));
        return a == b;
    }
};

class Solution {
public:
    bool CheckPermutation(string s1, string s2) {
        if (s1.size() != s2.size()) return false;
        unordered_set<char> a(begin(s1), end(s1));
        unordered_set<char> b(begin(s2), end(s2));
        return a == b;
    }
};

Using unordered_set in C++, you will have a O(N) complexity however if you are using set (which keeps the keys in order by tree), the complexity is O(N.Log(N)).

In Python, we can do the same, with less code:

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class Solution:
    def CheckPermutation(self, s1: str, s2: str) -> bool:
        if len(s1) != len(s2):
            return False
        return set(s1) == set(s2)

class Solution:
    def CheckPermutation(self, s1: str, s2: str) -> bool:
        if len(s1) != len(s2):
            return False
        return set(s1) == set(s2)

–EOF (The Ultimate Computing & Technology Blog) —