README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2506.Count%20Pairs%20Of%20Similar%20Strings/README_EN.md	1335	Weekly Contest 324 Q1	Bit Manipulation Array Hash Table String Counting

2506. Count Pairs Of Similar Strings

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Description

You are given a 0-indexed string array words.

Two strings are similar if they consist of the same characters.

• For example, "abca" and "cba" are similar since both consist of characters 'a', 'b', and 'c'.
• However, "abacba" and "bcfd" are not similar since they do not consist of the same characters.

Return the number of pairs (i, j) such that 0 <= i < j <= word.length - 1 and the two strings words[i] and words[j] are similar.

 

Example 1:

Input: words = ["aba","aabb","abcd","bac","aabc"]
Output: 2
Explanation: There are 2 pairs that satisfy the conditions:
- i = 0 and j = 1 : both words[0] and words[1] only consist of characters 'a' and 'b'. 
- i = 3 and j = 4 : both words[3] and words[4] only consist of characters 'a', 'b', and 'c'. 

Example 2:

Input: words = ["aabb","ab","ba"]
Output: 3
Explanation: There are 3 pairs that satisfy the conditions:
- i = 0 and j = 1 : both words[0] and words[1] only consist of characters 'a' and 'b'. 
- i = 0 and j = 2 : both words[0] and words[2] only consist of characters 'a' and 'b'.
- i = 1 and j = 2 : both words[1] and words[2] only consist of characters 'a' and 'b'.

Example 3:

Input: words = ["nba","cba","dba"]
Output: 0
Explanation: Since there does not exist any pair that satisfies the conditions, we return 0.

 

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 100
• words[i] consist of only lowercase English letters.

Solutions

Solution 1: Hash Table + Bit Manipulation

For each string, we can convert it into a binary number of length $26$, where the $i$-th bit being $1$ indicates that the string contains the $i$-th letter.

If two strings contain the same letters, their binary numbers are the same. Therefore, for each string, we use a hash table to count the occurrences of its binary number. Each time we add the count to the answer, then increment the count of its binary number by $1$.

The time complexity is $O(L)$, and the space complexity is $O(n)$. Here, $L$ is the total length of all strings, and $n$ is the number of strings.

Python3

class Solution:
    def similarPairs(self, words: List[str]) -> int:
        ans = 0
        cnt = Counter()
        for s in words:
            x = 0
            for c in map(ord, s):
                x |= 1 << (c - ord("a"))
            ans += cnt[x]
            cnt[x] += 1
        return ans

Java

class Solution {
    public int similarPairs(String[] words) {
        int ans = 0;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (var s : words) {
            int x = 0;
            for (char c : s.toCharArray()) {
                x |= 1 << (c - 'a');
            }
            ans += cnt.getOrDefault(x, 0);
            cnt.merge(x, 1, Integer::sum);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int similarPairs(vector<string>& words) {
        int ans = 0;
        unordered_map<int, int> cnt;
        for (const auto& s : words) {
            int x = 0;
            for (auto& c : s) {
                x |= 1 << (c - 'a');
            }
            ans += cnt[x]++;
        }
        return ans;
    }
};

Go

func similarPairs(words []string) (ans int) {
	cnt := map[int]int{}
	for _, s := range words {
		x := 0
		for _, c := range s {
			x |= 1 << (c - 'a')
		}
		ans += cnt[x]
		cnt[x]++
	}
	return
}

TypeScript

function similarPairs(words: string[]): number {
    let ans = 0;
    const cnt = new Map<number, number>();
    for (const s of words) {
        let x = 0;
        for (const c of s) {
            x |= 1 << (c.charCodeAt(0) - 97);
        }
        ans += cnt.get(x) || 0;
        cnt.set(x, (cnt.get(x) || 0) + 1);
    }
    return ans;
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn similar_pairs(words: Vec<String>) -> i32 {
        let mut ans = 0;
        let mut cnt: HashMap<i32, i32> = HashMap::new();
        for s in words {
            let mut x = 0;
            for c in s.chars() {
                x |= 1 << ((c as u8) - b'a');
            }
            ans += cnt.get(&x).unwrap_or(&0);
            *cnt.entry(x).or_insert(0) += 1;
        }
        ans
    }
}