2506. Count Pairs Of Similar Strings

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

Python Code
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class Solution:
    def similarPairs(self, words: List[str]) -> int:
        ans = 0
        cnt = Counter()
        for w in words:
            v = 0
            for c in w:
                v |= 1 << (ord(c) - ord("A"))
            ans += cnt[v]
            cnt[v] += 1
        return ans

Java Code
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class Solution {
    public int similarPairs(String[] words) {
        int ans = 0;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (var w : words) {
            int v = 0;
            for (int i = 0; i < w.length(); ++i) {
                v |= 1 << (w.charAt(i) - 'a');
            }
            ans += cnt.getOrDefault(v, 0);
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
        }
        return ans;
    }
}

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

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

TypeScript Code
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function similarPairs(words: string[]): number {
    let ans = 0;
    const cnt: Map<number, number> = new Map();
    for (const w of words) {
        let v = 0;
        for (let i = 0; i < w.length; ++i) {
            v |= 1 << (w.charCodeAt(i) - 'a'.charCodeAt(0));
        }
        ans += cnt.get(v) || 0;
        cnt.set(v, (cnt.get(v) || 0) + 1);
    }
    return ans;
}

Rust Code
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use std::collections::HashMap;

impl Solution {
    pub fn similar_pairs(words: Vec<String>) -> i32 {
        let mut ans = 0;
        let mut hash: HashMap<i32, i32> = HashMap::new();

        for w in words {
            let mut v = 0;

            for c in w.chars() {
                v |= 1 << ((c as u8) - b'a');
            }

            ans += hash.get(&v).unwrap_or(&0);
            *hash.entry(v).or_insert(0) += 1;
        }

        ans
    }
}