2506

2506. Count Pairs Of Similar Strings (Easy)

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.

Companies: Oracle

Related Topics:
Array, Hash Table, String

Similar Questions:

• Sort Characters By Frequency (Medium)
• Count the Number of Consistent Strings (Easy)
• Number of Good Paths (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/count-pairs-of-similar-strings
// Author: github.com/lzl124631x
// Time: O(NW)
// Space: O(N)
class Solution {
public:
    int similarPairs(vector<string>& A) {
        unordered_map<int, int> m;
        int ans = 0;
        for (auto &s : A) {
            int bits = 0;
            for (char c : s) bits |= (1 << (c - 'a'));
            ans += m[bits];
            m[bits]++;
        }
        return ans;
    }
};