Count Disjoint String Pairs

Count unique index pairs (i<j) of lowercase strings that share no common character; bitmask each word's 26-letter set and AND the masks. Count Disjoint String Pairs Count unique index pairs (i<j) of lowercase strings that share no common character; bitmask each word's 26-letter set and AND the masks. SWE bitmask string hashmap medium Frequency Low Last asked 2026-08-02 Stage phone-screen · onsite-coding Count Disjoint String Pairs Given an array of lowercase strings words, return the number of unique index pairs (i, j) such that i < j and the two strings share no common characters. Two strings are disjoint if there is no character that appears in both strings. Examples Example 1: Input: words = ["apple","banana","peach","kiwi"] Output: 3 Explanation: Valid pairs are ["apple","kiwi"], ["banana","kiwi"], and ["peach","kiwi"]. Example 2: Input: words = ["abc","def","gh"] Output: 3 Example 3: Input: words = ["a","aa","aaa"] Output: 0 Constraints 1 <= words.length <= 1000 1 <= words[i].length <= 1000 words[i] consists of lowercase English letters. Notes A phone-screen variant returns the actual index pairs instead of just a count: e.g. ["a", "ab", "b"] returns [[0, 2]] (only "a" and "b" share no character). Building the 26-bit mask per word and AND-ing every pair runs in O(n^2 * 26); this O(n^2) solution has been accepted without an optimization follow-up.