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2744. Find Maximum Number of String Pairs

Description

You are given a 0-indexed array words consisting of distinct strings.

The string words[i] can be paired with the string words[j] if:

  • The string words[i] is equal to the reversed string of words[j].
  • 0 <= i < j < words.length.

Return the maximum number of pairs that can be formed from the array words.

Note that each string can belong in at most one pair.

 

Example 1:

Input: words = ["cd","ac","dc","ca","zz"]
Output: 2
Explanation: In this example, we can form 2 pair of strings in the following way:
- We pair the 0th string with the 2nd string, as the reversed string of word[0] is "dc" and is equal to words[2].
- We pair the 1st string with the 3rd string, as the reversed string of word[1] is "ca" and is equal to words[3].
It can be proven that 2 is the maximum number of pairs that can be formed.

Example 2:

Input: words = ["ab","ba","cc"]
Output: 1
Explanation: In this example, we can form 1 pair of strings in the following way:
- We pair the 0th string with the 1st string, as the reversed string of words[1] is "ab" and is equal to words[0].
It can be proven that 1 is the maximum number of pairs that can be formed.

Example 3:

Input: words = ["aa","ab"]
Output: 0
Explanation: In this example, we are unable to form any pair of strings.

 

Constraints:

  • 1 <= words.length <= 50
  • words[i].length == 2
  • words consists of distinct strings.
  • words[i] contains only lowercase English letters.

Solutions

Solution 1: Hash Table

We can use a hash table $cnt$ to store the number of occurrences of each reversed string in the array $words$.

We iterate through the array $words$. For each string $w$, we add the number of occurrences of its reversed string to the answer, then increment the count of $w$ by $1$.

Finally, we return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $words$.

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class Solution:
    def maximumNumberOfStringPairs(self, words: List[str]) -> int:
        cnt = Counter()
        ans = 0
        for w in words:
            ans += cnt[w[::-1]]
            cnt[w] += 1
        return ans

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class Solution {
    public int maximumNumberOfStringPairs(String[] words) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int ans = 0;
        for (var w : words) {
            int a = w.charAt(0) - 'a', b = w.charAt(1) - 'a';
            ans += cnt.getOrDefault(b << 5 | a, 0);
            cnt.merge(a << 5 | b, 1, Integer::sum);
        }
        return ans;
    }
}

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class Solution {
public:
    int maximumNumberOfStringPairs(vector<string>& words) {
        unordered_map<int, int> cnt;
        int ans = 0;
        for (auto& w : words) {
            int a = w[0] - 'a', b = w[1] - 'a';
            ans += cnt[b << 5 | a];
            cnt[a << 5 | b]++;
        }
        return ans;
    }
};

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func maximumNumberOfStringPairs(words []string) (ans int) {
    cnt := map[int]int{}
    for _, w := range words {
        a, b := int(w[0]-'a'), int(w[1]-'a')
        ans += cnt[b<<5|a]
        cnt[a<<5|b]++
    }
    return
}

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function maximumNumberOfStringPairs(words: string[]): number {
    const cnt: { [key: number]: number } = {};
    let ans = 0;
    for (const w of words) {
        const [a, b] = [w.charCodeAt(0) - 97, w.charCodeAt(w.length - 1) - 97];
        ans += cnt[(b << 5) | a] || 0;
        cnt[(a << 5) | b] = (cnt[(a << 5) | b] || 0) + 1;
    }
    return ans;
}

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