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3838. Weighted Word Mapping

Description

You are given an array of strings words, where each string represents a word containing lowercase English letters.

You are also given an integer array weights of length 26, where weights[i] represents the weight of the ith lowercase English letter.

The weight of a word is defined as the sum of the weights of its characters.

For each word, take its weight modulo 26 and map the result to a lowercase English letter using reverse alphabetical order (0 -> 'z', 1 -> 'y', ..., 25 -> 'a').

Return a string formed by concatenating the mapped characters for all words in order.

Β 

Example 1:

Input: words = ["abcd","def","xyz"], weights = [5,3,12,14,1,2,3,2,10,6,6,9,7,8,7,10,8,9,6,9,9,8,3,7,7,2]

Output: "rij"

Explanation:

  • The weight of "abcd" is 5 + 3 + 12 + 14 = 34. The result modulo 26 is 34 % 26 = 8, which maps to 'r'.
  • The weight of "def" is 14 + 1 + 2 = 17. The result modulo 26 is 17 % 26 = 17, which maps to 'i'.
  • The weight of "xyz" is 7 + 7 + 2 = 16. The result modulo 26 is 16 % 26 = 16, which maps to 'j'.

Thus, the string formed by concatenating the mapped characters is "rij".

Example 2:

Input: words = ["a","b","c"], weights = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Output: "yyy"

Explanation:

Each word has weight 1. The result modulo 26 is 1 % 26 = 1, which maps to 'y'.

Thus, the string formed by concatenating the mapped characters is "yyy".

Example 3:

Input: words = ["abcd"], weights = [7,5,3,4,3,5,4,9,4,2,2,7,10,2,5,10,6,1,2,2,4,1,3,4,4,5]

Output: "g"

Explanation:​​​​​​​

The weight of "abcd" is 7 + 5 + 3 + 4 = 19. The result modulo 26 is 19 % 26 = 19, which maps to 'g'.

Thus, the string formed by concatenating the mapped characters is "g".

Β 

Constraints:

  • 1 <= words.length <= 100
  • 1 <= words[i].length <= 10
  • weights.length == 26
  • 1 <= weights[i] <= 100
  • words[i] consists of lowercase English letters.

Solution 1: Simulation

We iterate through each word \(w\) in \(\textit{words}\), calculate its weight \(s\), which is the sum of the weights of all characters in the word. Then we calculate \(s\) modulo 26, map the result to a lowercase English letter, and finally concatenate all the mapped characters and return.

The time complexity is \(O(L)\), where \(L\) is the sum of the lengths of all words in \(\textit{words}\). The space complexity is \(O(W)\), where \(W\) is the length of \(\textit{words}\).

Solution 1

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class Solution:
    def mapWordWeights(self, words: List[str], weights: List[int]) -> str:
        ans = []
        for w in words:
            s = sum(weights[ord(c) - ord('a')] for c in w)
            ans.append(ascii_lowercase[25 - s % 26])
        return ''.join(ans)

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class Solution {
    public String mapWordWeights(String[] words, int[] weights) {
        var ans = new StringBuilder();
        for (var w : words) {
            int s = 0;
            for (char c : w.toCharArray()) {
                s = (s + weights[c - 'a']) % 26;
            }
            ans.append((char) ('a' + (25 - s)));
        }
        return ans.toString();
    }
}

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class Solution {
public:
    string mapWordWeights(vector<string>& words, vector<int>& weights) {
        string ans;
        for (const string& w : words) {
            int s = 0;
            for (char c : w) {
                s = (s + weights[c - 'a']) % 26;
            }
            ans.push_back(char('a' + (25 - s)));
        }
        return ans;
    }
};

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func mapWordWeights(words []string, weights []int) string {
    ans := make([]byte, 0, len(words))
    for _, w := range words {
        s := 0
        for i := 0; i < len(w); i++ {
            s = (s + weights[int(w[i]-'a')]) % 26
        }
        ans = append(ans, byte('a'+(25-s)))
    }
    return string(ans)
}

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function mapWordWeights(words: string[], weights: number[]): string {
    const ans: string[] = [];
    for (const w of words) {
        let s = 0;
        for (const c of w) {
            s = (s + weights[c.charCodeAt(0) - 97]) % 26;
        }
        ans.push(String.fromCharCode(97 + (25 - s)));
    }
    return ans.join('');
}

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