2559. Count Vowel Strings in Ranges

Description

You are given a 0-indexed array of strings words and a 2D array of integers queries.

Each query queries[i] = [l_i, r_i] asks us to find the number of strings present in the range l_i to r_i (both inclusive) of words that start and end with a vowel.

Return an array ans of size queries.length, where ans[i] is the answer to the i^th query.

Note that the vowel letters are 'a', 'e', 'i', 'o', and 'u'.

Example 1:

Input: words = ["aba","bcb","ece","aa","e"], queries = [[0,2],[1,4],[1,1]]
Output: [2,3,0]
Explanation: The strings starting and ending with a vowel are "aba", "ece", "aa" and "e".
The answer to the query [0,2] is 2 (strings "aba" and "ece").
to query [1,4] is 3 (strings "ece", "aa", "e").
to query [1,1] is 0.
We return [2,3,0].

Example 2:

Input: words = ["a","e","i"], queries = [[0,2],[0,1],[2,2]]
Output: [3,2,1]
Explanation: Every string satisfies the conditions, so we return [3,2,1].

Constraints:

1 <= words.length <= 10⁵
1 <= words[i].length <= 40
words[i] consists only of lowercase English letters.
sum(words[i].length) <= 3 * 10⁵
1 <= queries.length <= 10⁵
0 <= l_i <= r_i < words.length

Solutions

Solution 1: Preprocessing + Binary Search

We can preprocess all the indices of the strings that start and end with a vowel, and record them in order in the array $nums$.

Next, we iterate through each query $(l, r)$, and use binary search to find the first index $i$ in $nums$ that is greater than or equal to $l$, and the first index $j$ that is greater than $r$. Therefore, the answer to the current query is $j - i$.

The time complexity is $O(n + m \times \log n)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of the arrays $words$ and $queries$, respectively.

Solution 2: Prefix Sum

We can create a prefix sum array $s$ of length $n+1$, where $s[i]$ represents the number of strings that start and end with a vowel in the first $i$ strings of the array $words$. Initially, $s[0] = 0$.

Next, we iterate through the array $words$. If the current string starts and ends with a vowel, then $s[i+1] = s[i] + 1$, otherwise $s[i+1] = s[i]$.

Finally, we iterate through each query $(l, r)$. Therefore, the answer to the current query is $s[r+1] - s[l]$.

The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of the arrays $words$ and $queries$, respectively.

class Solution {
    private List<Integer> nums = new ArrayList<>();

    public int[] vowelStrings(String[] words, int[][] queries) {
        Set<Character> vowels = Set.of('a', 'e', 'i', 'o', 'u');
        for (int i = 0; i < words.length; ++i) {
            char a = words[i].charAt(0), b = words[i].charAt(words[i].length() - 1);
            if (vowels.contains(a) && vowels.contains(b)) {
                nums.add(i);
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            ans[i] = search(r + 1) - search(l);
        }
        return ans;
    }

    private int search(int x) {
        int l = 0, r = nums.size();
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums.get(mid) >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

class Solution {
public:
    vector<int> vowelStrings(vector<string>& words, vector<vector<int>>& queries) {
        unordered_set<char> vowels = {'a', 'e', 'i', 'o', 'u'};
        vector<int> nums;
        for (int i = 0; i < words.size(); ++i) {
            char a = words[i][0], b = words[i].back();
            if (vowels.count(a) && vowels.count(b)) {
                nums.push_back(i);
            }
        }
        vector<int> ans;
        for (auto& q : queries) {
            int l = q[0], r = q[1];
            int cnt = upper_bound(nums.begin(), nums.end(), r) - lower_bound(nums.begin(), nums.end(), l);
            ans.push_back(cnt);
        }
        return ans;
    }
};

class Solution:
    def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
        vowels = set("aeiou")
        nums = [i for i, w in enumerate(words) if w[0] in vowels and w[-1] in vowels]
        return [bisect_right(nums, r) - bisect_left(nums, l) for l, r in queries]

func vowelStrings(words []string, queries [][]int) []int {
	vowels := map[byte]bool{'a': true, 'e': true, 'i': true, 'o': true, 'u': true}
	nums := []int{}
	for i, w := range words {
		if vowels[w[0]] && vowels[w[len(w)-1]] {
			nums = append(nums, i)
		}
	}
	ans := make([]int, len(queries))
	for i, q := range queries {
		l, r := q[0], q[1]
		ans[i] = sort.SearchInts(nums, r+1) - sort.SearchInts(nums, l)
	}
	return ans
}

function vowelStrings(words: string[], queries: number[][]): number[] {
    const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
    const nums: number[] = [];
    for (let i = 0; i < words.length; ++i) {
        if (vowels.has(words[i][0]) && vowels.has(words[i][words[i].length - 1])) {
            nums.push(i);
        }
    }
    const search = (x: number): number => {
        let l = 0,
            r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    return queries.map(([l, r]) => search(r + 1) - search(l));
}

2559 - Count Vowel Strings in Ranges

2559. Count Vowel Strings in Ranges

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2559. Count Vowel Strings in Ranges

Description

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