Question

2559. Count Vowel Strings in Ranges

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Description

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

Each query queries[i] = [li, ri] asks us to find the number of strings present in the range li to ri (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 <= li <= ri < 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.

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]

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;
  }
};

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));
}

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:
  def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
    vowels = set("aeiou")
    s = list(
      accumulate(
        (int(w[0] in vowels and w[-1] in vowels) for w in words), initial=0
      )
    )
    return [s[r + 1] - s[l] for l, r in queries]

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

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

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

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