Question

1170. Compare Strings by Frequency of the Smallest Character

中文文档

Description

Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest character is 'c', which has a frequency of 2.

You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words.

Return _an integer array _answer_, where each _answer[i]_ is the answer to the _ith_ query_.

 

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

 

Constraints:

• 1 <= queries.length <= 2000
• 1 <= words.length <= 2000
• 1 <= queries[i].length, words[i].length <= 10
• queries[i][j], words[i][j] consist of lowercase English letters.

Solutions

Solution 1: Sorting + Binary Search

First, according to the problem description, we implement a function $f(s)$, which returns the frequency of the smallest letter in the string $s$ in lexicographical order.

Next, we calculate $f(w)$ for each string $w$ in $words$, sort them, and store them in an array $nums$.

Then, we traverse each string $q$ in $queries$, and binary search in $nums$ for the first position $i$ that is greater than $f(q)$. Then, the elements at index $i$ and after in $nums$ all satisfy $f(q) < f(W)$, so the answer to the current query is $n - i$.

The time complexity is $O((n + q) \times M)$, and the space complexity is $O(n)$. Here, $n$ and $q$ are the lengths of the arrays $words$ and $queries$ respectively, and $M$ is the maximum length of the strings.

class Solution:
  def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
    def f(s: str) -> int:
      cnt = Counter(s)
      return next(cnt[c] for c in ascii_lowercase if cnt[c])

    n = len(words)
    nums = sorted(f(w) for w in words)
    return [n - bisect_right(nums, f(q)) for q in queries]

class Solution {
  public int[] numSmallerByFrequency(String[] queries, String[] words) {
    int n = words.length;
    int[] nums = new int[n];
    for (int i = 0; i < n; ++i) {
      nums[i] = f(words[i]);
    }
    Arrays.sort(nums);
    int m = queries.length;
    int[] ans = new int[m];
    for (int i = 0; i < m; ++i) {
      int x = f(queries[i]);
      int l = 0, r = n;
      while (l < r) {
        int mid = (l + r) >> 1;
        if (nums[mid] > x) {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      ans[i] = n - l;
    }
    return ans;
  }

  private int f(String s) {
    int[] cnt = new int[26];
    for (int i = 0; i < s.length(); ++i) {
      ++cnt[s.charAt(i) - 'a'];
    }
    for (int x : cnt) {
      if (x > 0) {
        return x;
      }
    }
    return 0;
  }
}

class Solution {
public:
  vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
    auto f = [](string s) {
      int cnt[26] = {0};
      for (char c : s) {
        cnt[c - 'a']++;
      }
      for (int x : cnt) {
        if (x) {
          return x;
        }
      }
      return 0;
    };
    int n = words.size();
    int nums[n];
    for (int i = 0; i < n; i++) {
      nums[i] = f(words[i]);
    }
    sort(nums, nums + n);
    vector<int> ans;
    for (auto& q : queries) {
      int x = f(q);
      ans.push_back(n - (upper_bound(nums, nums + n, x) - nums));
    }
    return ans;
  }
};

func numSmallerByFrequency(queries []string, words []string) (ans []int) {
  f := func(s string) int {
    cnt := [26]int{}
    for _, c := range s {
      cnt[c-'a']++
    }
    for _, x := range cnt {
      if x > 0 {
        return x
      }
    }
    return 0
  }
  n := len(words)
  nums := make([]int, n)
  for i, w := range words {
    nums[i] = f(w)
  }
  sort.Ints(nums)
  for _, q := range queries {
    x := f(q)
    ans = append(ans, n-sort.SearchInts(nums, x+1))
  }
  return
}

function numSmallerByFrequency(queries: string[], words: string[]): number[] {
  const f = (s: string): number => {
    const cnt = new Array(26).fill(0);
    for (const c of s) {
      cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }
    return cnt.find(x => x > 0);
  };
  const nums = words.map(f).sort((a, b) => a - b);
  const ans: number[] = [];
  for (const q of queries) {
    const x = f(q);
    let l = 0,
      r = nums.length;
    while (l < r) {
      const mid = (l + r) >> 1;
      if (nums[mid] > x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    ans.push(nums.length - l);
  }
  return ans;
}