1170. Compare Strings by Frequency of the Smallest Character

Problem

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.

Solution (Java)

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] queriesMinFrequecies = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            queriesMinFrequecies[i] = computeLowestFrequency(queries[i]);
        }
        int[] wordsMinFrequecies = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            wordsMinFrequecies[i] = computeLowestFrequency(words[i]);
        }
        Arrays.sort(wordsMinFrequecies);
        int[] result = new int[queries.length];
        for (int i = 0; i < result.length; i++) {
            result[i] = search(wordsMinFrequecies, queriesMinFrequecies[i]);
        }
        return result;
    }

    private int search(int[] nums, int target) {
        int count = 0;
        for (int i = nums.length - 1; i >= 0; i--) {
            if (nums[i] > target) {
                count++;
            } else {
                break;
            }
        }
        return count;
    }

    private int computeLowestFrequency(String string) {
        char[] str = string.toCharArray();
        Arrays.sort(str);
        String sortedString = new String(str);
        int frequency = 1;
        for (int i = 1; i < sortedString.length(); i++) {
            if (sortedString.charAt(i) == sortedString.charAt(0)) {
                frequency++;
            } else {
                break;
            }
        }
        return frequency;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).