1170. Compare Strings by Frequency of the Smallest Character

Let’s define a function f(s) over a non-empty string s, which calculates the frequency of the smallest character in s. For example, if s = "dcce" then f(s) = 2 because the smallest character is "c" and its frequency is 2.

Now, given string arrays queries and words, return an integer array answer, where each answer[i] is the number of words such that f(queries[i]) < f(W), where W is a word in words.

 

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] are English lowercase letters.

 

Straight forward solution O(q * w)

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] result = new int[queries.length];
        for (int i = 0 ; i < queries.length; i ++){
            int score = s(queries[i]);
            
            for (String w : words){
                if (s(w) > score){
                    result[i] ++;
                }
            }
        }
        return result;
    }
    
    public static int s(String s) {
        char min = 'z';
        char[] ca = s.toCharArray();
        for(char c : ca){
            if (c < min){
                min = c;
            }
            if (c == 'a'){
                break;
            }
        }
        int r = 0;
        for(char c : ca){
            if (c == min)
                r ++;
        }
        return r;
    }
}

Optimized solution O (q + w)

since we have 1 <= queries[i].length, words[i].length <= 10

which means query will have only 10 possible values (limited), so for the result.

so we can use counting.

 

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        
		int[] freqCounter = new int[11];
		int[] result = new int[queries.length];
		for (int i = 0; i < words.length; i++) {
			int freq = s(words[i]);
			freqCounter[freq]++;
		}

		for (int i = 1; i < freqCounter.length; i++) {
			freqCounter[i] += freqCounter[i - 1];
		}
		for (int i = 0; i < queries.length; i++) {
			int freq = s(queries[i]);
			result[i] = words.length - freqCounter[freq];
		}
        
        return result;
    }
    
    public static int s(String s) {
        char min = 'z';
        int r = 0;
        char[] ca = s.toCharArray();
        for(char c : ca){
            if (c < min){
                r = 1;
                min = c;
            } else if(c == min) {
                r ++;
            }

        }
        return r;
    }
    
}

 

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