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### by Sumit Chourasia | May 04, 2020 | Category :coding | Tags : stringडेटा-संरचनाleetcodealgorithmआसान #### Compare Strings by Frequency of the Smallest Character - मिनी टीवी

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: 

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.``````

Solution:

``````using System;
using System.Collections.Generic;
using System.Text;

{
class NumSmallerByFrequencySol
{
public void execute()
{
var queries = new string[] { "bbb", "cc" };
var words = new string[]{"a", "aa", "aaa", "aaaa"};
var res = NumSmallerByFrequency(queries, words);
}

public int[] NumSmallerByFrequency(string[] queries, string[] words)
{
int[] queries_count = new int[queries.Length];
int[] words_count = new int[words.Length];
int[] ans = new int[queries.Length];

for (int i = 0; i < queries.Length; i++)
{
queries_count[i] = -1;
ans[i] = 0;
}

for (int i = 0; i < words.Length; i++)
{
words_count[i] = -1;
}

for(int i = 0; i < queries.Length; i++)
{
if (queries_count[i] == -1)
{
queries_count[i] = CalculateCount(queries[i]);
}

for (int j = 0; j < words.Length; j++)
{

if(words_count[j] == -1)
{
words_count[j] = CalculateCount(words[j]);
}

if (queries_count[i] < words_count[j])
{
ans[i]++;
}
}
}

return ans;
}

private int CalculateCount(string v)
{
if (v == "") return 0;

char minChar = v;
int count = 1;
for(int i= 1; i < v.Length; i++)
{
if (v[i] < minChar)
{
minChar = v[i];
count = 1;
}
else if(v[i] == minChar)
{
count++;
}
}
return count;
}
}
}
``````

Time Complexity: Approx. O(n^2) Considering the size of given words are finite.

Space Complexity: O(n)