Frequency of smallest character in first sentence less than that of second sentence

Given two array of strings, arr1[] and arr2[], the task is to count the number of string in arr2[] whose frequency of smallest characters is less than frequency of smallest character for each string in arr1[].

Examples:

Input: arr1[] = {“cbd”}, arr2[] = {“zaaaz”}
Output: 1
Explanation:
Frequency of smallest characters in “cbd” is 1 which is less than the frequency of smallest characters in “zaaaz” which is 2.
Therefore the total count is 1 for string “cbd”.

Input: arr1[] = {“yyy”,”zz”}, arr2[] = {“x”,”xx”,”xxx”,”xxxx”}
Output: 1 2
Explanation:
1. frequency of smallest characters in “yyy” is 3 which is less than the frequency of smallest characters in “xxxx” which is 4.
Therefore the total count is 1 for string “yyy”.
2. frequency of smallest characters in “zz” is 2 which is less than the frequency of smallest characters in “xxx” and “xxxx” which is 3 and 4 respectively.
Therefore the total count is 2 for string “zz”.

Approach: This problem can be solved using Greedy Approach. Below are the steps:



  1. For each string in the array arr2[] count the frequency of smallest characters and store it in the array (say freq[]).
  2. Sort the frequency array freq[].
  3. Now for each string in the array arr1[] count the frequency of smallest characters in the string (say X).
  4. For each X, find the number of elements in greater than X in freq[] using Binary Search by using the approach discussed in this article.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the frequency of
// minimum character
int countMinFreq(string s)
{
  
    // Sort the string s
    sort(s.begin(), s.end());
  
    // Return the count with smallest
    // character
    return count(s.begin(), s.end(), s[0]);
}
  
// Function to count number of frequency
// of smallest character of string arr1[]
// is less than the string in arr2[]
void countLessThan(vector<string>& arr1,
                   vector<string>& arr2)
{
    // To store the frequency of smallest
    // character in each string of arr2
    vector<int> freq;
  
    // Traverse the arr2[]
    for (string s : arr2) {
  
        // Count the frequency of smallest
        // character in string s
        int f = countMinFreq(s);
  
        // Append the frequency to freq[]
        freq.push_back(f);
    }
  
    // Sort the frequency array
    sort(freq.begin(), freq.end());
  
    // Traverse the array arr1[]
    for (string s : arr1) {
  
        // Count the frequency of smallest
        // character in string s
        int f = countMinFreq(s);
  
        // find the element greater than f
        auto it = upper_bound(freq.begin(),
                              freq.end(), f);
  
        // Find the count such that
        // arr1[i] < arr2[j]
        int cnt = freq.size()
                  - (it - freq.begin());
  
        // Print the count
        cout << cnt << ' ';
    }
}
  
// Driver Code
int main()
{
  
    vector<string> arr1, arr2;
    arr1 = { "yyy", "zz" };
    arr2 = { "x", "xx", "xxx", "xxxx" };
  
    // Function Call
    countLessThan(arr1, arr2);
    return 0;
}

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# Python3 program for the above approach 


from bisect import bisect_right as upper_bound 


  


# Function to count the frequency  


# of minimum character 


def countMinFreq(s): 


  


    # Sort the string s 


    s = sorted(s) 


  


    # Return the count with smallest 


    # character 


    x = 0


    for i in s: 


        if i == s[0]: 


            x += 1


    return x 


  


# Function to count number of frequency 


# of smallest character of string arr1[] 


# is less than the string in arr2[] 


def countLessThan(arr1, arr2): 


      


    # To store the frequency of smallest 


    # character in each string of arr2 


    freq = [] 


  


    # Traverse the arr2[] 


    for s in arr2: 


  


        # Count the frequency of smallest 


        # character in string s 


        f = countMinFreq(s) 


  


        # Append the frequency to freq[] 


        freq.append(f) 


  


    # Sort the frequency array 


    feq = sorted(freq) 


  


    # Traverse the array arr1[] 


    for s in arr1: 


  


        # Count the frequency of smallest 


        # character in string s 


        f = countMinFreq(s); 


  


        # find the element greater than f 


        it = upper_bound(freq,f) 


  


        # Find the count such that 


        # arr1[i] < arr2[j] 


        cnt = len(freq)-it 


  


        # Print the count 


        print(cnt, end = " ") 


  


# Driver Code 


if __name__ == '__main__': 


  


    arr1 = ["yyy", "zz"] 


    arr2 = [ "x", "xx", "xxx", "xxxx"] 


  


    # Function Call 


    countLessThan(arr1, arr2); 


  


# This code is contributed by Mohit Kumar  



















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Output:


1 2





Time Complexity: O(N + M*log M), where N and M is the length of given arrays respectively.




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