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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 

chevron_right


Output:

1 2

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29