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

Difficulty Level : Easy
Last Updated : 21 May, 2021

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

Examples:

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. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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

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

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

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

Below is the implementation of the above approach:

C++

// 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;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to count the frequency of
  // minimum character
  static int countMinFreq(String s)
  {
 
    // Sort the string s
    char[] tempArray = s.toCharArray();
    Arrays.sort(tempArray);
    s = new String(tempArray);
 
    // Return the count with smallest
    // character
    int x = 0;
    for (int i = 0; i < s.length(); i++)
      if (s.charAt(i) == s.charAt(0))
        x++;
    return x;
  }
 
  // Function to count number of frequency
  // of smallest character of string arr1[]
  // is less than the string in arr2[]
  static void countLessThan(List<String> arr1,
                            List<String> arr2)
  {
 
    // To store the frequency of smallest
    // character in each string of arr2
    List<Integer> freq = new ArrayList<Integer>();
 
    // 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.add(f);
    }
 
    // Sort the frequency array
    Collections.sort(freq);
 
    // 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
      int it = upper_bound(freq, f);
 
      // Find the count such that
      // arr1[i] < arr2[j]
      int cnt = freq.size() - it;
 
      // Print the count
      System.out.print(cnt + " ");
    }
  }
 
  static int upper_bound(List<Integer> freq, int f)
  {
    int low = 0, high = freq.size() - 1;
 
    while (low < high) {
      int mid = (low + high) / 2;
      if (freq.get(mid) > f)
        high = mid;
      else
        low = mid + 1;
    }
 
    return (freq.get(low) < f) ? low++ : low;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    List<String> arr1, arr2;
    arr1 = Arrays.asList(new String[] { "yyy", "zz" });
    arr2 = Arrays.asList(
      new String[] { "x", "xx", "xxx", "xxxx" });
 
    // Function Call
    countLessThan(arr1, arr2);
  }
}
 
// This code is contributed by jithin.

Python3

# 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

Javascript

<script>
 
// JS program for the above approach
function upper_bound(freq, f)
{
    let low = 0, high = freq.length - 1;
 
    while (low < high) {
        let mid = Math.floor((low + high) / 2);
        if (freq[mid] > f)
            high = mid;
          else
            low = mid + 1;
    }
 
    return (freq[low] < f) ? low++ : low;
}
// Function to count the frequency of
// minimum character
function countMinFreq(s)
{
 
    // Sort the string s
    s = s.split('').sort().join('');
    // Return the count with smallest
    // character
    let x = 0;
    for(let i=0;i<s.length;i++){
        if(s[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[]
function countLessThan(arr1,arr2)
{
    // To store the frequency of smallest
    // character in each string of arr2
    let freq = [];
 
    // Traverse the arr2[]
    for (let i = 0;i< arr2.length;i++) {
 
        // Count the frequency of smallest
        // character in string s
        let f = countMinFreq(arr2[i]);
 
        // Append the frequency to freq[]
        freq.push(f);
    }
 
    // Sort the frequency array
    freq= freq.sort(function(a,b){return a-b});
    // Traverse the array arr1[]
    for (let i = 0;i< arr1.length;i++) {
 
        // Count the frequency of smallest
        // character in string s
        let f = countMinFreq(arr1[i]);
 
        // find the element greater than f
        let it = upper_bound(freq, f);
 
        // Find the count such that
        // arr1[i] < arr2[j]
        let cnt = freq.length
                  - (it);
 
        // Print the count
        document.write(cnt,' ');
    }
}
 
// Driver Code
let arr1, arr2;
arr1 = [ "yyy", "zz" ];
arr2 = [ "x", "xx", "xxx", "xxxx" ];
 
// Function Call
countLessThan(arr1, arr2);
 
</script>

Output:

1 2

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

My Personal Notes

Related Articles

Related Articles

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

C++

Java

Python3

Javascript