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:
- For each string in the array arr2[] count the frequency of 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 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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the frequency of// minimum characterint 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 Codeint 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 approachimport 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 approachfrom bisect import bisect_right as upper_bound# Function to count the frequency # of minimum characterdef 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 Codeif __name__ == '__main__': arr1 = ["yyy", "zz"] arr2 = [ "x", "xx", "xxx", "xxxx"] # Function Call countLessThan(arr1, arr2);# This code is contributed by Mohit Kumar |
1 2
Time Complexity: O(N + M*log M), where N and M is the length of given arrays respectively.
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