Count of strings in the first array which are smaller than every string in the second array
Given two arrays A[] and B[] which consists of N and M strings respectively. A string S1 is said to be smaller than string S2 if the frequency of the smallest character in the S1 is smaller than the frequency of the smallest character in S2. The task is to count the number of strings in A[] which are smaller than B[i] for every i.
Examples:
Input: A[] = {“aaa”, “aa”, “bdc”}, B[] = {“cccch”, “cccd”}
Output: 3 2
“cccch” has frequency of the smallest character as 4, and all the strings
in A[] have frequencies of the smallest characters less than 4.
“cccd” has frequency of the smallest character as 3 and only “aa” and “bdc”
have frequencies of the smallest character less than 3.
Input: A[] = {“abca”, “jji”}, B[] = {“jhgkki”, “aaaa”, “geeks”}
Output: 0 2 1
A naive approach is to take every string in B[] and then count the number of strings in A[] which will satisfy the condition.
An efficient approach is to solve it using Binary Search and some pre-calculations as mentioned below:
- Initially count the frequency of the smallest character of every string and store in the vector/array.
- Sort the vector/array in ascending order.
- Now for every string in B[i], find the frequency of the smallest character.
- Using lower_bound function in C++, or by doing a binary search in the vector/array, find the count of numbers which has frequency smaller than the frequency of the smallest character for every B[i].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAX 26 // Function to count the number of smaller // strings in A[] for every string in B[] vector<int> findCount(string a[], string b[], int n, int m) { // Count the frequency of all characters int freq[MAX] = { 0 }; vector<int> smallestFreq; // Iterate for all possible strings in A[] for (int i = 0; i < n; i++) { string s = a[i]; memset(freq, 0, sizeof freq); // Increase the frequency of every character for (int j = 0; j < s.size(); j++) { freq[s[j] - 'a']++; } // Check for the smallest character's frequency for (int j = 0; j < MAX; j++) { // Get the smallest character frequency if (freq[j]) { // Insert it in the vector smallestFreq.push_back(freq[j]); break; } } } // Sort the count of all the frequency of the smallest // character in every string sort(smallestFreq.begin(), smallestFreq.end()); vector<int> ans; // Iterate for every string in B[] for (int i = 0; i < m; i++) { string s = b[i]; // Hash set every frequency 0 memset(freq, 0, sizeof freq); // Count the frequency of every character for (int j = 0; j < s.size(); j++) { freq[s[j] - 'a']++; } int frequency = 0; // Find the frequency of the smallest character for (int j = 0; j < MAX; j++) { if (freq[j]) { frequency = freq[j]; break; } } // Count the number of strings in A[] // which has the frequency of the smaller // character less than the frequency of the // smaller character of the string in B[] int ind = lower_bound(smallestFreq.begin(), smallestFreq.end(), frequency) - smallestFreq.begin(); // Store the answer ans.push_back(ind); } return ans; } // Function to print the answer void printAnswer(string a[], string b[], int n, int m) { // Get the answer vector<int> ans = findCount(a, b, n, m); // Print the number of strings // for every answer for (auto it : ans) { cout << it << " "; } } // Driver code int main() { string A[] = { "aaa", "aa", "bdc" }; string B[] = { "cccch", "cccd" }; int n = sizeof(A) / sizeof(A[0]); int m = sizeof(B) / sizeof(B[0]); printAnswer(A, B, n, m); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { static int MAX = 26; // Function to count the number of smaller // strings in A[] for every String in B[] static Vector<Integer> findCount(String a[], String b[], int n, int m) { // Count the frequency of all characters int []freq = new int[MAX]; Vector<Integer> smallestFreq = new Vector<Integer>(); // Iterate for all possible strings in A[] for (int i = 0; i < n; i++) { String s = a[i]; Arrays.fill(freq, 0); // Increase the frequency of every character for (int j = 0; j < s.length(); j++) { freq[s.charAt(j) - 'a']++; } // Check for the smallest character's frequency for (int j = 0; j < MAX; j++) { // Get the smallest character frequency if (freq[j] > 0) { // Insert it in the vector smallestFreq.add(freq[j]); break; } } } // Sort the count of all the frequency of // the smallest character in every string Collections.sort(smallestFreq); Vector<Integer> ans = new Vector<Integer>(); // Iterate for every String in B[] for (int i = 0; i < m; i++) { String s = b[i]; // Hash set every frequency 0 Arrays.fill(freq, 0); // Count the frequency of every character for (int j = 0; j < s.length(); j++) { freq[s.charAt(j) - 'a']++; } int frequency = 0; // Find the frequency of the smallest character for (int j = 0; j < MAX; j++) { if (freq[j] > 0) { frequency = freq[j]; break; } } // Count the number of strings in A[] // which has the frequency of the smaller // character less than the frequency of the // smaller character of the String in B[] int [] array = new int[smallestFreq.size()]; int k = 0; for(Integer val:smallestFreq) { array[k] = val; k++; } int ind = lower_bound(array, 0, smallestFreq.size(), frequency); // Store the answer ans.add(ind); } return ans; } static int lower_bound(int[] a, int low, int high, int element) { while(low < high) { int middle = low + (high - low) / 2; if(element > a[middle]) low = middle + 1; else high = middle; } return low; } // Function to print the answer static void printAnswer(String a[], String b[], int n, int m) { // Get the answer Vector<Integer> ans = findCount(a, b, n, m); // Print the number of strings // for every answer for (Integer it : ans) { System.out.print(it + " "); } } // Driver code public static void main(String[] args) { String A[] = { "aaa", "aa", "bdc" }; String B[] = { "cccch", "cccd" }; int n = A.length; int m = B.length; printAnswer(A, B, n, m); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach from bisect import bisect_left as lower_bound MAX = 26 # Function to count the number of smaller # strings in A for every in B def findCount(a, b, n, m): # Count the frequency of all characters freq=[0 for i in range(MAX)] smallestFreq=[] # Iterate for all possible strings in A for i in range(n): s = a[i] for i in range(MAX): freq[i]=0 # Increase the frequency of every character for j in range(len(s)): freq[ord(s[j]) - ord('a')]+= 1 # Check for the smallest character's frequency for j in range(MAX): # Get the smallest character frequency if (freq[j]): # Insert it in the vector smallestFreq.append(freq[j]) break # Sort the count of all the frequency of the smallest # character in every string smallestFreq=sorted(smallestFreq) ans=[] # Iterate for every in B for i in range(m): s = b[i] # Hash set every frequency 0 for i in range(MAX): freq[i]=0 # Count the frequency of every character for j in range(len(s)): freq[ord(s[j]) - ord('a')]+= 1 frequency = 0 # Find the frequency of the smallest character for j in range(MAX): if (freq[j]): frequency = freq[j] break # Count the number of strings in A # which has the frequency of the smaller # character less than the frequency of the # smaller character of the in B ind = lower_bound(smallestFreq,frequency) # Store the answer ans.append(ind) return ans # Function to print the answer def printAnswer(a, b, n, m): # Get the answer ans = findCount(a, b, n, m) # Print the number of strings # for every answer for it in ans: print(it,end=" ") # Driver code A = ["aaa", "aa", "bdc"] B = ["cccch", "cccd"] n = len(A) m = len(B) printAnswer(A, B, n, m) # This code is contributed by mohit kumar 29 |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; public class GFG { static int MAX = 26; // Function to count the number of smaller // strings in A[] for every String in B[] static List<int> findCount(String []a, String []b, int n, int m) { // Count the frequency of all characters int []freq = new int[MAX]; List<int> smallestFreq = new List<int>(); // Iterate for all possible strings in A[] for (int i = 0; i < n; i++) { String s = a[i]; for (int l = 0; l < freq.Length; l++) freq[l]=0; // Increase the frequency of every character for (int j = 0; j < s.Length; j++) { freq[s[j] - 'a']++; } // Check for the smallest character's frequency for (int j = 0; j < MAX; j++) { // Get the smallest character frequency if (freq[j] > 0) { // Insert it in the vector smallestFreq.Add(freq[j]); break; } } } // Sort the count of all the frequency of // the smallest character in every string smallestFreq.Sort(); List<int> ans = new List<int>(); // Iterate for every String in B[] for (int i = 0; i < m; i++) { String s = b[i]; // Hash set every frequency 0 for (int l = 0; l < freq.Length; l++) freq[l]=0; // Count the frequency of every character for (int j = 0; j < s.Length; j++) { freq[s[j] - 'a']++; } int frequency = 0; // Find the frequency of the smallest character for (int j = 0; j < MAX; j++) { if (freq[j] > 0) { frequency = freq[j]; break; } } // Count the number of strings in A[] // which has the frequency of the smaller // character less than the frequency of the // smaller character of the String in B[] int [] array = new int[smallestFreq.Count]; int k = 0; foreach (int val in smallestFreq) { array[k] = val; k++; } int ind = lower_bound(array, 0, smallestFreq.Count, frequency); // Store the answer ans.Add(ind); } return ans; } static int lower_bound(int[] a, int low, int high, int element) { while(low < high) { int middle = low + (high - low) / 2; if(element > a[middle]) low = middle + 1; else high = middle; } return low; } // Function to print the answer static void printAnswer(String []a, String []b, int n, int m) { // Get the answer List<int> ans = findCount(a, b, n, m); // Print the number of strings // for every answer foreach (int it in ans) { Console.Write(it + " "); } } // Driver code public static void Main(String[] args) { String []A = { "aaa", "aa", "bdc" }; String []B = { "cccch", "cccd" }; int n = A.Length; int m = B.Length; printAnswer(A, B, n, m); } } // This code is contributed by Princi Singh |
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Output:
3 2
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