Count of strings to be concatenated with a character having frequency greater than sum of others
Given an array arr[] containing N strings, the task is to find the maximum number of strings that can be concatenated such that one character has a frequency greater than the sum of the frequencies of all others.
Example:
Input: arr[]: {“qpr, “pppsp, “t”}
Output: 3
Explanation: Concatenate all 3 strings to get: “aprpppspt”. The frequency of character ‘p’ is 5 and sum of all other frequencies is 4.Input: arr[]: {“bcdba”, “abaa”, “acc”, “abcbc”}
Output: 2
Approach: To solve this problem follow the below steps:
- Iterate for all characters, i.e. from ‘a’ to ‘z’ and in each iteration find the net frequency of that character in all strings. Here net frequency can be calculated by subtracting all other frequencies from it, which means if the net frequency is greater than 0 then that character’s frequency is more than the sum of all other frequencies. Store these frequencies in a vector v.
- Now, sort v in decreasing order.
- And then find for each character, the maximum number of strings that can be combined, so that the frequency of that character is greater than the sum of other frequencies.
- Return the maximum possible answer.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach #include <bits/stdc++.h> using namespace std; // Function of find the // frequency of all characters after // reducing the sum of // other frequencies in strings vector< int > frequency(vector<string>& arr, int ch) { // Vector to store frequency vector< int > v; // Iterate over the array of strings for ( int i = 0; i < arr.size(); i++) { string s = arr[i]; // Variable to store frequencies int net_freq = 0; // Iterate over the string for ( auto x : s) { // If x is equal // to current character // increment net_freq by 1 if (x == ch) net_freq++; // Else decrement net_freq by 1 else net_freq--; } // After the iteration of string // store the frequency in vector v.push_back(net_freq); } return v; } // Function to find // the longest string that // can be made from // a given vector of strings int longestConcatenatedStr( vector<string>& arr) { // Variable to store maximum count int mx = 0; // Iterate over all alphabets for ( char ch = 'a' ; ch <= 'z' ; ch++) { // Vector to store the // net_frequency of character // ch after reducing // the sum of all other // frequencies in all strings vector< int > v = frequency(arr, ch); // Sort the vector in decreasing order sort(v.begin(), v.end(), greater< int >()); // Variable to store answer int ans = 0; int sum = 0; for ( auto x : v) { sum += x; // If sum is greater than 0 // increment ans by 1 if (sum > 0) { ans++; } } // Keep track of the maximum one mx = max(mx, ans); } // Return the maximum value return mx; } // Driver Code int main() { vector<string> arr = { "abac" , "bacbc" , "aacab" }; cout << longestConcatenatedStr(arr); return 0; } |
Java
// Java implementation for the above approach import java.util.*; class GFG { static String[] arr = { "abac" , "bacbc" , "aacab" }; // Function of find the // frequency of all characters after // reducing the sum of // other frequencies in Strings static Vector<Integer> frequency( int ch) { // Vector to store frequency Vector<Integer> v = new Vector<>(); // Iterate over the array of Strings for ( int i = 0 ; i < arr.length; i++) { String s = arr[i]; // Variable to store frequencies int net_freq = 0 ; // Iterate over the String for ( char x : s.toCharArray()) { // If x is equal // to current character // increment net_freq by 1 if (x == ch) net_freq++; // Else decrement net_freq by 1 else net_freq--; } // After the iteration of String // store the frequency in vector v.add(net_freq); } return v; } // Function to find // the longest String that // can be made from // a given vector of Strings static int longestConcatenatedStr() { // Variable to store maximum count int mx = 0 ; // Iterate over all alphabets for ( char ch = 'a' ; ch <= 'z' ; ch++) { // Vector to store the // net_frequency of character // ch after reducing // the sum of all other // frequencies in all Strings Vector<Integer> v = frequency(ch); // Sort the vector in decreasing order Collections.sort(v); Collections.reverse(v); // Variable to store answer int ans = 0 ; int sum = 0 ; for ( int x : v) { sum += x; // If sum is greater than 0 // increment ans by 1 if (sum > 0 ) { ans++; } } // Keep track of the maximum one mx = Math.max(mx, ans); } // Return the maximum value return mx; } // Driver Code public static void main(String[] args) { System.out.print(longestConcatenatedStr()); } } // This code is contributed by Rajput-Ji |
Python3
# python implementation for the above approach # Function of find the # frequency of all characters after # reducing the sum of # other frequencies in strings def frequency(arr, ch): # Vector to store frequency v = [] # Iterate over the array of strings for i in range ( 0 , len (arr)): s = arr[i] # Variable to store frequencies net_freq = 0 # Iterate over the string for x in s: # If x is equal # to current character # increment net_freq by 1 if (x = = ch): net_freq + = 1 # Else decrement net_freq by 1 else : net_freq - = 1 # After the iteration of string # store the frequency in vector v.append(net_freq) return v # Function to find # the longest string that # can be made from # a given vector of strings def longestConcatenatedStr(arr): # Variable to store maximum count mx = 0 # Iterate over all alphabets for ch in range ( 0 , 26 ): # Vector to store the # net_frequency of character # ch after reducing # the sum of all other # frequencies in all strings v = frequency(arr, chr (ch + ord ( 'a' ))) # Sort the vector in decreasing order v.sort(reverse = True ) # Variable to store answer ans = 0 sum = 0 for x in v: sum + = x # If sum is greater than 0 # increment ans by 1 if ( sum > 0 ): ans + = 1 # Keep track of the maximum one mx = max (mx, ans) # Return the maximum value return mx # Driver Code if __name__ = = "__main__" : arr = [ "abac" , "bacbc" , "aacab" ] print (longestConcatenatedStr(arr)) # This code is contributed by rakeshsahni |
C#
using System; using System.Collections.Generic; // C# implementation for the above approach class GFG { // Function of find the // frequency of all characters after // reducing the sum of // other frequencies in strings static int [] frequency( string [] arr, int ch) { // List to store frequency List< int > v = new List< int >(); // Iterate over the array of strings for ( int i = 0; i < arr.Length; i++) { string s = arr[i]; // Variable to store frequencies int net_freq = 0; // Iterate over the string for ( int j = 0; j < s.Length; j++) { // If x is equal // to current character // increment net_freq by 1 char x = s[j]; if (x == ch) net_freq++; // Else decrement net_freq by 1 else net_freq--; } // After the iteration of string // store the frequency in vector v.Add(net_freq); } int [] u = v.ToArray(); return u; } // Function to find // the longest string that // can be made from // a given vector of strings static int longestConcatenatedStr( string [] arr) { // Variable to store maximum count int mx = 0; // Iterate over all alphabets for ( char ch = 'a' ; ch <= 'z' ; ch++) { // Array to store the // net_frequency of character // ch after reducing // the sum of all other // frequencies in all strings int [] v = frequency(arr, ch); // Sort the vector in decreasing order Array.Sort(v); Array.Reverse(v); // Variable to store answer int ans = 0; int sum = 0; for ( int i = 0; i < v.Length; i++) { int x = v[i]; sum += x; // If sum is greater than 0 // increment ans by 1 if (sum > 0) { ans++; } } // Keep track of the maximum one mx = Math.Max(mx, ans); } // Return the maximum value return mx; } static void Main() { string [] arr = { "abac" , "bacbc" , "aacab" }; Console.Write(longestConcatenatedStr(arr)); } } // This code is contributed by garg28harsh. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function of find the // frequency of all characters after // reducing the sum of // other frequencies in strings function frequency(arr, ch) { // Vector to store frequency let v = []; // Iterate over the array of strings for (let i = 0; i < arr.length; i++) { let s = arr[i]; // Variable to store frequencies let net_freq = 0; // Iterate over the string for (let x of s) { // If x is equal // to current character // increment net_freq by 1 if (x == ch) net_freq++; // Else decrement net_freq by 1 else net_freq--; } // After the iteration of string // store the frequency in vector v.push(net_freq); } return v; } // Function to find // the longest string that // can be made from // a given vector of strings function longestConcatenatedStr( arr) { // Variable to store maximum count let mx = 0; // Iterate over all alphabets for (let ch = 'a' ; ch <= 'z' ; ch++) { // Vector to store the // net_frequency of character // ch after reducing // the sum of all other // frequencies in all strings let v = frequency(arr, ch); // Sort the vector in decreasing order v.sort( function (a, b) { return b - a }) // Variable to store answer let ans = 0; let sum = 0; for (let x of v) { sum += x; // If sum is greater than 0 // increment ans by 1 if (sum > 0) { ans++; } } // Keep track of the maximum one mx = Math.max(mx, ans); } // Return the maximum value return mx; } // Driver Code let arr = [ "abac" , "bacbc" , "aacab" ]; document.write(longestConcatenatedStr(arr)); // This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(NlogN)
Auxiliary Space: O(N)