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Sort array of strings after sorting each string after removing characters whose frequencies are not a powers of 2
  • Last Updated : 19 Apr, 2021

Given an array arr[] consisting of N strings, the task is to sort the array in ascending order after modifying each string by removing all characters that are not perfect power of 2 and then sort the modified string in decreasing order.

Examples:

Input: arr[] = {“aaacbb”, “geeks”, “aaa”}
Output: cbb skgee
Explanation:
Following are the modified strings in the array arr[]:

  1. For the string “aaacbb”: The frequency of a is not the power of 2. Therefore, removing ‘a’ modifies the string to “cbb”. Now, sorting the string “cbb” in increasing order of frequency modifies the string to “cbb”.
  2. For the string “aaacbb”: The frequency of every character is a power of 2. Now, sorting the string “geeks” in increasing order of frequency modifies the string to “skgee”.
  3. For the string “aaacbb”: The frequency of a is not the power of 2. Therefore, removing ‘a’ modifies the string to “”.

Therefore, sorting the above strings in increasing order gives {“cbb”, “skgee”}.

Input: S[] = {“c”, “a”, “b”}
Output: a b c



Approach: The given problem can be solved by using the Hashing to store the frequencies of all the characters for each string and then perform the given operations. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if N is power of
// 2 or not
bool isPowerOfTwo(int n)
{
    // Base Case
    if (n == 0)
        return false;
 
    // Return true if N is power of 2
    return (ceil(log2(n))
            == floor(log2(n)));
}
 
// Function to print array of strings
// in ascending order
void printArray(vector<string> res)
{
    // Sort strings in ascending order
    sort(res.begin(), res.end());
 
    // Print the array
    for (int i = 0; i < res.size(); i++) {
        cout << res[i] << " ";
    }
}
 
// Function to sort the strings after
// modifying each string according to
// the given conditions
void sortedStrings(string S[], int N)
{
    // Store the frequency of each
    // characters of the string
    unordered_map<char, int> freq;
 
    // Stores the required
    // array of strings
    vector<string> res;
 
    // Traverse the array of strings
    for (int i = 0; i < N; i++) {
 
        // Temporary string
        string st = "";
 
        // Stores frequency of each
        // alphabet of the string
        for (int j = 0;
             j < S[i].size(); j++) {
 
            // Update frequency of S[i][j]
            freq[S[i][j]]++;
        }
 
        // Traverse the map freq
        for (auto i : freq) {
 
            // Check if the frequency
            // of i.first is a power of 2
            if (isPowerOfTwo(i.second)) {
 
                // Update string st
                for (int j = 0;
                     j < i.second; j++) {
                    st += i.first;
                }
            }
        }
 
        // Clear the map
        freq.clear();
 
        // Null string
        if (st.size() == 0)
            continue;
 
        // Sort the string in
        // descending order
        sort(st.begin(), st.end(),
             greater<char>());
 
        // Update res
        res.push_back(st);
    }
 
    // Print the array of strings
    printArray(res);
}
 
// Driver Code
int main()
{
    string arr[] = { "aaacbb", "geeks", "aaa" };
    int N = sizeof(arr) / sizeof(arr[0]);
    sortedStrings(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to check if N is power of
// 2 or not
static boolean isPowerOfTwo(int n)
{
     
    // Base Case
    if (n == 0)
        return false;
 
    // Return true if N is power of 2
    return (Math.ceil(Math.log(n) / Math.log(2)) ==
           Math.floor(Math.log(n) / Math.log(2)));
}
 
// Function to print array of strings
// in ascending order
static void printArray(ArrayList<String> res)
{
     
    // Sort strings in ascending order
    Collections.sort(res);
 
    // Print the array
    for(int i = 0; i < res.size(); i++)
    {
        System.out.print(res.get(i) + " ");
    }
}
 
// Function to sort the strings after
// modifying each string according to
// the given conditions
static void sortedStrings(String S[], int N)
{
     
    // Store the frequency of each
    // characters of the string
    HashMap<Character, Integer> freq = new HashMap<>();
 
    // Stores the required
    // array of strings
    ArrayList<String> res = new ArrayList<>();
 
    // Traverse the array of strings
    for(int i = 0; i < N; i++)
    {
         
        // Temporary string
        String st = "";
 
        // Stores frequency of each
        // alphabet of the string
        for(int j = 0; j < S[i].length(); j++)
        {
             
            // Update frequency of S[i][j]
            freq.put(S[i].charAt(j),
                freq.getOrDefault(S[i].charAt(j), 0) + 1);
        }
 
        // Traverse the map freq
        for(char ch : freq.keySet())
        {
             
            // Check if the frequency
            // of i.first is a power of 2
            if (isPowerOfTwo(freq.get(ch)))
            {
 
                // Update string st
                for(int j = 0; j < freq.get(ch); j++)
                {
                    st += ch;
                }
            }
        }
 
        // Clear the map
        freq.clear();
 
        // Null string
        if (st.length() == 0)
            continue;
 
        // Sort the string in
        // descending order
        char myCharArr[] = st.toCharArray();
        Arrays.sort(myCharArr);
        String ns = "";
         
        for(int j = myCharArr.length - 1; j >= 0; --j)
            ns += myCharArr[j];
 
        // Update res
        res.add(ns);
    }
     
    // Print the array of strings
    printArray(res);
}
 
// Driver Code
public static void main(String[] args)
{
    String arr[] = { "aaacbb", "geeks", "aaa" };
    int N = arr.length;
     
    sortedStrings(arr, N);
}
}
 
// This code is contributed by Kingash

Python3




# Python3 program for the above approach
from collections import defaultdict
import math
 
# Function to check if N is power of
# 2 or not
def isPowerOfTwo(n):
 
    # Base Case
    if (n == 0):
        return False
 
    # Return true if N is power of 2
    return (math.ceil(math.log2(n)) ==
            math.floor(math.log2(n)))
 
# Function to print array of strings
# in ascending order
def printArray(res):
 
    # Sort strings in ascending order
    res.sort()
 
    # Print the array
    for i in range(len(res)):
        print(res[i], end = " ")
 
# Function to sort the strings after
# modifying each string according to
# the given conditions
def sortedStrings(S, N):
 
    # Store the frequency of each
    # characters of the string
    freq = defaultdict(int)
 
    # Stores the required
    # array of strings
    res = []
 
    # Traverse the array of strings
    for i in range(N):
 
        # Temporary string
        st = ""
 
        # Stores frequency of each
        # alphabet of the string
        for j in range(len(S[i])):
 
            # Update frequency of S[i][j]
            freq[S[i][j]] += 1
 
        # Traverse the map freq
        for i in freq:
 
            # Check if the frequency
            # of i.first is a power of 2
            if (isPowerOfTwo(freq[i])):
 
                # Update string st
                for j in range(freq[i]):
                    st += i
 
        # Clear the map
        freq.clear()
 
        # Null string
        if (len(st) == 0):
            continue
 
        # Sort the string in
        # descending order
        st = list(st)
        st.sort(reverse=True)
        st = ''.join(st)
 
        # Update res
        res.append(st)
 
    # Print the array of strings
    printArray(res)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ "aaacbb", "geeks", "aaa" ]
    N = len(arr)
     
    sortedStrings(arr, N)
 
# This code is contributed by ukasp
Output: 
cbb skgee

 

Time Complexity: O(N * log N + M * log M), where M is the maximum length of a string in S[]
Auxiliary Space: O(N)

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