Rearrange the given string such that all prime multiple indexes have same character

Given a string str of size N. The task is to find out whether it is possible to rearrange characters in string str so that for any prime number p <= N and for any integer i ranging from 1 to N/p the condition str_p = str_p*i must be satisfied. If it is not possible to for any such rearrangement then print -1.

Examples:

Input : str = “aabaaaa”
Output : baaaaaa
Size of the string is 7.
Indexes 2, 4, 6 contains same character.
Indexes 3, 6 contains the same character.

Input : str = “abcd”
Output : -1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

All positions except the first and those whose number is a prime greater N/2 must have the same symbol.
Remaining positions can have any symbol. This positions kept using the sieve.
If the most occurred element in the string is less than these positions then print -1.

Below is the implementation of the above approach:

C++

// CPP program to rearrange the given  
// string such that all prime multiple 
// indexes have same character 
#include <bits/stdc++.h> 
using namespace std; 
#define N 100005 
  
// To store answer 
char ans[N]; 
int sieve[N]; 
  
// Function to rearrange the given string 
// such that all prime multiple indexes 
// have the same character. 
void Rearrange(string s, int n) 
{ 
    // Initially assume that we can kept 
    // any symbol at any positions. 
    // If at any index contains one then it is not 
    // counted in our required positions 
    fill(sieve + 1, sieve + n + 1, 1); 
  
    // To store number of positions required 
    // to store elements of same kind 
    int sz = 0; 
  
    // Start sieve 
    for (int i = 2; i <= n / 2; i++) { 
        if (sieve[i]) { 
            // For all multiples of i 
            for (int j = 1; i * j <= n; j++) { 
                if (sieve[i * j]) 
                    sz++; 
                sieve[i * j] = 0; 
            } 
        } 
    } 
  
    // map to store frequency of each character 
    map<char, int> m; 
    for (auto it : s) 
        m[it]++; 
  
    // Store all characters in the vector and 
    // sort the vector to find the character with 
    // highest frequency 
    vector<pair<int, char> > v; 
    for (auto it : m) 
        v.push_back({ it.second, it.first }); 
    sort(v.begin(), v.end()); 
  
    // If most occured character is less than 
    // required positions 
    if (v.back().first < sz) { 
        cout << -1; 
        return; 
    } 
  
    // In all required positions keep 
    // character which occured most times 
    for (int i = 2; i <= n; i++) { 
        if (!sieve[i]) { 
  
            ans[i] = v.back().second; 
        } 
    } 
  
    // Fill all other indexes with 
    // remaining characters 
    int idx = 0; 
    for (int i = 1; i <= n; i++) { 
        if (sieve[i]) { 
            ans[i] = v[idx].second; 
            v[idx].first--; 
            // If character frequency becomes 
            // zero then go to next character 
            if (v[idx].first == 0) 
                idx++; 
        } 
        cout << ans[i]; 
    } 
} 
  
// Driver code 
int main() 
{ 
    string str = "aabaaaa"; 
  
    int n = str.size(); 
  
    // Function call 
    Rearrange(str, n); 
  
    return 0; 
} 

Python3

# Python3 program to rearrange the given  
# string such that all prime multiple  
# indexes have same character  
  
N = 100005
  
# To store answer  
ans = [0]*N;  
# sieve = [1]*N;  
  
# Function to rearrange the given string  
# such that all prime multiple indexes  
# have the same character.  
def Rearrange(s, n) :  
  
    # Initially assume that we can kept  
    # any symbol at any positions.  
    # If at any index contains one then it is not  
    # counted in our required positions  
    sieve = [1]*(N+1); 
      
    # To store number of positions required  
    # to store elements of same kind 
    sz = 0; 
      
    # Start sieve 
    for i in range(2, n//2 + 1) : 
        if (sieve[i]) : 
              
            # For all multiples of i 
            for j in range(1, n//i + 1) : 
                if (sieve[i * j]) : 
                    sz += 1; 
                sieve[i * j] = 0; 
                  
    # map to store frequency of each character  
    m = dict.fromkeys(s,0); 
      
    for it in s : 
        m[it] += 1; 
          
    # Store all characters in the vector and  
    # sort the vector to find the character with  
    # highest frequency 
    v = []; 
    for key,value in m.items() : 
        v.append([ value, key] ); 
      
    v.sort(); 
      
    # If most occured character is less than 
    # required positions 
    if (v[-1][0] < sz) : 
        print(-1,end=""); 
        return; 
          
    # In all required positions keep 
    # character which occured most times 
    for i in range(2, n + 1) : 
        if (not sieve[i]) : 
            ans[i] = v[-1][1]; 
              
    # Fill all other indexes with  
    # remaining characters  
    idx = 0; 
      
    for i in range(1, n + 1) : 
        if (sieve[i]): 
            ans[i] = v[idx][1]; 
            v[idx][0] -= 1; 
              
            # If character frequency becomes  
            # zero then go to next character 
            if (v[idx][0] == 0) : 
                idx += 1; 
                  
        print(ans[i],end= "");  
          
          
# Driver code  
if __name__ == "__main__" :  
  
    string = "aabaaaa";  
  
    n = len(string);  
  
    # Function call  
    Rearrange(string, n);  
  
# This code is contributed by AnkitRai01 

Output:

baaaaaa

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3

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