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

  • Last Updated : 04 Jul, 2019

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 strp = strp*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

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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pawan_asipu
@pawan_asipu
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