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

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 occurred character is less than
    // required positions
    if (v.back().first < sz) {
        cout << -1;
        return;
    }

    // In all required positions keep
    // character which occurred 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 occurred character is less than
    # required positions
    if (v[-1][0] < sz) :
        print(-1,end="");
        return;

    # In all required positions keep
    # character which occurred 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

Javascript

// Javascript program for the above approach

const N = 100005;
const ans = new Array(N).fill(0);

function Rearrange(s, n) {
  const sieve = new Array(N + 1).fill(1);
  let sz = 0;

  for (let i = 2; i <= Math.floor(n / 2); i++) {
    if (sieve[i]) {
      for (let j = 1; j <= Math.floor(n / i); j++) {
        if (sieve[i * j]) {
          sz += 1;
        }
        sieve[i * j] = 0;
      }
    }
  }

  const m = {};
  for (const c of s) {
    m = (m || 0) + 1;
  }

  const v = [];
  for (const [key, value] of Object.entries(m)) {
    v.push([value, key]);
  }

  v.sort();

  if (v[v.length - 1][0] < sz) {
    console.log(-1);
    return;
  }

  for (let i = 2; i <= n; i++) {
    if (!sieve[i]) {
      ans[i] = v[v.length - 1][1];
    }
  }

  let idx = 0;
  for (let i = 1; i <= n; i++) {
    if (sieve[i]) {
      ans[i] = v[idx][1];
      v[idx][0] -= 1;
      if (v[idx][0] === 0) {
        idx += 1;
      }
    }
    process.stdout.write(ans[i] || '');
  }
}

const string = 'aabaaaa';
const n = string.length;
Rearrange(string, n);



// contributed by adityasharmadev01

Output:

baaaaaa