# 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:

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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 ``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 > 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 = *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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