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Rearrange the given string such that all Prime Multiple indexes have Same Character
A string is a sequence of characters, numbers, Alphanumeric characters and special characters. The length of a string is the number of characters that are present in it. A prime number is the number which is divisible by 1 and the number itself.
In this artice, we are given a sample input string of length n. We are going to develop a code where the idea is to substitute the same character at any position p, such that the the characters at positions ranging from 1 to n/p should coincide.
Some of the examples illustrating the problem statement are as follows −
Sample Example
Example 1 - alphaaa
Output - haaalap
Explanation - The prime indexes 2,4,6 and the indexes 3,6 contain the same character ‘a’.
Example 1 - alphaa
Output - -1
Explanation - As soon as we reduce one character ‘a’ from the string, all the required
positions cannot accommodate the same character. Therefore, the output returned is -1.
Algorithm
Step 1 − A map is initialised to keep a count of the characters occurring within the string.
Step 2 − A map contains a tuple of the type
Step 3 − Every time a character is encountered, its frequency is increased by 1.
Step 4 − A vector vec, is also initialized in order to keep map entries in the reversed form, of the nature
Step 5 − The vector is then sorted in order to arrange the characters in the order of their frequencies.
Step 6 − The position array pos is declared with a size equivalent to maximum value possible.
Step 7 − It is initialised with a value 1 using the fill() method available in C++ STL.
Step 8 − An iteration of the string is initiated, using the pointer i, beginning with the integer value 2, and if the pos array for this position contains a value, then all the prime multiples using a pointer j, are evaluated.
Step 9 − For all the valid positions, the counter of the positions to be filled with the same character is incremented.
Step 10 − Otherwise, the pos value of the i*j index is evaluated.
Step 11 − The frequency of the most occurring character is then stored, consider it to be much.
Step 12 − If the number of positions to be occupied by the same character is greater than the max occurring character frequency, then -1 is returned, since this is an impossible operation.
Step 13 − Else, initially all the prime multiple indexes, which are the desirable positions are filled out by the maximum occurring character.
Step 14 − The remaining positions are then filled.
Step 15 − Every time a character from the vector is taken out and filled at the remaining position.
Step 16 − The frequency of this character is decremented.
Step 17 − When all the occurrences of a particular character have been exhausted, then next character is fetched from the vector.
Step 18 − The final output array string is then printed after swapping the characters.
Example
The following C++ code snippet takes a sample input string and then places the same character on all the prime multiple indices, if possible.
//including the required libraries
#include <bits/stdc++.h>
using namespace std;
//rearranging prime indexes to store same character
void primeindx(string str){
// To store the result after transformation
char res[100005];
int pos[100005];
//calculating the size of string
int len = str.size();
//counter of same elements
int cnt = 0;
//maintaining map to store the number of occurrences of characters in string
map<char, int> map;
for (auto ch : str){
//incrementing count of character
map[ch]+=1;
}
//maintaining a vector to store the different characters in map and their respective counts to sort them according to the frequency
vector<pair<int, char> > vec;
for (auto entry : map)
vec.push_back({ entry.second, entry.first });
//sorting the vector according to the frequencies
sort(vec.begin(), vec.end());
//fill the positions of the array pos with 1
fill(pos + 1, pos + len + 1, 1);
//storing indices correctly in the pos array
for (int i = 2; i <= len / 2; i++) {
//check if the value at pos[i] is equivalent to 1
if (pos[i]) {
//getting all multiples of i
for (int j = 1; i*j<= len; j++) {
int val = i*j;
if (pos[val])
//incrementing the number of same characters is value is contained in the multiple
cnt++;
//else storing 0 in the pos vector at that position
pos[val] = 0;
}
}
}
//getting the occurrence of the character occurring the maximum number of times
int mch = vec.back().first;
if ( mch < cnt) {
//not possible
cout << -1;
return;
}
for (int i = 2; i <= len; i++) {
if (!pos[i]) {
//storing the character at the prime indexes is the count of the same character satisfies
res[i] = vec.back().second;
}
}
//fetching characters after prime indexes have been filled out
int k = 0;
for (int i = 1; i <= len; i++) {
//if ith index contains 1
if (pos[i]) {
//get the character in the result array
res[i] = vec[k].second;
//decrement the count available for the character that has been substitutes
vec[k].first-=1;
//if there are no more occurrences of that particular character, skip to next one
if (vec[k].first == 0)
k++;
}
}
//printing the final result array
for(int i =1;i<=len;i++){
cout << res[i];
}
}
//calling the main method
int main(){
//declaring a sample string
string str = "codeeee";
//prime multiple indexes arrangement of character
primeindx(str);
return 0;
}
Output
ceeedeo
Conclusion
Character counting is an important aspect of C++ strings. A large number of decisions can be taken based on the frequency of characters available in the string. Maps are an efficient way of storing a data pair, such as characters and their respective counts in a very user-friendly manner.
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