Minimum addition/removal of characters to be done to make frequency of each character prime

Given a string S of length N, the task is to find the minimum operations required to make the frequency of every distinct character prime. Frequency of a character can be increased or decreased by 1 in a single operation.

Examples:

Input: S = “abba”
Output: 0
Explanation: There are two characters in the string S and the frequency of ‘a’ is 2 which is prime. The frequency of ‘b’ is 2 which is also a prime number.Hence no operations are required for this case.

Input: S = “aaaaaaaaa”
Output: 2
Explanation: The frequency of ‘a’ in the string is 9. Either remove 2 a’s from the string or add 2 a’s in the string to make the frequency of ‘a’ prime. Thus minimum number of operations needed is 2.

Naive Approach:
Find the frequency of each character in the string. Let the frequency be X. Find the prime number greater than X and less than X. Compare the difference between X and the two prime numbers found and choose the closest prime number. Add the difference between the closest prime number and X to the number of operations. Thus the minimum number of operations will be obtained. It is an inefficient approach as the lower and upper bound to find the prime number is not known.



Efficient Approach :

  1. Use the sieve algorithm to find all the prime numbers up to N and store them in an array.
  2. Find out the nearest lower prime number for all the numbers from i = 1 to N and store the difference between i and its nearest lower prime number in an array say arr1[].
  3. Find out the nearest higher prime number for all the numbers from i = 1 to N and store the difference between i and its nearest higher prime number in an array say arr2[].
  4. Traverse the string and find the frequency of all the distinct characters in the string and use an unordered_map to save the frequencies of these distinct characters.
  5. Let the frequency of any distinct character be X then find out the distance between X and nearest prime number from the arrays arr1[] and arr2[].
  6. Choose the distance which is lesser between the two and add this distance to the number of operations.
  7. Do this for all the distinct characters and print the number of operations finally.

Below is the implementation of the above approach:

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// C++ code for the above approach
  
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
  
// using the seive to find
// the prime factors.
void spf_array(int spf[], int d)
{
    spf[1] = 1;
    int i = 0;
  
    // setting every number
    // as prime number initailly
    for (i = 2; i <= d; i++) {
        spf[i] = i;
    }
  
    // removing the even numbers as
    // they cannot be prime except 2
    for (i = 2; i <= d; i = i + 2) {
        spf[i] = 2;
    }
  
    for (i = 3; i * i <= d; i++) {
        // if they are prime
        if (spf[i] == i) {
            int j;
  
            // iterating the loop for
            // prime and eliminate all
            // the multiples of three
            for (j = i * i; j <= d; j = j + i) {
                if (spf[j] == j) {
                    spf[j] = i;
                }
            }
        }
    }
}
  
// function to find the closest
// prime number of every
// number upto N (size of the string)
void closest_prime_spf(int spf[],
                       int cspf[], int d)
{
    // for the base case
    cspf[1] = 1;
  
    int i = 0, j = 0, k = 0;
  
    // iterating to find the
    // distance from the
    // lesser nearest prime
    for (i = 2; i < d; i++) {
        if (spf[i] != i) {
            cspf[i] = abs(i - k);
        }
        else {
            cspf[i] = -1;
            k = i;
        }
    }
  
    // iterating to find the
    // distance from the
    // lesser nearest prime
    for (i = d - 1; i >= 2; i--) {
        if (spf[i] != i) {
            if (cspf[i] > abs(k - i)) {
                cspf[i] = abs(k - i);
            }
        }
        else {
            k = i;
        }
    }
}
  
// function to find the
// minimum operation
int minimum_operation(int cspf[],
                      string s)
{
  
    // created map to find the
    // frequency of distinct characters
    unordered_map<char, int> m;
  
    // variable to iterate and
    // holding the minimum operation
    int i = 0, k = 0;
  
    // loop for calculation frequency
    for (i = 0; i < s.length(); i++) {
        m[s[i]] = m[s[i]] + 1;
    }
  
    // iterate over frequency
    // if we not get a chcarter
    // frequency prime
    // then we find the closest
    // prime and add
    for (auto x : m) {
        int h = x.second;
        if (cspf[h] != -1) {
            k = k + cspf[h];
        }
    }
    return k;
}
  
// Function to find the
// minimum number of operations
void minOper(string s)
{
  
    int spf[s.length() + 1];
    int cspf[s.length() + 1];
  
    // function called to create
    // the spf
    spf_array(spf, s.length() + 1);
  
    // function called to
    // create the cspf
    closest_prime_spf(spf, cspf,
                      s.length() + 1);
  
    cout << minimum_operation(cspf, s)
         << endl;
}
  
// Driver Code
int main()
{
    // input string
    string s = "aaaaaaaaabbcccccc";
  
    minOper(s);
  
    return 0;
}

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Output:

3


Time complexity: O(N * log(log(N)))
Auxillary Space complexity: O(N)

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