Minimize Cost to sort a String in Increasing Order of Frequencies of Characters

Given a string S, the task is to calculate the minimum cost to sort the string in increasing order of their frequencies by swapping a block of repeated characters with another block of different repeated characters. The cost of each operation is the absolute difference of the two blocks.
Examples: 

Input: S = “aabbcccdeffffggghhhhhii” 
Output: 5 
Explanation

  • Swap ‘d’ with ‘aa’. The Cost of this Operation is 1
  • Swap ‘e’ with ‘bb’. The Cost of this Operation is 1
  • Swap ‘ii’ with ‘ccc’. The Cost of this Operation is cost is 1
  • Swap ‘ccc’ with ‘ffff’. The Cost of this Operation is 1
  • Swap ‘ffff’ with ‘hhhhh’. The Cost of this Operation is 1

Input : S = “aaaa” 
Output : 0 

 

Approach: 
 Follow the steps below to solve the problem:

Below is the implementation of the above approach: 

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#include 
using namespace std;
 
int sortString(string S)
{
 
    vector sorted, original;
 
    bool insert = false;
 
    // For a single character
    if (S.length() == 1) {
        cout << 0 << endl;
    }
 
    // Stores count of repetitions
    // of a character
    int curr = 1;
 
    for (int i = 0; i < (S.length() - 1); i++) {
        // If repeating character
        if (S[i] == S[i + 1]) {
            curr += 1;
            insert = false;
        }
 
        // Otherwise
        else {
            // Store frequency
            sorted.push_back(curr);
            original.push_back(curr);
 
            // Reset count
            curr = 1;
            insert = true;
        }
    }
    // Insert the last character block
    if ((S[(S.length() - 1)] != S[(S.length() - 2)])
        || insert == false) {
        sorted.push_back(curr);
        original.push_back(curr);
    }
 
    // Sort the frequencies
    sort(sorted.begin(), sorted.end());
 
    // Stores the minimum cost of all operations
    int t_cost = 0;
 
    for (int i = 0; i < sorted.size(); i++) {
        // Store the absolute difference of
        // i-th frequencies of ordered and
        // unordered sequences
        t_cost += abs(sorted[i] - original[i]);
    }
 
    // Return the minimum cost
    return (t_cost / 2);
}
 
// Driver Code
int main()
{
    string S = "aabbcccdeffffggghhhhhii";
    cout << sortString(S);
 
    return 0;
}
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// Java program to implement
// above approach
import java.util.*;
class GFG{
     
public static int sortString(String S)
{
 
    Vector sorted = new Vector();
    Vector original = new Vector();
 
    boolean insert = false;
 
    // For a single character
    if (S.length() == 1)
    {
        System.out.println(0);
    }
 
    // Stores count of repetitions
    // of a character
    int curr = 1;
 
    for(int i = 0; i < (S.length() - 1); i++)
    {
         
        // If repeating character
        if (S.charAt(i) == S.charAt(i + 1))
        {
            curr += 1;
            insert = false;
        }
 
        // Otherwise
        else
        {
             
            // Store frequency
            sorted.add(curr);
            original.add(curr);
 
            // Reset count
            curr = 1;
            insert = true;
        }
    }
    // Insert the last character block
    if ((S.charAt(S.length() - 1) !=
         S.charAt(S.length() - 2)) ||
         insert == false)
    {
        sorted.add(curr);
        original.add(curr);
    }
 
    // Sort the frequencies
    Collections.sort(sorted);
 
    // Stores the minimum cost of all operations
    int t_cost = 0;
 
    for(int i = 0; i < sorted.size(); i++)
    {
         
        // Store the absolute difference of
        // i-th frequencies of ordered and
        // unordered sequences
        t_cost += Math.abs(sorted.get(i) -
                         original.get(i));
    }
 
    // Return the minimum cost
    return (t_cost / 2);
}
 
// Driver code
public static void main(String[] args)
{
    String S = "aabbcccdeffffggghhhhhii";
    System.out.print(sortString(S));
}
}
 
// This code is contributed by divyeshrabadiya07
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# Python program to implement
# the above approach
def sortString(S):
    sorted1 = []
    original = []
    insert = False
 
    # For a single character
    if (len(S) == 1):
        print(0)
 
    # Stores count of repetitions
    # of a character
    curr = 1
 
    for i in range(len(S) - 1):
        # If repeating character
        if (S[i] == S[i + 1]):
            curr += 1
            insert = False
 
        # Otherwise
        else:
            # Store frequency
            sorted1.append(curr)
            original.append(curr)
 
            # Reset count
            curr = 1
            insert = True
 
    # Insert the last character block
    if ((S[(len(S) - 1)] != S[(len(S) - 2)]) or
         insert == False):
        sorted1.append(curr)
        original.append(curr)
 
    # Sort the frequencies
    sorted1.sort()
 
    # Stores the minimum cost of all operations
    t_cost = 0
 
    for i in range(len(sorted1)):
       
        # Store the absolute difference of
        # i-th frequencies of ordered and
        # unordered sequences
        t_cost += abs(sorted1[i] - original[i])
 
    # Return the minimum cost
    return (t_cost // 2)
 
# Driver Code
if __name__ == "__main__":
    S = "aabbcccdeffffggghhhhhii"
    print(sortString(S))
 
# This code is contributed by Chitranayal
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Output: 
5


 

Time Complexity: O(NlogN) 
Auxiliary Space: O(N)

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Competitive Programmer | Python Developer

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