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:

  • Store the frequency of each character present in the String.
  • Sort the frequencies.
  • Calculate the difference between each frequency in the sorted and original sequence.
  • Half of the total sum of all the differences is the required answer. This is because, for each swap, the difference is calculated twice in the above step.

Below is the implementation of the above approach: 

C++

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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

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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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Python3

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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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