Minimum operations to make frequency of all characters equal K

Given a string S of length N. The task is to find the minimum number of steps required on strings, so that it has exactly K different alphabets all with the same frequency.

Note: In one step, we can change a letter to any other letter.

Examples:



Input: S = "abbc", N = 4, K = 2
Output: 1 
In one step convert 'c' to 'a'. Hence string has 
two different letters a and b both 
occurring 2 times. 

Approach:

  1. Check if K divides N, then only it is possible to convert the given string, otherwise not.
  2. Maintain the count of all alphabets present in string S, in an array A.
  3. Evaluate E = N/K, the frequency with which alphabets will be present in the final string.
  4. Separate the alphabets with frequency more than or equal to E and less than E in two parts.
  5. Maintain the number of steps required for each alphabet to convert its count to E, sort these vector obtained in above step.
  6. Lastly, take all possibility to pick:
    Set 1 : 0   Set 2 : K
    Set 1 : 1   Set 2 : K-1 .... so on
    
  7. Keep a ans variable to calculate minimum number of steps among all possibility in step 6.
  8. Say L1 is the number of operation required on Set 1, L2 is the number of operations required on set 2. Then total operations required is maximum of L1, L2 . As suppose ‘a’ is required one less in string while ‘b’ is required one more than we can change ‘a’ to ‘b’, thus reducing number of steps.

Below is the implementation of the above appraoch:

C++

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// C++ program to convert the given string
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the minimum number of
// operations to convert the given string
void minOperation(string S, int N, int K)
{
    // Check if N is divisible by K
    if (N % K) {
        cout << "Not Possible" << endl;
        return;
    }
  
    // Array to store frequency of characters
    // in given string
    int count[26] = { 0 };
    for (int i = 0; i < N; i++) {
        count[S[i] - 97]++;
    }
  
    int E = N / K;
  
    vector<int> greaterE;
    vector<int> lessE;
  
    for (int i = 0; i < 26; i++) {
  
        // Two arrays with number of operations
        // required
        if (count[i] < E)
            lessE.push_back(E - count[i]);
        else
            greaterE.push_back(count[i] - E);
    }
  
    sort(greaterE.begin(), greaterE.end());
    sort(lessE.begin(), lessE.end());
  
    int mi = INT_MAX;
  
    for (int i = 0; i <= K; i++) {
  
        // Checking for all possibility
        int set1 = i;
        int set2 = K - i;
  
        if (greaterE.size() >= set1 && lessE.size() >= set2) {
  
            int step1 = 0;
            int step2 = 0;
  
            for (int j = 0; j < set1; j++)
                step1 += greaterE[j];
  
            for (int j = 0; j < set2; j++)
                step2 += lessE[j];
  
            mi = min(mi, max(step1, step2));
        }
    }
  
    cout << mi << endl;
}
  
// Driver Code
int main()
{
    string S = "accb";
    int N = S.size();
    int K = 2;
  
    minOperation(S, N, K);
  
    return 0;
}

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Python3

# Python3 program to convert the given string

# Function to find the minimum number of
# operations to convert the given string
def minOperation(S, N, K):



# Check if N is divisible by K
if N % K:
print(“Not Possible”)
return

# Array to store frequency of
# characters in given string
count = [0] * 26
for i in range(0, N):
count[ord(S[i]) – 97] += 1

E = N // K
greaterE = []
lessE = []

for i in range(0, 26):

# Two arrays with number of
# operations required
if count[i] < E: lessE.append(E - count[i]) else: greaterE.append(count[i] - E) greaterE.sort() lessE.sort() mi = float('inf') for i in range(0, K + 1): # Checking for all possibility set1, set2 = i, K - i if (len(greaterE) >= set1 and
len(lessE) >= set2):

step1, step2 = 0, 0

for j in range(0, set1):
step1 += greaterE[j]

for j in range(0, set2):
step2 += lessE[j]

mi = min(mi, max(step1, step2))

print(mi)

# Driver Code
if __name__ == “__main__”:

S = “accb”
N = len(S)
K = 2

minOperation(S, N, K)

# This code is contributed by Rituraj Jain

Output:

1


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Improved By : rituraj_jain