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:
- CPP
#include <bits/stdc++.h>
**using** **namespace** std;
**int** sortString(string S)
{
vector<``**int**``> 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;
}
Output:
5
Time Complexity:_ O(NlogN)_
Auxiliary Space:_ O(N)_
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#greedy #mathematical #sorting #strings #frequency-counting #arrays
