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Problem solution · C++

Minimum Operations to Make Character Frequencies Equal

Minimum Operations to Make Character Frequencies Equal: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
49 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Minimum Operations to Make Character Frequencies Equal, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 49 lines of C++ from the credited upstream file 3389.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 3 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Operations to Make Character Frequencies Equal · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int makeStringGood(string s) {    int ans = s.length();    vector<int> count(26);     for (const char c : s)      ++count[c - 'a'];     const int mx = ranges::max(count);    for (int target = 1; target <= mx; ++target)      ans = min(ans, getMinOperations(count, target));     return ans;  }  private:  int getMinOperations(const vector<int>& count, int target) {    // dp[i] := the minimum number of operations to make the frequency of    // (i..25)-th (0-indexed) letters equal to `target`.    vector<int> dp(27);     for (int i = 25; i >= 0; --i) {      // 1. Delete all the i-th letters.      const int deleteAllToZero = count[i];      // 2. Insert/delete the i-th letters to have `target` number of letters.      const int deleteOrInsertToTarget = abs(target - count[i]);      dp[i] = min(deleteAllToZero, deleteOrInsertToTarget) + dp[i + 1];      if (i + 1 < 26 && count[i + 1] < target) {        const int nextDeficit = target - count[i + 1];        // Make the frequency of the i-th letter equal to the `target` or 0.        const int needToChange =            count[i] > target ? count[i] - target : count[i];        const int changeToTarget =            nextDeficit > needToChange                // 3. Change all the i-th letters to the next letter and then                // insert the remaining deficit for the next letter.                ? needToChange + (nextDeficit - needToChange)                // 4. Change `nextDeficit` i-th letters to the next letter and                // then delete the remaining i-th letters.                : nextDeficit + (needToChange - nextDeficit);        dp[i] = min(dp[i], changeToTarget + dp[i + 2]);      }    }     return dp[0];  }};

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