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Problem solution · C++

ABC428 B — Most Frequent Substrings

ABC428 B — Most Frequent Substrings: a C++ solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to michimani AtCoder Solutions.

Technique
Sorting and greedy selection
Source
michimani AtCoder Solutions
Length
39 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

Sorting and greedy selection

For ABC428 B — Most Frequent Substrings, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 39 lines of C++ from the credited upstream file abc428_b.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 4 loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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

Source

Code and credit

This code comes from michimani AtCoder Solutions by michimani and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeABC428 B — Most Frequent Substrings · C++C++
Use this to learn the idea, then write your own version.
#include <algorithm>#include <cmath>#include <iostream>#include <map>#include <vector> using namespace std;using ui = unsigned int; int main() {    ui n, k;    cin >> n >> k;    string s;    cin >> s;     map<string, ui> tm = {};    string t = "";    ui mx = 0;    for (ui i = 0; i <= n - k; i++) {        t = "";        for (ui l = 0; l < k; l++) t += s[i + l];        tm[t]++;        mx = max(mx, tm[t]);    }     vector<string> tl = {};    for (auto m : tm)        if (m.second == mx) tl.push_back(m.first);     cout << mx << endl;    sort(tl.begin(), tl.end());    for (ui i = 0; i < tl.size(); i++) {        if (i > 0) cout << " ";        cout << tl[i];    }    cout << endl;     return 0;}

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