AtCoder · abc449_c

Comfortable Distance

This C++ solution uses prefix sums for AtCoder abc449_c Comfortable Distance. Read the reasoning, inspect the code, or try your own test case below.

abc449_cMathPrefix sumsC++36 lines
Solution048of 248
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Approach

Prefix sums

Comfortable Distance asks for equal-character pairs whose distance lies in [L,R]; the code counts them with per-letter prefix sums.

Math

Problem and code

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Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

C_Comfortable_Distance.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1e18; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int n, l, r; cin >> n >> l >> r;
        string s; cin >> s;
        // int len = r - l;
        ll cnt = 0;
        // for (int i = 0; i < s.size(); i++) {
        //     char c = s[i];
        //     for (int j = min(n - 1, i + l); j <= min(n - 1, j + r); j++) {
        //         if (j >= i + l && j <= i + r && s[j] == c) {
        //             cnt++;
        //         }
        //     }
        // }
        vector<vector<int>> psa(26, vector<int>(n));
        psa[s[0] - 'a'][0] = 1;
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < 26; j++) {
                psa[j][i] = psa[j][i - 1];
            }
            psa[s[i] - 'a'][i]++;
        }
        for (int i = 0; i < n; i++) {
            int lft = i + l, rit = min(n - 1, i + r);
            if (lft > rit) continue;
            cnt += (lft > 0 ? psa[s[i] - 'a'][rit] - psa[s[i] - 'a'][lft - 1] : psa[s[i] - 'a'][rit]);
        }
        cout << cnt << '\n';
        return 0;
    }
        

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