Hard
Minimum Number of Operations to Make String Sorted — C++
Full explanation · Time O(n) · Space O(maxn)
// Time: O(26 * n) = O(n)
// Space: O(max_n) = O(max_n)
class Solution {
public:
int makeStringSorted(string s) { // count of prev_permutation
static const int MOD = 1e9 + 7;
array<int, 26> count = {0};
int result = 0, comb_total = 1;
for (int i = size(s) - 1; i >= 0; --i) {
int num = s[i] - 'a';
comb_total = ((comb_total * int64_t(size(s) - i) % MOD) * inverse(++count[num], MOD)) % MOD;
result = (result + ((comb_total * accumulate(cbegin(count), cbegin(count) + num, 0LL) % MOD) * inverse(size(s) - i, MOD) % MOD)) % MOD;
}
return result;
}
private:
int inverse(int n, int m) {
static vector<int> inv = {0, 1};
for (int i = size(inv); i <= n; ++i) { // lazy initialization
inv.emplace_back(inv[m % i] * int64_t(m - m / i) % m); // https://cp-algorithms.com/algebra/module-inverse.html
}
return inv[n];
}
};