第 368 场 LeetCode 周赛题解
A 元素和最小的山形三元组 I
前后缀操作：求出前后缀上的最小值数组，然后枚举 j j j
class Solution {
public:
int minimumSum(vector<int> &nums) {
int n = nums.size();
vector<int> l(n), r(n);//l[i]=min{nums[0],...,nums[i]}, r[i]=min{nums[i],...,nums[n-1]}
l[0] = nums[0];
for (int i = 1; i < n; i++)
l[i] = min(l[i - 1], nums[i]);
r[n - 1] = nums[n - 1];
for (int i = n - 2; i >= 0; i--)
r[i] = min(r[i + 1], nums[i]);
int res = INT32_MAX;
for (int j = 1; j < n - 1; j++)
if (l[j - 1] < nums[j] && r[j + 1] < nums[j])
res = min(res, l[j - 1] + nums[j] + r[j + 1]);
return res == INT32_MAX ? -1 : res;
}
};
B 元素和最小的山形三元组 II
同 A …
class Solution {
public:
int minimumSum(vector<int> &nums) {
int n = nums.size();
vector<int> l(n), r(n);//l[i]=min{nums[0],...,nums[i]}, r[i]=min{nums[i],...,nums[n-1]}
l[0] = nums[0];
for (int i = 1; i < n; i++)
l[i] = min(l[i - 1], nums[i]);
r[n - 1] = nums[n - 1];
for (int i = n - 2; i >= 0; i--)
r[i] = min(r[i + 1], nums[i]);
int res = INT32_MAX;
for (int j = 1; j < n - 1; j++)
if (l[j - 1] < nums[j] && r[j + 1] < nums[j])
res = min(res, l[j - 1] + nums[j] + r[j + 1]);
return res == INT32_MAX ? -1 : res;
}
};
C 合法分组的最少组数
枚举：枚举分组中有最多下标的组的下标数 i i i ，设一个数 v a l val val 在 n u m s nums nums 的出现次数为 c i ci ci ，若 v a l val val 可以满足条件的被分为若干组，则有 c i = a × i + b × ( i − 1 ) , a ≥ 0 , b ≥ 0 ci=a\times i+b\times(i-1), a\ge0,b\ge 0 ci=a×i+b×(i−1),a≥0,b≥0，等价于 c i = a b × i − b , a b ≥ b ≥ 0 ci=ab\times i - b,ab\ge b\ge0 ci=ab×i−b,ab≥b≥0，即 ⌈ c i i ⌉ ≥ ⌈ c i i ⌉ × i − c i \left \lceil \frac {ci} i \right \rceil \ge \left \lceil \frac {ci} i \right \rceil\times i-ci ⌈ici⌉≥⌈ici⌉×i−ci
class Solution {
public:
int minGroupsForValidAssignment(vector<int> &nums) {
unordered_map<int, int> cnt;
for (auto x: nums)
cnt[x]++;
unordered_map<int, int> can;
int m = cnt.size();//nums中不同的数val的个数
for (auto &[_, ci]: cnt) {
for (int i = 1; i <= ci + 1; i++) {
int ab = (ci - 1) / i + 1;
int b = ab * i - ci;
if (ab >= b)
can[i]++;
}
}
int mx = 1;
for (auto &[gs, ci]: can)
if (ci == m)//对nums中每个数val都可以划分为若干大小为gs-1和若干大小为g的组
mx = max(mx, gs);
int res = 0;
for (auto &[_, ci]: cnt)
res += (ci - 1) / mx + 1;
return res;
}
};
D 得到 K 个半回文串的最少修改次数
动态规划：首先需要预处理求出 c o s t cost cost 数组（ c o s t [ i ] [ j ] cost[i][j] cost[i][j] 为将字符串 s [ i , j ] s[i,j] s[i,j] 修改为半回文串的最少修改次数），然后设 p [ i ] [ j ] p[i][j] p[i][j] 为使 s [ 0 , i ] s[0,i] s[0,i] 可分为 j j j 个半回文串的最少操作数，枚举其第 j j j 个半回文串可能的长度来进行状态转移
class Solution {
public:
vector<vector<int>> cost;
string s;
int n;
void comp_cost(int i, int j, int d) {//以d为模数(题目描述中的d)更新cost[i][j]
int g = (j - i + 1) / d;//分组大小
int tot = 0;
for (int ind = 0; ind < d; ind++) {//枚举d个分组，若s[i,j]是半回文串则每个分组都是回文串
for (int l = i + ind, r = i + ind + d * (g - 1); l <= r; l += d, r -= d)
if (s[l] != s[r])
tot++;
}
cost[i][j] = min(cost[i][j], tot);//更新cost[i][j]
}
int minimumChanges(string s, int k) {
this->s = s;
n = s.size();
cost = vector<vector<int>>(n, vector<int>(n, INT32_MAX));
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {//s[i,j]
int len = j - i + 1;
for (int d = 1; d * d <= len; d++)
if (len % d == 0) {//len的因子d
comp_cost(i, j, d);
if (d != len / d && d != 1)//len的因子len/d
comp_cost(i, j, len / d);
}
}
}
int p[n][n + 1];// loc, k
for (int i = 0; i < n; i++)
for (int j = 0; j <= n; j++)
p[i][j] = INT32_MAX;//无效状态标志
for (int i = 1; i < n; i++) {
for (int j = 1; j <= (i + 1) / 2; j++) {
if (j == 1)
p[i][j] = cost[0][i];
for (int pre = 0; i - pre > 1; pre++)
if (p[pre][j - 1] != INT32_MAX)
p[i][j] = min(p[i][j], p[pre][j - 1] + cost[pre + 1][i]);//考虑最后一个半回文串为s[pre+1][i]的情况
}
}
return p[n - 1][k];
}
};