思路: 模拟即可

class Solution {public int maxFrequencyElements(int[] nums) {Map<Integer, Integer> hash = new HashMap<>();for (int v: nums) {hash.merge(v, 1, Integer::sum);}int fre = hash.values().stream().mapToInt(Integer::valueOf).max().getAsInt();return (int)hash.values().stream().filter(x -> x == fre).count() * fre;}
}

和T4一起讲

思路: 数位DP+二分

这题题目真的绕，有点头痛

其二分阶段性还是非常的明显的，就是如何写check的问题。

数位DP，可以递推着写，就是沿着边界计算，因为这是二进制，反而比 传统的数位DP（记忆化搜索）更简洁。

其核心为

$$定义左侧1的有效位数acc1，右侧的长度left$$

$$s1=前置1的个数\ast 右侧可构造的个数=acc1\ast 2^{left}$$

$$s2=右侧可构造的个数中包含1的总个数=2^{left}\ast (left/x)/2$$

class Solution {boolean check(long m, long k, int x) {String s = new StringBuilder(Long.toString(m, 2)).reverse().toString();char[] str = s.toCharArray();long res = 0;int acc = 0;for (int i = str.length - 1; i >= 0; i--) {int p = str[i] - '0';if (p == 1) {int left = i;res += (1l << left) * acc;res += (1l << left) * (left / x) / 2;}if ((i + 1) % x == 0 && p == 1) acc++;}res += acc;return res <= k;}// 二分+数位DPpublic long findMaximumNumber(long k, int x) {long l = 0, r = 1l << 50;while (l <= r) {long m = l + (r - l) / 2;if (check(m, k, x)) {l = m + 1;} else {r = m - 1;}}return r;}}

如果非要用记忆化的方式，感觉状态应该是

$$(第i位，前缀1的有效个数)$$

思路： KMP/StringHash

因为手头有StringHash板子，所以用了StringHash求解

当然用了双Hash, 避免被Hack.

class Solution {public List<Integer> beautifulIndices(String s, String a, String b, int k) {long mod = (long)1e9 + 7;int p1 = 13, p2 = 17;StringHash h1 = new StringHash(s, p1, mod);StringHash h2 = new StringHash(s, p2, mod);List<Integer> lists = new ArrayList<>();long t1 = StringHash.evaluate(a, p1, mod);long t2 = StringHash.evaluate(a, p2, mod);for (int i = 0; i <= s.length() - a.length(); i++) {if (h1.query(i, i + a.length() - 1) == t1 && h2.query(i, i + a.length() - 1) == t2) {lists.add(i);}}TreeSet<Integer> ts = new TreeSet<>();long t3 = StringHash.evaluate(b, p1, mod);long t4 = StringHash.evaluate(b, p2, mod);for (int i = 0; i <= s.length() - b.length(); i++) {if (h1.query(i, i + b.length() - 1) == t3&& h2.query(i, i + b.length() - 1) == t4) {ts.add(i);}}List<Integer> res = new ArrayList<>();for (int i: lists) {if (ts.contains(i)) {res.add(i);} else {Integer k1 = ts.floor(i);Integer k2 = ts.ceiling(i);if (k1 != null && Math.abs(k1 - i) <= k) res.add(i);else if (k2 != null && Math.abs(k2 - i) <= k) res.add(i);}}return res;}staticclass StringHash {char[] str;long p, mod;int n;long[] pre; // hash前缀和long[] pow; // p的幂次public StringHash(String s, int p, long mod) {this.str = s.toCharArray();this.p = p;this.mod = mod;this.n = str.length;pre = new long[n + 1];pow = new long[n + 1];pow[0] = 1;for (int i = 1; i <= n; i++) {pow[i] = pow[i - 1] * p % mod;}for (int i = 0; i < str.length; i++) {pre[i + 1] = (pre[i] * p % mod + str[i]) % mod;}}long query(int l, int r) {long res = pre[r + 1] - pre[l] * pow[r - l + 1] % mod;return (res % mod + mod) % mod;}long rotate(int l) {if (l < 0 || l >= str.length - 1) {return query(0, str.length - 1);} else {long h1 = query(0, l);long h2 = query(l + 1, str.length - 1);return (h2 * pow[l + 1] % mod + h1) % mod;}}static long evaluate(String s, int p, long mod) {long h = 0;for (char c: s.toCharArray()) {h = (h * p % mod + c) % mod;}return h;}}
}