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package 01.mca_test;
public class Problem_0395_LongestSubstringWithAtLeastKRepeatingCharacters {
public static int longestSubstring1(String s, int k) {
char[] str = s.toCharArray();
int N = str.length;
int max = 0;
for (int i = 0; i < N; i++) {
int[] count = new int[256];
int collect = 0;
int satisfy = 0;
for (int j = i; j < N; j++) {
if (count[str[j]] == 0) {
collect++;
}
if (count[str[j]] == k - 1) {
satisfy++;
}
count[str[j]]++;
if (collect == satisfy) {
max = Math.max(max, j - i + 1);
}
}
}
return max;
}
public static int longestSubstring2(String s, int k) {
char[] str = s.toCharArray();
int N = str.length;
int max = 0;
for (int require = 1; require <= 26; require++) {
// 3种
// a~z 出现次数
int[] count = new int[26];
// 目前窗口内收集了几种字符了
int collect = 0;
// 目前窗口内出现次数>=k次的字符满足了几种
int satisfy = 0;
// 窗口右边界
int R = -1;
for (int L = 0; L < N; L++) { // L要尝试每一个窗口的最左位置
// [L..R] R+1
while (R + 1 < N && !(collect == require && count[str[R + 1] - 'a'] == 0)) {
R++;
if (count[str[R] - 'a'] == 0) {
collect++;
}
if (count[str[R] - 'a'] == k - 1) {
satisfy++;
}
count[str[R] - 'a']++;
}
// [L...R]
if (satisfy == require) {
max = Math.max(max, R - L + 1);
}
// L++
if (count[str[L] - 'a'] == 1) {
collect--;
}
if (count[str[L] - 'a'] == k) {
satisfy--;
}
count[str[L] - 'a']--;
}
}
return max;
}
// 会超时,但是思路的确是正确的
public static int longestSubstring3(String s, int k) {
return process(s.toCharArray(), 0, s.length() - 1, k);
}
public static int process(char[] str, int L, int R, int k) {
if (L > R) {
return 0;
}
int[] counts = new int[26];
for (int i = L; i <= R; i++) {
counts[str[i] - 'a']++;
}
char few = 0;
int min = Integer.MAX_VALUE;
for (int i = 0; i < 26; i++) {
if (counts[i] != 0 && min > counts[i]) {
few = (char) (i + 'a');
min = counts[i];
}
}
if (min >= k) {
return R - L + 1;
}
int pre = 0;
int max = Integer.MIN_VALUE;
for (int i = L; i <= R; i++) {
if (str[i] == few) {
max = Math.max(max, process(str, pre, i - 1, k));
pre = i + 1;
}
}
if (pre != R + 1) {
max = Math.max(max, process(str, pre, R, k));
}
return max;
}
}