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package class35;
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public class Problem_0395_LongestSubstringWithAtLeastKRepeatingCharacters {
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public static int longestSubstring1(String s, int k) {
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char[] str = s.toCharArray();
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int N = str.length;
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int max = 0;
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for (int i = 0; i < N; i++) {
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int[] count = new int[256];
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int collect = 0;
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int satisfy = 0;
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for (int j = i; j < N; j++) {
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if (count[str[j]] == 0) {
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collect++;
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}
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if (count[str[j]] == k - 1) {
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satisfy++;
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}
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count[str[j]]++;
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if (collect == satisfy) {
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max = Math.max(max, j - i + 1);
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}
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}
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}
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return max;
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}
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public static int longestSubstring2(String s, int k) {
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char[] str = s.toCharArray();
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int N = str.length;
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int max = 0;
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for (int require = 1; require <= 26; require++) {
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// 3种
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// a~z 出现次数
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int[] count = new int[26];
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// 目前窗口内收集了几种字符了
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int collect = 0;
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// 目前窗口内出现次数>=k次的字符，满足了几种
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int satisfy = 0;
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// 窗口右边界
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int R = -1;
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for (int L = 0; L < N; L++) { // L要尝试每一个窗口的最左位置
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// [L..R] R+1
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while (R + 1 < N && !(collect == require && count[str[R + 1] - 'a'] == 0)) {
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R++;
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if (count[str[R] - 'a'] == 0) {
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collect++;
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}
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if (count[str[R] - 'a'] == k - 1) {
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satisfy++;
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}
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count[str[R] - 'a']++;
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}
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// [L...R]
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if (satisfy == require) {
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max = Math.max(max, R - L + 1);
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}
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// L++
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if (count[str[L] - 'a'] == 1) {
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collect--;
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}
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if (count[str[L] - 'a'] == k) {
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satisfy--;
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}
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count[str[L] - 'a']--;
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}
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}
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return max;
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}
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// 会超时，但是思路的确是正确的
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public static int longestSubstring3(String s, int k) {
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return process(s.toCharArray(), 0, s.length() - 1, k);
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}
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public static int process(char[] str, int L, int R, int k) {
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if (L > R) {
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return 0;
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}
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int[] counts = new int[26];
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for (int i = L; i <= R; i++) {
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counts[str[i] - 'a']++;
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}
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char few = 0;
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int min = Integer.MAX_VALUE;
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for (int i = 0; i < 26; i++) {
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if (counts[i] != 0 && min > counts[i]) {
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few = (char) (i + 'a');
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min = counts[i];
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}
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}
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if (min >= k) {
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return R - L + 1;
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}
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int pre = 0;
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int max = Integer.MIN_VALUE;
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for (int i = L; i <= R; i++) {
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if (str[i] == few) {
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max = Math.max(max, process(str, pre, i - 1, k));
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pre = i + 1;
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}
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}
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if (pre != R + 1) {
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max = Math.max(max, process(str, pre, R, k));
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}
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return max;
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}
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}
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