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-rw-r--r--max_diff_even_odd_freq2.c120
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+// 3445. Maximum Difference Between Even and Odd Frequency II
+#include "leetcode.h"
+
+/*
+ * you are given a string 's' and an integer 'k'. your task is to find the
+ * maximum difference between the frequency of two characters 'freq[a], -
+ * freq[b]', in a substring 'subs' of 's' such that 'subs' has a size of at
+ * least 'k', character 'a' has an odd frequency in 'subs', character 'b' has an
+ * even frequency in 'subs'. return the maximum difference. note that 'subs' can
+ * contain more than 2 distinct characters.
+ */
+
+static int upper_bound(const int *arr, int n, int key) {
+ int low = 0, high = n;
+ while (low < high) {
+ int mid = low + ((high - low) >> 1);
+ if (arr[mid] <= key)
+ low = mid + 1;
+ else
+ high = mid;
+ }
+ return low;
+}
+
+int maxDifference(char *s, int k) {
+ int n = strlen(s);
+ if (!n || k > n)
+ return -1;
+ int (*freq)[5] = malloc(sizeof(int[5]) * n), present[5] = {0};
+ if (!freq)
+ return -1;
+ for (int i = 0; i < n; i++) {
+ int c = s[i] - '0';
+ if (!i)
+ memset(freq[i], 0, sizeof(int) * 5);
+ else
+ memcpy(freq[i], freq[i - 1], sizeof(int) * 5);
+ freq[i][c]++;
+ present[c] = 1;
+ }
+ int ans = INT_MIN;
+ for (int a = 0; a < 5; a++) {
+ if (!present[a])
+ continue;
+ for (int b = 0; b < 5; b++) {
+ if (!present[b] || a == b)
+ continue;
+ int *idx_list[4], *val_list[4], size[4];
+ for (int i = 0; i < 4; i++) {
+ idx_list[i] = (int *)malloc(sizeof(int) * n);
+ val_list[i] = (int *)malloc(sizeof(int) * n);
+ size[i] = 0;
+ }
+ int odd = 0, even = 0;
+ for (int i = 0; i < n; i++) {
+ int c = s[i] - '0';
+ if (c == a)
+ odd++;
+ else if (c == b)
+ even++;
+ if ((odd & 1) && (even & 1) == 0 && (i + 1) >= k && even > 0) {
+ int diff = odd - even;
+ if (diff > ans)
+ ans = diff;
+ }
+ int key1 = ((odd % 2 == 0) ? 1 : 0) << 1 | ((even % 2 == 0) ? 0 : 1);
+ if (size[key1] > 0) {
+ int limit = i - k;
+ if (limit >= 0) {
+ int pos = upper_bound(idx_list[key1], size[key1], limit);
+ if (pos > 0) {
+ int prev = pos - 1, l = 0, r = prev, curr_min = INT_MAX;
+ while (l <= r) {
+ int mid = (l + r) >> 1, j = idx_list[key1][mid];
+ if (freq[i][b] != freq[j][b]) {
+ int v = val_list[key1][mid];
+ if (v < curr_min)
+ curr_min = v;
+ l = mid + 1;
+ } else
+ r = mid - 1;
+ }
+ if (curr_min != INT_MAX) {
+ int cur = odd - even - curr_min;
+ if (cur > ans)
+ ans = cur;
+ }
+ }
+ }
+ }
+ int key2 = (odd % 2) << 1 | (even % 2), curr_diff = odd - even;
+ if (!size[key2]) {
+ idx_list[key2][0] = i;
+ val_list[key2][0] = curr_diff;
+ size[key2] = 1;
+ } else {
+ int prev_min = val_list[key2][size[key2] - 1];
+ int new_min = (prev_min < curr_diff) ? prev_min : curr_diff;
+ idx_list[key2][size[key2]] = i;
+ val_list[key2][size[key2]] = new_min;
+ size[key2]++;
+ }
+ }
+ for (int i = 0; i < 4; i++) {
+ free(idx_list[i]);
+ free(val_list[i]);
+ }
+ }
+ }
+
+ free(freq);
+ return (ans == INT_MIN ? -1 : ans);
+}
+
+int main() {
+ char *s1 = "12233", *s2 = "1122211", *s3 = "110";
+ printf("%d\n", maxDifference(s1, 4)); // expect: -1
+ printf("%d\n", maxDifference(s2, 3)); // expect: 1
+ printf("%d\n", maxDifference(s3, 3)); // expect: -1
+}