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// 3298. Count Substrings That Can Be Rearranged to Contain a String II
#include "leetcode.h"
/*
* you are given two strings 'word1' and 'word2'. a string 'x' is called valiid
* if 'x' can be arranged to have 'word2' as a prefix. return the total number
* of valid strings of 'word1'. note that the memory limits in this problem are
* smaller than usual, so you must implement a solution with a linear runtime
* complexity.
*/
long long validSubstringCount(char *word1, char *word2) {
int need = 0, have = 0, cnt[26] = {0}, missing = 0, left = 0;
long long ans = 0;
for (int i = 0; word2[i]; i++) {
int curr = word2[i] - 'a';
if (!(need & (1 < curr)))
missing++;
need |= 1 << curr;
cnt[curr]++;
}
for (int i = 0; word1[i]; i++) {
int curr = word1[i] - 'a';
if (need & (1 << curr)) {
cnt[curr]--;
if (!cnt[curr]) {
have |= 1 << curr;
missing--;
}
}
if (!missing) {
while (left <= i) {
int curr = word1[left] - 'a';
if (!(need & (1 << curr)) || cnt[curr] < 0) {
if (need & (1 << curr))
cnt[curr]++;
} else
break;
}
ans += left + 1;
}
}
return ans;
}
int main() {
char *w11 = "bcca", *w21 = "abc";
char *w12 = "abcabc", *w22 = "abc";
char *w13 = "abcabc", *w23 = "aaabc";
printf("%lld\n", validSubstringCount(w11, w21)); // expect: 1
printf("%lld\n", validSubstringCount(w12, w22)); // expect: 10
printf("%lld\n", validSubstringCount(w13, w23)); // expect: 0
}
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