| comments | difficulty | edit_url | rating | source | tags | ||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
true |
中等 |
1480 |
第 380 场周赛 Q2 |
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给你一个下标从 0 开始的字符串 s 、字符串 a 、字符串 b 和一个整数 k 。
如果下标 i 满足以下条件，则认为它是一个 美丽下标：
0 <= i <= s.length - a.lengths[i..(i + a.length - 1)] == a- 存在下标
j使得：0 <= j <= s.length - b.lengths[j..(j + b.length - 1)] == b|j - i| <= k
以数组形式按 从小到大排序 返回美丽下标。
示例 1：
输入：s = "isawsquirrelnearmysquirrelhouseohmy", a = "my", b = "squirrel", k = 15 输出：[16,33] 解释：存在 2 个美丽下标：[16,33]。 - 下标 16 是美丽下标，因为 s[16..17] == "my" ，且存在下标 4 ，满足 s[4..11] == "squirrel" 且 |16 - 4| <= 15 。 - 下标 33 是美丽下标，因为 s[33..34] == "my" ，且存在下标 18 ，满足 s[18..25] == "squirrel" 且 |33 - 18| <= 15 。 因此返回 [16,33] 作为结果。
示例 2：
输入：s = "abcd", a = "a", b = "a", k = 4 输出：[0] 解释：存在 1 个美丽下标：[0]。 - 下标 0 是美丽下标，因为 s[0..0] == "a" ，且存在下标 0 ，满足 s[0..0] == "a" 且 |0 - 0| <= 4 。 因此返回 [0] 作为结果。
提示：
1 <= k <= s.length <= 1051 <= a.length, b.length <= 10s、a、和b只包含小写英文字母。
class Solution:
def beautifulIndices(self, s: str, a: str, b: str, k: int) -> List[int]:
def build_prefix_function(pattern):
prefix_function = [0] * len(pattern)
j = 0
for i in range(1, len(pattern)):
while j > 0 and pattern[i] != pattern[j]:
j = prefix_function[j - 1]
if pattern[i] == pattern[j]:
j += 1
prefix_function[i] = j
return prefix_function
def kmp_search(pattern, text, prefix_function):
occurrences = []
j = 0
for i in range(len(text)):
while j > 0 and text[i] != pattern[j]:
j = prefix_function[j - 1]
if text[i] == pattern[j]:
j += 1
if j == len(pattern):
occurrences.append(i - j + 1)
j = prefix_function[j - 1]
return occurrences
prefix_a = build_prefix_function(a)
prefix_b = build_prefix_function(b)
resa = kmp_search(a, s, prefix_a)
resb = kmp_search(b, s, prefix_b)
res = []
print(resa, resb)
i = 0
j = 0
while i < len(resa):
while j < len(resb):
if abs(resb[j] - resa[i]) <= k:
res.append(resa[i])
break
elif j + 1 < len(resb) and abs(resb[j + 1] - resa[i]) < abs(
resb[j] - resa[i]
):
j += 1
else:
break
i += 1
return respublic class Solution {
public void computeLPS(String pattern, int[] lps) {
int M = pattern.length();
int len = 0;
lps[0] = 0;
int i = 1;
while (i < M) {
if (pattern.charAt(i) == pattern.charAt(len)) {
len++;
lps[i] = len;
i++;
} else {
if (len != 0) {
len = lps[len - 1];
} else {
lps[i] = 0;
i++;
}
}
}
}
public List<Integer> KMP_codestorywithMIK(String pat, String txt) {
int N = txt.length();
int M = pat.length();
List<Integer> result = new ArrayList<>();
int[] lps = new int[M];
computeLPS(pat, lps);
int i = 0; // Index for text
int j = 0; // Index for pattern
while (i < N) {
if (pat.charAt(j) == txt.charAt(i)) {
i++;
j++;
}
if (j == M) {
result.add(i - j); // Pattern found at index i-j+1 (If you have to return 1 Based
// indexing, that's why added + 1)
j = lps[j - 1];
} else if (i < N && pat.charAt(j) != txt.charAt(i)) {
if (j != 0) {
j = lps[j - 1];
} else {
i++;
}
}
}
return result;
}
private int lowerBound(List<Integer> list, int target) {
int left = 0, right = list.size() - 1, result = list.size();
while (left <= right) {
int mid = left + (right - left) / 2;
if (list.get(mid) >= target) {
result = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
public List<Integer> beautifulIndices(String s, String a, String b, int k) {
int n = s.length();
List<Integer> i_indices = KMP_codestorywithMIK(a, s);
List<Integer> j_indices = KMP_codestorywithMIK(b, s);
List<Integer> result = new ArrayList<>();
for (int i : i_indices) {
int left_limit = Math.max(0, i - k); // To avoid out of bound -> I used max(0, i-k)
int right_limit
= Math.min(n - 1, i + k); // To avoid out of bound -> I used min(n-1, i+k)
int lowerBoundIndex = lowerBound(j_indices, left_limit);
if (lowerBoundIndex < j_indices.size()
&& j_indices.get(lowerBoundIndex) <= right_limit) {
result.add(i);
}
}
return result;
}
}class Solution {
public:
vector<int> beautifulIndices(string s, string patternA, string patternB, int k) {
vector<int> beautifulIndicesA = kmpSearch(s, patternA);
vector<int> beautifulIndicesB = kmpSearch(s, patternB);
sort(beautifulIndicesB.begin(), beautifulIndicesB.end());
vector<int> result;
for (int indexA : beautifulIndicesA) {
int left = lower_bound(beautifulIndicesB.begin(), beautifulIndicesB.end(), indexA - k) - beautifulIndicesB.begin();
int right = lower_bound(beautifulIndicesB.begin(), beautifulIndicesB.end(), indexA + k + patternB.length()) - beautifulIndicesB.begin();
left = (left >= 0) ? left : -(left + 1);
right = (right >= 0) ? right : -(right + 1);
for (int indexB = left; indexB < right; indexB++) {
if (abs(beautifulIndicesB[indexB] - indexA) <= k) {
result.push_back(indexA);
break;
}
}
}
return result;
}
private:
vector<int> kmpSearch(string text, string pattern) {
vector<int> indices;
vector<int> pi = computePrefixFunction(pattern);
int q = 0;
for (int i = 0; i < text.length(); i++) {
while (q > 0 && pattern[q] != text[i]) {
q = pi[q - 1];
}
if (pattern[q] == text[i]) {
q++;
}
if (q == pattern.length()) {
indices.push_back(i - q + 1);
q = pi[q - 1];
}
}
return indices;
}
vector<int> computePrefixFunction(string pattern) {
int m = pattern.length();
vector<int> pi(m, 0);
int k = 0;
for (int q = 1; q < m; q++) {
while (k > 0 && pattern[k] != pattern[q]) {
k = pi[k - 1];
}
if (pattern[k] == pattern[q]) {
k++;
}
pi[q] = k;
}
return pi;
}
};func beautifulIndices(s string, a string, b string, k int) []int {
s_len := len(s)
a_len := len(a)
b_len := len(b)
final := make([]int, 0)
lps_a := make([]int, a_len)
lps_b := make([]int, b_len)
a_index := make([]int, 0)
b_index := make([]int, 0)
var pat func(lps []int, s_l int, pattern string)
pat = func(lps []int, s_l int, pattern string) {
l := 0
lps[0] = 0
i := 1
for i < s_l {
if pattern[i] == pattern[l] {
l++
lps[i] = l
i++
} else {
if l != 0 {
l = lps[l-1]
} else {
lps[i] = l
i++
}
}
}
}
pat(lps_a, a_len, a)
pat(lps_b, b_len, b)
var kmp func(pat string, pat_l int, lps []int, index *[]int)
kmp = func(pat string, pat_l int, lps []int, index *[]int) {
i := 0
j := 0
for s_len-i >= pat_l-j {
if s[i] == pat[j] {
i++
j++
}
if j == pat_l {
*index = append(*index, i-pat_l)
j = lps[j-1]
} else if s[i] != pat[j] {
if j != 0 {
j = lps[j-1]
} else {
i++
}
}
}
}
kmp(a, a_len, lps_a, &a_index)
kmp(b, b_len, lps_b, &b_index)
// fmt.Println(a_index, b_index)
i := 0
j := 0
for i < len(a_index) && j < len(b_index) {
if a_index[i]+k >= b_index[j] && a_index[i]-k <= b_index[j] {
final = append(final, a_index[i])
i++
} else if a_index[i]-k > b_index[j] {
j++
} else {
i++
}
}
return final
}