kmp Archives - Huahua's Tech Road

花花酱 LeetCode 1764. Form Array by Concatenating Subarrays of Another Array

By zxi on February 20, 2021

You are given a 2D integer array groups of length n. You are also given an integer array nums.

You are asked if you can choose n disjoint subarrays from the array nums such that the i^th subarray is equal to groups[i] (0-indexed), and if i > 0, the (i-1)^th subarray appears before the i^th subarray in nums (i.e. the subarrays must be in the same order as groups).

Return true if you can do this task, and false otherwise.

Note that the subarrays are disjoint if and only if there is no index k such that nums[k] belongs to more than one subarray. A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: groups = [[1,-1,-1],[3,-2,0]], nums = [1,-1,0,1,-1,-1,3,-2,0]
Output: true
Explanation: You can choose the 0^th subarray as [1,-1,0,1,-1,-1,3,-2,0] and the 1^st one as [1,-1,0,1,-1,-1,3,-2,0].
These subarrays are disjoint as they share no common nums[k] element.

Example 2:

Input: groups = [[10,-2],[1,2,3,4]], nums = [1,2,3,4,10,-2]
Output: false
Explanation: Note that choosing the subarrays [1,2,3,4,10,-2] and [1,2,3,4,10,-2] is incorrect because they are not in the same order as in groups.
[10,-2] must come before [1,2,3,4].

Example 3:

Input: groups = [[1,2,3],[3,4]], nums = [7,7,1,2,3,4,7,7]
Output: false
Explanation: Note that choosing the subarrays [7,7,1,2,3,4,7,7] and [7,7,1,2,3,4,7,7] is invalid because they are not disjoint.
They share a common elements nums[4] (0-indexed).

Constraints:

groups.length == n
1 <= n <= 10³
1 <= groups[i].length, sum(groups[i].length) <= 10³
1 <= nums.length <= 10³
-10⁷ <= groups[i][j], nums[k] <= 10⁷

Solution: Brute Force

Time complexity: O(n^2?)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool canChoose(vector<vector<int>>& groups, vector<int>& nums) {

const int n = nums.size();

int s = 0;

for (const auto& g : groups) {

bool found = false;

for (int i = s; i <= n - g.size(); ++i)

if (equal(begin(g), end(g), begin(nums) + i)) {

s = i + g.size();

found = true;

break;

}

if (!found) return false;

}

return true;

}

};

KMP Algorithm SP19

By zxi on March 31, 2020

KMP Algorithm, KMP 字符串搜索算法

Time complexity: O(m+n)
Space complexity: O(m)

Implementation

C++

#include <iostream>

#include <vector>

using namespace std;

// Author: Huahua

namespace KMP {

vector<int> Build(const string& p) {

const int m = p.length();

vector<int> nxt{0, 0};

for (int i = 1, j = 0; i < m; ++i) {

while (j > 0 && p[i] != p[j])

j = nxt[j];

if (p[i] == p[j])

++j;

nxt.push_back(j);

}

return nxt;

}

vector<int> Match(const string& s, const string& p) {

vector<int> nxt(Build(p));

vector<int> ans;

const int n = s.length();

const int m = p.length();

for (int i = 0, j = 0; i < n; ++i) {

while (j > 0 && s[i] != p[j])

j = nxt[j];

if (s[i] == p[j])

++j;

if (j == m) {

ans.push_back(i - m + 1);

j = nxt[j];

}

return ans;

}

}; // namespace KMP

void CheckEQ(const vector<int>& actual, const vector<int>& expected) {

if (actual != expected) {

std::cout << "expected:";

for (int v : expected)

std::cout << " " << v;

std::cout << " actual:";

for (int v : actual)

std::cout << " " << v;

std::cout << std::endl;

} else {

std::cout << "PASS" << std::endl;

}

int main(int argc, char** argv) {

CheckEQ(KMP::Build("ABCDABD"), {0, 0, 0, 0, 0, 1, 2, 0});

CheckEQ(KMP::Build("AB"), {0, 0, 0});

CheckEQ(KMP::Build("A"), {0, 0});

CheckEQ(KMP::Build("AA"), {0, 0, 1});

CheckEQ(KMP::Build("AAA"), {0, 0, 1, 2});

CheckEQ(KMP::Build("AABA"), {0, 0, 1, 0, 1});

CheckEQ(KMP::Match("ABC ABCDAB ABCDABCDABDE", "ABCDABD"), {15});

CheckEQ(KMP::Match("ABC ABCDAB ABCDABCDABDE", "AB"), {0, 4, 8, 11, 15, 19});

CheckEQ(KMP::Match("ABC ABCDAB ABCDABCDABDE", "B"), {1, 5, 9, 12, 16, 20});

CheckEQ(KMP::Match("AAAAA", "A"), {0, 1, 2, 3, 4});

CheckEQ(KMP::Match("AAAAA", "AA"), {0, 1, 2, 3});

CheckEQ(KMP::Match("AAAAA", "AAA"), {0, 1, 2});

CheckEQ(KMP::Match("ABC", "ABC"), {0});

return 0;

}

Python3

#!/usr/bin/env python3

from typing import List

# Author: Huahua

def Build(p: str) -> List[int]:

m = len(p)

nxt = [0, 0]

j = 0

for i in range(1, m):

while j > 0 and p[i] != p[j]:

j = nxt[j]

if p[i] == p[j]:

j += 1

nxt.append(j)

return nxt

def Match(s: str, p: str) -> List[int]:

n, m = len(s), len(p)

nxt = Build(p)

ans = []

j = 0

for i in range(n):

while j > 0 and s[i] != p[j]:

j = nxt[j]

if s[i] == p[j]:

j += 1

if j == m:

ans.append(i - m + 1)

j = nxt[j]

return ans

def CheckEQ(actual, expected):

if actual != expected:

print('actual: %s, expected: %s' % (actual, expected))

else:

print('Pass')

if __name__ == "__main__":

CheckEQ(Build("ABCDABD"), [0, 0, 0, 0, 0, 1, 2, 0])

CheckEQ(Build("AB"), [0, 0, 0])

CheckEQ(Build("A"), [0, 0])

CheckEQ(Build("AA"), [0, 0, 1])

CheckEQ(Build("AAAA"), [0, 0, 1, 2, 3])

CheckEQ(Build("AABA"), [0, 0, 1, 0, 1])

CheckEQ(Match("ABC ABCDAB ABCDABCDABDE", "ABCDABD"), [15])

CheckEQ(Match("ABC ABCDAB ABCDABCDABDE", "AB"), [0, 4, 8, 11, 15, 19])

CheckEQ(Match("ABC ABCDAB ABCDABCDABDE", "B"), [1, 5, 9, 12, 16, 20])

CheckEQ(Match("AAAAA", "A"), [0, 1, 2, 3, 4])

CheckEQ(Match("AAAAA", "AA"), [0, 1, 2, 3])

CheckEQ(Match("AAAAA", "AAAA"), [0, 1])

CheckEQ(Match("AAAAA", "AAAAA"), [0])

CheckEQ(Match("AABAABA", "AABA"), [0, 3])

Applications

LeetCode 28. strStr()

C++

// Author: Huahua

class Solution {

public:

int strStr(string haystack, string needle) {

if (needle.empty()) return 0;

auto matches = KMP::Match(haystack, needle);

return matches.empty() ? -1 : matches[0];

}

};

LeetCode 459. Repeated Substring Pattern

C++

// Author: Huahua

class Solution {

public:

bool repeatedSubstringPattern(string str) {

const int n = str.length();

auto nxt = KMP::Build(str);

return nxt[n] && nxt[n] % (n - nxt[n]) == 0;

}

};

1392. Longest Happy Prefix

C++

// Author: Huahua

class Solution {

public:

string longestPrefix(const string& s) {

return s.substr(0, KMP::Build(s).back());

}

};

花花酱 LeetCode 1392. Longest Happy Prefix

By zxi on March 21, 2020

A string is called a happy prefix if is a non-empty prefix which is also a suffix (excluding itself).

Given a string s. Return the longest happy prefix of s .

Return an empty string if no such prefix exists.

Example 1:

Input: s = "level"
Output: "l"
Explanation: s contains 4 prefix excluding itself ("l", "le", "lev", "leve"), and suffix ("l", "el", "vel", "evel"). The largest prefix which is also suffix is given by "l".

Example 2:

Input: s = "ababab"
Output: "abab"
Explanation: "abab" is the largest prefix which is also suffix. They can overlap in the original string.

Example 3:

Input: s = "leetcodeleet"
Output: "leet"

Example 4:

Input: s = "a"
Output: ""

Constraints:

1 <= s.length <= 10^5
s contains only lowercase English letters.

Solution: Rolling Hash

Time complexity: O(n) / worst case: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

string longestPrefix(string_view s) {

const int kMod = 16769023;

const int kBase = 128;

const int n = s.length();

int p = 0;

int q = 0;

int a = 1;

int ans = 0;

for (int i = 1; i < n; ++i) {

p = (p + a * s[i - 1]) % kMod;

q = (q * kBase + s[n - i]) % kMod;

a = (a * kBase) % kMod;

if (p == q && s.substr(0, i) == s.substr(n - i))

ans = i;

}

return string(s.substr(0, ans));

}

};

Posts tagged as “kmp”

花花酱 LeetCode 1764. Form Array by Concatenating Subarrays of Another Array

Solution: Brute Force

C++

KMP Algorithm SP19

Implementation

C++

Python3

Applications

C++

C++

C++

花花酱 LeetCode 1392. Longest Happy Prefix

Solution: Rolling Hash

C++