Posts tagged as “easy”

花花酱 LeetCode 438. Find All Anagrams in a String

By zxi on May 28, 2018

Problem

题目大意：在s中找出所有p的Anagrams。

https://leetcode.com/problems/find-all-anagrams-in-a-string/description/

Given a string s and a non-empty string p, find all the start indices of p‘s anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

Solution: HashTable + Sliding Window

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 35 ms

class Solution {

public:

vector<int> findAnagrams(string s, string p) {

int n = s.length();

int l = p.length();

vector<int> ans;

vector<int> vp(26, 0);

vector<int> vs(26, 0);

for (char c : p) ++vp[c - 'a'];

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

if (i >= l) --vs[s[i - l] - 'a'];

++vs[s[i] - 'a'];

if (vs == vp) ans.push_back(i + 1 - l);

}

return ans;

}

};

花花酱 LeetCode 830. Positions of Large Groups

By zxi on May 13, 2018

Problem

In a string S of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".

Call a group large if it has 3 or more characters. We would like the starting and ending positions of every large group.

The final answer should be in lexicographic order.

Example 1:

Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the single large group with starting 3 and ending positions 6.

Example 2:

Input: "abc"
Output: []
Explanation: We have "a","b" and "c" but no large group.

Example 3:

Input: "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]

Note: 1 <= S.length <= 1000

Solution: Brute Force

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

vector<vector<int>> largeGroupPositions(string S) {

constexpr int kLargeGroupSize = 3;

const int n = S.size();

vector<vector<int>> ans;

int c = 0;

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

if (i == n || S[i] != S[i - 1]) {

if (c >= kLargeGroupSize) ans.push_back({i - c, i - 1});

c = 0;

}

return ans;

}

};

花花酱 LeetCode 832. Flipping an Image

By zxi on May 13, 2018

Problem

Given a binary matrix A, we want to flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed. For example, flipping [1, 1, 0] horizontally results in [0, 1, 1].

To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0. For example, inverting [0, 1, 1] results in [1, 0, 0].

Example 1:

Input: [[1,1,0],[1,0,1],[0,0,0]]
Output: [[1,0,0],[0,1,0],[1,1,1]]
Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]

Example 2:

Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]

Notes:

1 <= A.length = A[0].length <= 20
0 <= A[i][j] <= 1

Solution 1: Brute Force

Time complexity: O(m*n)

Space complexity: O(m*n)

C++

// Author: Huahua

// Running time: 15 ms

class Solution {

public:

vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {

auto B = A;

int m = A.size();

int n = A[0].size();

for (int i = 0; i < m; ++i)

for (int j = 0; j < n; ++j)

B[i][j] = 1 - A[i][n - j - 1];

return B;

}

};

花花酱 LeetCode 824. Goat Latin

By zxi on April 29, 2018

Problem

A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and uppercase letters only.

We would like to convert the sentence to “Goat Latin” (a made-up language similar to Pig Latin.)

The rules of Goat Latin are as follows:

If a word begins with a vowel (a, e, i, o, or u), append "ma" to the end of the word.
For example, the word ‘apple’ becomes ‘applema’.
If a word begins with a consonant (i.e. not a vowel), remove the first letter and append it to the end, then add "ma".
For example, the word "goat" becomes "oatgma".
Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1.
For example, the first word gets "a" added to the end, the second word gets "aa" added to the end and so on.

Return the final sentence representing the conversion from S to Goat Latin.

Example 1:

Input: "I speak Goat Latin"
Output: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"

Example 2:

Input: "The quick brown fox jumped over the lazy dog"
Output: "heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa"

Notes:

S contains only uppercase, lowercase and spaces. Exactly one space between each word.
1 <= S.length <= 100.

Solution: String

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

string toGoatLatin(string S) {

const string vowls = "aeiouAEIOU";

string ans;

string word;

int index = 0;

istringstream s(S);

while (s >> word) {

if (vowls.find(word[0]) == string::npos)

word = word.substr(1) + word[0];

ans += " " + word + "ma" + string(++index, 'a');

}

return ans.substr(1);

}

};

花花酱 LeetCode 821. Shortest Distance to a Character

By zxi on April 22, 2018

Problem

题目大意：输出字符串的每个字符到一个指定字符的对短距离。

https://leetcode.com/problems/shortest-distance-to-a-character/description/

Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.

Example 1:

Input: S = "loveleetcode", C = 'e'
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]

Note:

S string length is in [1, 10000].
C is a single character, and guaranteed to be in string S.
All letters in S and C are lowercase.

Solution: Two Pass

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 14 ms

class Solution {

public:

vector<int> shortestToChar(string S, char C) {

const int n = S.length();

vector<int> ans(n, INT_MAX);

int index = -1;

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

if (S[i] == C) index = i;

if (index < 0) continue;

ans[i] = abs(i - index);

}

index = -1;

for (int i = n - 1; i >= 0; --i) {

if (S[i] == C) index = i;

if (index < 0) continue;

ans[i] = min(ans[i], abs(i - index));

}

return ans;

}

};

// Author: Huahua

// Running time: 14 ms

class Solution {

public:

vector<int> shortestToChar(string S, char C) {

const int n = S.length();

vector<int> indices(2 * n); // 0, 1, ..., n - 1, n - 1, n - 2, ..., 0

std::iota(indices.begin(), indices.begin() + n, 0);

std::iota(indices.rbegin(), indices.rbegin() + n, 0);

vector<int> ans(n, INT_MAX);

int index = -n;

for (int i : indices) {

if (S[i] == C) index = i;

ans[i] = min(ans[i], abs(i - index));

}

return ans;

}

};