Huahua's Tech Road

花花酱 LeetCode 1314. Matrix Block Sum

By zxi on January 11, 2020

Given a m * n matrix mat and an integer K, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for i - K <= r <= i + K, j - K <= c <= j + K, and (r, c) is a valid position in the matrix.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]

Example 2:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n, K <= 100
1 <= mat[i][j] <= 100

Solution: 2D range query

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

C++

// Author: Huahua, 20 ms

class Solution {

public:

vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int K) {

const int n = mat.size();

const int m = mat[0].size();

vector<vector<int>> dp(n + 1, vector<int>(m + 1));

for (int y = 1; y <= n; ++y)

for (int x = 1; x <= m; ++x)

dp[y][x] = dp[y - 1][x] + dp[y][x - 1] + mat[y - 1][x - 1] - dp[y - 1][x - 1];

auto ans = mat;

for (int y = 1; y <= n; ++y)

for (int x = 1; x <= m; ++x) {

int x1 = max(1, x - K);

int x2 = min(m, x + K);

int y1 = max(1, y - K);

int y2 = min(n, y + K);

ans[y - 1][x - 1] = dp[y2][x2] - dp[y1 - 1][x2] - dp[y2][x1 - 1] + dp[y1 - 1][x1 - 1];

}

return ans;

}

};

花花酱 LeetCode 1315. Sum of Nodes with Even-Valued Grandparent

By zxi on January 11, 2020

Given a binary tree, return the sum of values of nodes with even-valued grandparent. (A grandparent of a node is the parent of its parent, if it exists.)

If there are no nodes with an even-valued grandparent, return 0.

Example 1:

Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
Output: 18
Explanation: The red nodes are the nodes with even-value grandparent while the blue nodes are the even-value grandparents.

Constraints:

The number of nodes in the tree is between 1 and 10^4.
The value of nodes is between 1 and 100.

Solution: Recursion

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

C++

// Author: Huahua

class Solution {

public:

int sumEvenGrandparent(TreeNode* root, int p = 1, int gp = 1) {

if (!root) return 0;

return sumEvenGrandparent(root->left, root->val, p)

+ sumEvenGrandparent(root->right, root->val, p)

+ (gp & 1 ? 0 : root->val);

}

};

花花酱 LeetCode 1316. Distinct Echo Substrings

By zxi on January 11, 2020

Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself.

Example 1:

Input: text = "abcabcabc"
Output: 3
Explanation: The 3 substrings are "abcabc", "bcabca" and "cabcab".

Example 2:

Input: text = "leetcodeleetcode"
Output: 2
Explanation: The 2 substrings are "ee" and "leetcodeleetcode".

Constraints:

1 <= text.length <= 2000
text has only lowercase English letters.

Solution 1: Brute Force + HashSet

Try all possible substrings

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

int distinctEchoSubstrings(string text) {

const int n = text.length();

string_view t(text);

unordered_set<string_view> s;

for (int k = 1; k <= n / 2; ++k)

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

if (t.substr(i, k) == t.substr(i + k, k))

s.insert(t.substr(i, 2 * k));

return s.size();

}

};

花花酱 LeetCode 1310. XOR Queries of a Subarray

By zxi on January 5, 2020

Given the array arr of positive integers and the array queries where queries[i] = [L_i,R_i], for each query i compute the XOR of elements from L_i to Ri (that is, arr[L_i] xor arr[L_i+1] xor ... xor arr[R_i] ). Return an array containing the result for the given queries.

Example 1:

Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
Output: [2,7,14,8] 
Explanation: 
The binary representation of the elements in the array are:
1 = 0001 
3 = 0011 
4 = 0100 
8 = 1000 
The XOR values for queries are:
[0,1] = 1 xor 3 = 2 
[1,2] = 3 xor 4 = 7 
[0,3] = 1 xor 3 xor 4 xor 8 = 14 
[3,3] = 8

Example 2:

Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
Output: [8,0,4,4]

Constraints:

1 <= arr.length <= 3 * 10^4
1 <= arr[i] <= 10^9
1 <= queries.length <= 3 * 10^4
queries[i].length == 2
0 <= queries[i][0] <= queries[i][1] < arr.length

Solution: Prefix Sum

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

C++

// Author: Huahua

class Solution {

public:

vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) {

// xors[i + 1] = arr[0] ^ arr[1] ^ ... ^ arr[i]

vector<int> xors(arr.size() + 1);

for (int i = 0; i < arr.size(); ++i) {

xors[i + 1] = xors[i] ^ arr[i];

}

vector<int> ans;

for (const auto& q : queries) {

const int l = q[0];

const int r = q[1];

ans.push_back(xors[r + 1] ^ xors[l]);

}

return ans;

}

};

tby

花花酱 LeetCode 1312. Minimum Insertion Steps to Make a String Palindrome

By zxi on January 4, 2020

Given a string s. In one step you can insert any character at any index of the string.

Return the minimum number of steps to make s palindrome.

A Palindrome String is one that reads the same backward as well as forward.

Example 1:

Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we don't need any insertions.

Example 2:

Input: s = "mbadm"
Output: 2
Explanation: String can be "mbdadbm" or "mdbabdm".

Example 3:

Input: s = "leetcode"
Output: 5
Explanation: Inserting 5 characters the string becomes "leetcodocteel".

Example 4:

Input: s = "g"
Output: 0

Example 5:

Input: s = "no"
Output: 1

Constraints:

1 <= s.length <= 500
All characters of s are lower case English letters.

Solution: DP

dp[i][j] := min chars to insert
dp[j][j] = dp[i-1][j+1] if s[i] == s[j] else min(dp[i+1][j] , dp[i][j-1]) + 1
base case: dp[i][i] = 0
ans: dp[0][n-1]

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

C++

// Author: Huahua

class Solution {

public:

int minInsertions(string s) {

const int n = s.length();

vector<vector<int>> dp(n, vector<int>(n));

for (int l = 2; l <= n; ++l)

for (int i = 0, j = l - 1; j < n; ++i, ++j)

dp[i][j] = s[i] == s[j] ? dp[i + 1][j - 1] : min(dp[i + 1][j], dp[i][j - 1]) + 1;

return dp[0][n - 1];

}

};