Huahua’s Tech Road

花花酱 LeetCode 985. Sum of Even Numbers After Queries

By zxi on February 2, 2019

Problem

We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index]. Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries. Your answer array should have answer[i] as the answer to the i-th query.

Example 1:

Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: 
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Note:

1 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length

Solution: Simulation

Time complexity: O(n + |Q|)
Space complexity: O(n)

C++

// Author: Huahua, Running time: 100 ms / 6.5 MB

class Solution {

public:

vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {

int sum = 0;

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

if (A[i] % 2 == 0)

sum += A[i];

vector<int> ans;

for (const auto& query : queries) {

int& a = A[query[1]];

if (a % 2 == 0)

sum -= a;

a += query[0];

if (a % 2 == 0)

sum += a;

ans.push_back(sum);

}

return ans;

}

};

Python3

# Author: Huahua, running time: 188 ms / 16.4MB

class Solution:

def sumEvenAfterQueries(self, A: 'List[int]', queries: 'List[List[int]]') -> 'List[int]':

s = sum(x for x in A if x % 2 == 0)

ans = []

for val, index in queries:

if A[index] % 2 == 0: s -= A[index]

A[index] += val

if A[index] % 2 == 0: s += A[index]

ans.append(s)

return ans

花花酱 LeetCode 778. Swim in Rising Water

By zxi on January 31, 2019

On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Note:

2 <= N <= 50.
grid[i][j] is a permutation of [0, …, N*N – 1].

Solution 1: Dijkstra’s Algorithm

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

C++

// Author: Huahua, running time: 8 ms

class Solution {

public:

int swimInWater(vector<vector<int>>& grid) {

const int n = grid.size();

priority_queue<pair<int, int>> q; // {-time, y * N + x}

q.push({-grid[0][0], 0 * n + 0});

vector<int> seen(n * n);

vector<int> dirs{-1, 0, 1, 0, -1};

seen[0 * n + 0] = 1;

while (!q.empty()) {

auto node = q.top(); q.pop();

int t = -node.first;

int x = node.second % n;

int y = node.second / n;

if (x == n - 1 && y == n - 1) return t;

for (int i = 0; i < 4; ++i) {

int tx = x + dirs[i];

int ty = y + dirs[i + 1];

if (tx < 0 || ty < 0 || tx >= n || ty >= n) continue;

if (seen[ty * n + tx]) continue;

seen[ty * n + tx] = 1;

q.push({-max(t, grid[ty][tx]), ty * n + tx});

}

return -1;

}

};

Solution 2: Binary Search + BFS

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

C++

// Author: Huahua, running time: 8 ms

class Solution {

public:

int swimInWater(vector<vector<int>>& grid) {

const int n = grid.size();

auto hasPath = [&grid, n](int t) {

if (grid[0][0] > t) return false;

queue<int> q;

vector<int> seen(n * n);

vector<int> dirs{1, 0, -1, 0, 1};

q.push(0);

while (!q.empty()) {

const int x = q.front() % n;

const int y = q.front() / n;

q.pop();

if (x == n - 1 && y == n - 1) return true;

for (int i = 0; i < 4; ++i) {

const int tx = x + dirs[i];

const int ty = y + dirs[i + 1];

if (tx < 0 || ty < 0 || tx >= n || ty >= n || grid[ty][tx] > t) continue;

const int key = ty * n + tx;

if (seen[key]) continue;

seen[key] = 1;

q.push(key);

}

return false;

};

int l = 0;

int r = n * n;

while (l < r) {

int m = l + (r - l) / 2;

if (hasPath(m)) {

r = m;

} else {

l = m + 1;

}

return l;

}

};

花花酱 Segment Tree 线段树 SP14

By zxi on January 30, 2019

Segment tree is a balanced binary tree with O(logn) height given n input segments.
Segment tree supports fast range query O(logn + k), and update O(logn).
Building such a tree takes O(n) time if the input is an array of numbers.

Tree之所以大行其道就是因为其结构和人类组织结构非常接近。就拿公司来说好了，CEO统领全局(root)，下面CTO，CFO等各自管理一个部门，每个部门下面又正好有好两个VP，每个VP下面又有两个director，每个director下面又有2个manager，每个manager下面又有2个底层员工(leaf)，为什么是2？因为我们用二叉树啊~

故事还是要从LeetCode 307. Range Sum Query – Mutable说起。

题目大意是：给你一个数组，再给你一个范围，让你求这个范围内所有元素的和，其中元素的值是可变的，通过update(index, val)更新。

nums = [1, 3, 5]，
sumRange(0, 2) = 1+3+5 = 9
update(1, 2) => [1, 2, 5]
sumRange(0, 2) = 1 + 2 + 5 = 7

暴力求解就是扫描一下这个范围。

时间复杂度：Update O(1), Query O(n)。

如果数组元素不变的话（303题），我们可以使用动态规划求出前n个元素的和然后存在sums数组中。i到j所有元素的和等于0~j所有元素的和减去0~(i-1)所有元素的和，即：

sumRange(i, j) := sums[j] – sums[i – 1] if i > 0 else sums[j]

这样就可以把query的时间复杂度降低到O(1)。

但是这道题元素的值可变，那么就需要维护sums，虽然可以把query的时间复杂度降低到了O(1)，但update的时间复杂度是O(n)，并没有比暴力求解快。

这个时候就要请出我们今天的主人公Segment Tree了，可以做到

Update: O(logn)，Query: O(logn+k)

其实Segment Tree的思想还是很好理解的，比我们之前讲过的Binary Indexed Tree要容易理解的多（回复 SP3 获取视频），但是代码量又是另外一回事情了…

回到一开始讲的公司组织结构上面，假设一个公司有200个底层员工，

最原始的信息（元素的值）只掌握在他们工手里，层层上报。每个manager把他所管理的人的元素值求和之后继续上报，直到CEO。CEO知道但仅知道0-199的和。

当你问CEO，5到199的和是多少？他手上只有0-199的和，一下子慌了，赶紧找来CTO，CFO说你们把5到199的和给报上来，CFO一看报表，心中暗喜：还好我负责的这个区间(100~199)已经计算过了，就直接把之前的总和上报了。CTO一看报表，自己负责0-99这个区间，只知道0-99的和，但5-99的和，还是问下面的人吧… 最后结果再一层层返回给CEO。

说到这里大家应该已经能想象Segment Tree是怎么工作了吧：

每个leaf负责一个元素的值

每个parent负责的范围是他的children所负责的范围的union，并把所有范围内的元素值相加。

同一层的节点没有overlap。

root存储的是所有元素的和。

所以一个SegmentTreeNode需要记录以下信息

start #起始范围
end #终止范围
mid #拆分点，通常是 (start + end) // 2
val #所有子元素的和
left #左子树
right #右子树

Update: 由于每次只更新一个元素的值，所以CEO知道这个员工是哪个人管的，派发下去就行了，然后把新的结果再返回回来。

def update(root, index, val):

# 到底层员工了，直接更新

if root.start == index and root.end == index:

root.val = val

return

# 根据拆分点，更新左子树或右子树

if val <= root.mid:

update(root.left, index, val)

else:

update(root.right, index, val)

# 重新计算和

root.val = root.left?.val + root.right?.val

T(n) = T(n/2) + 1 => O(logn)

Query: query的range可以是任意的，就有以下三种情况：

case 1: 这个range正好和我负责的range完全相同。比如sumQuery(CTO, 0, 99)，这个时候CTO直接返回已经求解过的所有下属的和即可。

case 2: 这个range只由我其中一个下属负责。比如sumQuery(CEO, 0, 10)，CEO知道0~10全部由CFO负责，那么他就直接返回sumQuery(CTO, 0, 10)。

case 3: 这个range覆盖了我的两个下属，那么我就需要调用2次。比如sumQuery(CEO, 80, 120)，CEO知道0~99由CTO管，100~199由CFO管，所以他只需要返回：

sumQuery(CTO, 80, 99) + sumQuery(CFO, 100, 120)

由此可见sumQuery的时间复杂度是不确定的，运气好时O(1)，运气不好时是O(logn+k)，k是需要访问的节点的数量。

我做了一个实验，给定N，尝试所有的(i,j)组合，看sumQuery的最坏情况和平均情况，结果如下图：

Query需要访问到的节点数量的worst和average case。Worst case 大约访问 4*logn – 3 个节点，这个数字远远小于n。和n成对数关系。

虽然不像Binary Indexed Tree query是严格的O(logn)，但Segment tree query的worst case增长的非常慢，可以说是对数级别的。当N是2^20的时候，worst case也“只需要”访问77个节点。

最后我们再来讲一下这棵树是怎么创建的，其实方法有很多种，一分为二是比较常规的一种做法。

CEO管200人，那么就找2个人(CTO，CFO各管100人。CTO管100人，再找2个VP各管50人，以此类推，直到manager管2个人，2个人都是底层员工(leaf)，没有人管（双关）。

def buildTree(start, end):

# 超过范围返回空节点

if start > end: return None

boss.start = start

boss.end = end

# 如果是叶子节点，记录元素的值

if start == end: boss.val = nums[start]

boss.mid = (start + end) // 2

# 递归构建子树

boss.left = buildTree(start, mid)

boss.right = buildTree(mid + 1, end)

# 求和

boss.val += boss.left?.val + boss.right?.val

return boss

CEO = buildTree(0, 199)

CEO.left # CTO 负责0~99
CEO.right # CFO 负责100~199

Query: # of nodes visited: Average and worst are both ~O(logn)

Applications

LeetCode 307 Range Sum Query – Mutable

C++

// Author: Huahua, running time: 24 ms

class SegmentTreeNode {

public:

SegmentTreeNode(int start, int end, int sum,

SegmentTreeNode* left = nullptr,

SegmentTreeNode* right = nullptr):

start(start),

end(end),

sum(sum),

left(left),

right(right){}

SegmentTreeNode(const SegmentTreeNode&) = delete;

SegmentTreeNode& operator=(const SegmentTreeNode&) = delete;

~SegmentTreeNode() {

delete left;

delete right;

left = right = nullptr;

}

int start;

int end;

int sum;

SegmentTreeNode* left;

SegmentTreeNode* right;

};

class NumArray {

public:

NumArray(vector<int> nums) {

nums_.swap(nums);

if (!nums_.empty())

root_.reset(buildTree(0, nums_.size() - 1));

}

void update(int i, int val) {

updateTree(root_.get(), i, val);

}

int sumRange(int i, int j) {

return sumRange(root_.get(), i, j);

}

private:

vector<int> nums_;

std::unique_ptr<SegmentTreeNode> root_;

SegmentTreeNode* buildTree(int start, int end) {

if (start == end) {

return new SegmentTreeNode(start, end, nums_[start]);

}

int mid = start + (end - start) / 2;

auto left = buildTree(start, mid);

auto right = buildTree(mid + 1, end);

auto node = new SegmentTreeNode(start, end, left->sum + right->sum, left, right);

return node;

}

void updateTree(SegmentTreeNode* root, int i, int val) {

if (root->start == i && root->end == i) {

root->sum = val;

return;

}

int mid = root->start + (root->end - root->start) / 2;

if (i <= mid) {

updateTree(root->left, i, val);

} else {

updateTree(root->right, i, val);

}

root->sum = root->left->sum + root->right->sum;

}

int sumRange(SegmentTreeNode* root, int i, int j) {

if (i == root->start && j == root->end) {

return root->sum;

}

int mid = root->start + (root->end - root->start) / 2;

if (j <= mid) {

return sumRange(root->left, i, j);

} else if (i > mid) {

return sumRange(root->right, i, j);

} else {

return sumRange(root->left, i, mid) + sumRange(root->right, mid + 1, j);

}

};

Python3

# Author: Huahua, running time: 176 ms

class SegmentTreeNode:

def __init__(self, start, end, val, left=None, right=None):

self.start = start

self.end = end

self.mid = start + (end - start) // 2

self.val = val

self.left = left

self.right = right

class NumArray:

def __init__(self, nums):

self.nums = nums

if self.nums:

self.root = self._buildTree(0, len(nums) - 1)

def update(self, i, val):

self._updateTree(self.root, i, val)

def sumRange(self, i, j):

return self._sumRange(self.root, i, j)

def _buildTree(self, start, end):

if start == end: return SegmentTreeNode(start, end, self.nums[start])

mid = start + (end - start) // 2

left = self._buildTree(start, mid)

right = self._buildTree(mid + 1, end)

return SegmentTreeNode(start, end, left.val + right.val, left, right)

def _updateTree(self, root, i, val):

if root.start == i and root.end == i:

root.val = val

return

if i <= root.mid:

self._updateTree(root.left, i, val)

else:

self._updateTree(root.right, i, val)

root.val = root.left.val + root.right.val

def _sumRange(self, root, i, j):

if root.start == i and root.end == j:

return root.val

if j <= root.mid:

return self._sumRange(root.left, i, j)

elif i > root.mid:

return self._sumRange(root.right, i, j)

else:

return self._sumRange(root.left, i, root.mid) + self._sumRange(root.right, root.mid + 1, j)

花花酱 LeetCode 984. String Without AAA or BBB

By zxi on January 27, 2019

Given two integers A and B, return any string S such that:

S has length A + B and contains exactly A 'a' letters, and exactly B 'b' letters;
The substring 'aaa' does not occur in S;
The substring 'bbb' does not occur in S.

Example 1:

Input: A = 1, B = 2
Output: "abb"
Explanation: "abb", "bab" and "bba" are all correct answers.

Example 2:

Input: A = 4, B = 1
Output: "aabaa"

Note:

0 <= A <= 100
0 <= B <= 100
It is guaranteed such an S exists for the given A and B.

C++

class Solution {

public:

string strWithout3a3b(int A, int B) {

char a = 'a';

char b = 'b';

if (B > A) {

swap(A, B);

swap(a, b);

}

string ans;

while (A || B) {

if (A > 0) { ans += a; --A; }

if (A > B) { ans += a; --A; }

if (B > 0) { ans += b; --B; }

}

return ans;

}

};

C++ / overload

class Solution {

public:

string strWithout3a3b(int A, int B, char a = 'a', char b = 'b') {

if (B > A) return strWithout3a3b(B, A, b, a);

string ans;

while (A || B) {

if (A > 0) { ans += a; --A; }

if (A > B) { ans += a; --A; }

if (B > 0) { ans += b; --B; }

}

return ans;

}

};

花花酱 LeetCode Weekly Contest 121

By zxi on January 27, 2019

Huahua's Tech Road

花花酱 LeetCode 985. Sum of Even Numbers After Queries

Problem

Solution: Simulation

C++

Python3

花花酱 LeetCode 778. Swim in Rising Water

Solution 1: Dijkstra’s Algorithm

C++

Solution 2: Binary Search + BFS

C++

花花酱 Segment Tree 线段树 SP14

Applications

C++

Python3

花花酱 LeetCode 984. String Without AAA or BBB

C++

C++ / overload

花花酱 LeetCode Weekly Contest 121