head Archives - Huahua's Tech Road

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;

}

};

Problem:

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list “parts”.

The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.

Return a List of ListNode’s representing the linked list parts that are formed.

Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]

Example 1:

Input:

root = [1, 2, 3], k = 5

Output: [[1],[2],[3],[],[]]

Explanation:

The input and each element of the output are ListNodes, not arrays.

For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.

The first element output[0] has output[0].val = 1, output[0].next = null.

The last element output[4] is null, but it's string representation as a ListNode is [].

Example 2:

Input:

root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3

Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]

Explanation:

The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

Note:

The length of root will be in the range [0, 1000].
Each value of a node in the input will be an integer in the range [0, 999].
k will be an integer in the range [1, 50].

Idea:

List + Simulation

Solution:

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

vector<ListNode*> splitListToParts(ListNode* root, int k) {

int len = 0;

for (ListNode* head = root; head; head = head->next) ++len;

vector<ListNode*> ans(k, nullptr);

int l = len / k;

int r = len % k;

ListNode* head = root;

ListNode* prev = nullptr;

for (int i = 0; i < k; ++i, --r) {

ans[i] = head;

for (int j = 0; j < l + (r > 0); ++j) {

prev = head;

head = head->next;

}

if (prev) prev->next = nullptr;

}

return ans;

}

};

Java

// Author: Huahua

// Runtime: 4 ms

class Solution {

public ListNode[] splitListToParts(ListNode root, int k) {

ListNode[] ans = new ListNode[k];

int len = 0;

for (ListNode head = root; head != null; head = head.next) ++len;

int l = len / k;

int r = len % k;

ListNode prev = null;

ListNode head = root;

for (int i = 0; i < k; ++i, --r) {

ans[i] = head;

int part_len = l + ((r > 0) ? 1 : 0);

for (int j = 0; j < part_len; ++j) {

prev = head;

head = head.next;

}

if (prev != null) prev.next = null;

}

return ans;

}

Python

"""

Author: Huahua

Runtime: 79 ms

"""

class Solution:

def splitListToParts(self, root, k):

total_len = 0

head = root

while head:

total_len += 1

head = head.next

ans = [None for _ in range(k)]

l, r = total_len // k, total_len % k

prev, head = None, root

for i in range(k):

ans[i] = head

for j in range(l + (1 if r > 0 else 0)):

prev, head = head, head.next

if prev: prev.next = None

r -= 1

return ans

Posts tagged as “head”

花花酱 LeetCode 778. Swim in Rising Water

Solution 1: Dijkstra’s Algorithm

C++

Solution 2: Binary Search + BFS

C++

Posts tagged as “head”

花花酱 LeetCode 778. Swim in Rising Water

Solution 1: Dijkstra’s Algorithm

C++

Solution 2: Binary Search + BFS

C++

花花酱 LeetCode 725. Split Linked List in Parts