2021-06-20 Daily-Challenge
Today is Sunday, I gonna review the tasks I've done this week, and finish today's leetcode's June LeetCoding Challenge with cpp.
LeetCode Review
K Inverse Pairs Array
try to optimize by template metaprogramming
const int MOD = 1e9 + 7;
const int SZ = 1000;
template<int N>
struct Table{
constexpr Table() : values() {
if(N < 1) return;
values[0][0] = 1;
for(int i = 1; i <= N; ++i) {
int nextJ = i * (i + 1) / 2;
int J = i * (i - 1) / 2;
J = J > SZ ? SZ : J;
nextJ = nextJ > SZ ? SZ : nextJ;
for(int j = 0; j <= J; ++j) {
values[i][j] = (j ? values[i][j - 1] : 0) + values[i - 1][j] - (i > j ? 0 : values[i - 1][j - i]);
values[i][j] = (values[i][j] % MOD + MOD) % MOD;
}
for(int j = J + 1; j <= nextJ; ++j) {
values[i][j] = values[i][j - 1];
}
}
}
int values[N + 1][SZ + 1] = {};
};
constexpr auto table = Table<333>();
int values[2][SZ + 1] = {};
void compute(int n) {
for(int i = 0; i <= SZ; ++i) values[1][i] = table.values[333][i];
for(int i = 334; i <= n; ++i) {
int parity = i & 1;
int J = i * (i - 1) / 2;
J = J > SZ ? SZ : J;
for(int j = 0; j <= J; ++j) {
values[parity][j] = (j ? values[parity][j - 1] : 0) + values[!parity][j] - (i > j ? 0 : values[!parity][j - i]);
values[parity][j] = (values[parity][j] % MOD + MOD) % MOD;
}
for(int j = J + 1; j <= SZ; ++j) {
values[parity][j] = values[parity][j - 1];
}
}
}
class Solution {
public:
int kInversePairs(int n, int k) {
if(!k) return 1;
if(k == 1) return n - 1;
int answer;
if(n <= 333) {
answer = table.values[n][k] - table.values[n][k - 1];
answer = (answer % MOD + MOD) % MOD;
} else {
compute(n);
int parity = n & 1;
answer = values[parity][k] - values[parity][k - 1];
answer = (answer % MOD + MOD) % MOD;
}
return answer;
}
};
// Accepted
// 80/80 cases passed (40 ms)
// Your runtime beats 51.87 % of cpp submissions
// Your memory usage beats 87.17 % of cpp submissions (6.6 MB)
then found out plain solution get best result:
const int N = 1001;
const int MOD = 1e9 + 7;
int values[N][N];
struct Solution {
Solution() {
if (values[0][0]) return;
values[0][0] = 1;
for (int i = 1; i < N; ++i) {
int sum = 0;
int end = min(N - 1, i * (i - 1) / 2);
for (int j = 0; j <= end; ++j) {
sum += values[i - 1][j];
if (j >= i) sum -= values[i - 1][j - i];
if (sum >= MOD) sum -= MOD;
else if (sum < 0) sum += MOD;
values[i][j] = sum;
}
}
}
int kInversePairs(int n, int k) const {
return values[n][k];
}
};
// Accepted
// 80/80 cases passed (4 ms)
// Your runtime beats 99.47 % of cpp submissions
// Your memory usage beats 85.03 % of cpp submissions (9.7 MB)
June LeetCoding Challenge 20
Description
Swim in Rising Water
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
simple bfs
const int moves[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
using ti = tuple<int, int, int>;
class Solution {
public:
int swimInWater(vector<vector<int>>& grid) {
int size = grid.size();
int maxTime = size * size - 1;
if(grid.front().front() == maxTime || grid.back().back() == maxTime) return maxTime;
vector<vector<bool>> vis(size, vector<bool>(size));
priority_queue<ti, vector<ti>, greater<ti>> pq;
pq.push({grid[0][0], 0, 0});
vis[0][0] = true;
while(pq.size()) {
auto [time, row, col] = pq.top();
pq.pop();
if(row == size - 1 && col == size - 1) return time;
for(int i = 0; i < 4; ++i) {
int newRow = row + moves[i][0];
int newCol = col + moves[i][1];
if(newRow < 0 || newRow >= size || newCol < 0 || newCol >= size) continue;
if(vis[newRow][newCol]) continue;
vis[newRow][newCol] = true;
pq.push({max(time, grid[newRow][newCol]), newRow, newCol});
}
}
return -1;
}
};
// 41 / 41 test cases passed.
// Status: Accepted
// Runtime: 12 ms
// Memory Usage: 8.5 MB
using pair make no difference.
using pi = tuple<int, int>;
class Solution {
const int moves[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public:
int swimInWater(vector<vector<int>>& grid) {
int size = grid.size();
int maxTime = size * size - 1;
if(grid.front().front() == maxTime || grid.back().back() == maxTime) return maxTime;
vector<vector<bool>> vis(size, vector<bool>(size));
priority_queue<pi, vector<pi>, greater<pi>> pq;
pq.push({grid[0][0], 0});
vis[0][0] = true;
while(pq.size()) {
auto [time, index] = pq.top();
int row = index / size;
int col = index % size;
pq.pop();
if(row == size - 1 && col == size - 1) return time;
for(int i = 0; i < 4; ++i) {
int newRow = row + moves[i][0];
int newCol = col + moves[i][1];
if(newRow < 0 || newRow >= size || newCol < 0 || newCol >= size) continue;
if(vis[newRow][newCol]) continue;
vis[newRow][newCol] = true;
pq.push({max(time, grid[newRow][newCol]), newRow * size + newCol});
}
}
return -1;
}
};
// 41 / 41 test cases passed.
// Status: Accepted
// Runtime: 16 ms
// Memory Usage: 8.6 MB