Posts tagged as “dp”

花花酱 LeetCode 871. Minimum Number of Refueling Stops

By zxi on July 14, 2018

Problem

A car travels from a starting position to a destination which is target miles east of the starting position.

Along the way, there are gas stations. Each station[i] represents a gas station that is station[i][0] miles east of the starting position, and has station[i][1] liters of gas.

The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses 1 liter of gas per 1 mile that it drives.

When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.

What is the least number of refueling stops the car must make in order to reach its destination? If it cannot reach the destination, return -1.

Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. If the car reaches the destination with 0 fuel left, it is still considered to have arrived.

Example 1:

Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.

Example 2:

Input: target = 100, startFuel = 1, stations = [[10,100]]
Output: -1
Explanation: We can't reach the target (or even the first gas station).

Example 3:

Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]]
Output: 2
Explanation: 
We start with 10 liters of fuel.
We drive to position 10, expending 10 liters of fuel.  We refuel from 0 liters to 60 liters of gas.
Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
and refuel from 10 liters to 50 liters of gas.  We then drive to and reach the target.
We made 2 refueling stops along the way, so we return 2.

Note:

1 <= target, startFuel, stations[i][1] <= 10^9
0 <= stations.length <= 500
0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target

Solution1: DP

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

const int n = stations.size();

// dp[i]: max distance to go with i stops.

vector<long> dp(n + 1, startFuel);

// For each station

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

// for j stops, start from high to low to avoid reusing station i

for (int j = i + 1; j >= 1; --j)

if (dp[j - 1] >= stations[i][0])

// if we can reach station i with j - 1 stops,

// we can reach dp[j - 1] + station[i][1] with j stops.

dp[j] = max(dp[j], dp[j - 1] + stations[i][1]);

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

if (dp[i] >= target) return i;

return -1;

}

};

Solution2: Priority Queue

Time complexity: O(nlogn)

Space complexity: O(n)

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

int cur = startFuel;

int stops = 0;

int i = 0;

priority_queue<int> q; // gas (high to low) of reachable stations.

while (true) {

if (cur >= target) return stops;

while (i < stations.size() && stations[i][0] <= cur)

q.push(stations[i++][1]);

if (q.empty()) break;

cur += q.top(); q.pop();

++stops;

}

return -1;

}

};

V2: Iterator

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

int cur = startFuel;

int stops = 0;

priority_queue<int> q; // gas (high to low) of reachable stations.

auto it = begin(stations);

while (true) {

if (cur >= target) return stops;

while (it != end(stations) && it->at(0) <= cur)

q.push(it++->at(1));

if (q.empty()) break;

cur += q.top(); q.pop();

++stops;

}

return -1;

}

};

Problem

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:
Input:

0 0 0
0 1 0
0 0 0

Output:

0 0 0
0 1 0
0 0 0

Example 2:
Input:

0 0 0
0 1 0
1 1 1

Output:

0 0 0
0 1 0
1 2 1

Note:

The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
The cells are adjacent in only four directions: up, down, left and right.

Solution 1: DP

Two passes:

down, right
up, left

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 132 ms

class Solution {

public:

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {

int m = matrix.size();

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

vector<vector<int>> ans(m, vector<int>(n, INT_MAX - m * n));

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

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

if (matrix[i][j]) {

if (i > 0) ans[i][j] = min(ans[i][j], ans[i - 1][j] + 1);

if (j > 0) ans[i][j] = min(ans[i][j], ans[i][j - 1] + 1);

} else {

ans[i][j] = 0;

}

for (int i = m - 1; i >= 0; --i)

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

if (i < m - 1) ans[i][j] = min(ans[i][j], ans[i + 1][j] + 1);

if (j < n - 1) ans[i][j] = min(ans[i][j], ans[i][j + 1] + 1);

}

return ans;

}

};

Solution 2: BFS

Start from all 0 cells and find shortest paths to rest of the cells.

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 136 ms

class Solution {

public:

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {

const int m = matrix.size();

const int n = matrix[0].size();

queue<pair<int, int>> q;

vector<vector<int>> seen(m, vector<int>(n, 0));

vector<vector<int>> ans(m, vector<int>(n, INT_MAX));

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

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

if (matrix[i][j] == 0) {

q.push({i, j});

seen[i][j] = 1;

}

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

int steps = 0;

while (!q.empty()) {

int size = q.size();

while (size--) {

auto pair = q.front(); q.pop();

int i = pair.first;

int j = pair.second;

ans[i][j] = steps;

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

int x = j + dirs[k];

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

if (x < 0 || x >= n || y < 0 || y >= m || seen[y][x]) continue;

seen[y][x] = 1;

q.push({y, x});

}

++steps;

}

return ans;

}

};

花花酱 LeetCode 410. Split Array Largest Sum

By zxi on July 3, 2018

Problem

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.

Note:
If n is the length of array, assume the following constraints are satisfied:

1 ≤ n ≤ 1000
1 ≤ m ≤ min(50, n)

Examples:

Input:
nums = [7,2,5,10,8]
m = 2

Output:
18

Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [7,2,5] and [10,8],
where the largest sum among the two subarrays is only 18.

Solution: DP

Time complexity: O(n^2*m)

Space complexity: O(n*m)

C++ / Recursion + Memorization

// Author: Huahua

// Running time: 111 ms

class Solution {

public:

int splitArray(vector<int>& nums, int m) {

const int n = nums.size();

sums_ = vector<int>(n);

mem_ = vector<vector<int>>(n, vector<int>(m + 1, INT_MAX));

sums_[0] = nums[0];

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

sums_[i] = sums_[i - 1] + nums[i];

return splitArray(nums, n - 1, m);

}

private:

vector<vector<int>> mem_;

vector<int> sums_;

// min of largest sum of spliting nums[0] ~ nums[k] into m groups

int splitArray(const vector<int>& nums, int k, int m) {

if (m == 1) return sums_[k];

if (m > k + 1) return INT_MAX;

if (mem_[k][m] != INT_MAX) return mem_[k][m];

int ans = INT_MAX;

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

ans = min(ans, max(splitArray(nums, i, m - 1), sums_[k] - sums_[i]));

return mem_[k][m] = ans;

}

};

C++ / DP

// Author: Huahua

// Running time: 90 ms

class Solution {

public:

int splitArray(vector<int>& nums, int m) {

const int n = nums.size();

vector<int> sums(n);

// dp[i][j] := min of largest sum of splitting nums[0] ~ nums[j] into i groups.

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

sums[0] = nums[0];

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

sums[i] = sums[i - 1] + nums[i];

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

dp[1][i] = sums[i];

for (int i = 2; i <= m; ++i)

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

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

dp[i][j] = min(dp[i][j], max(dp[i - 1][k], sums[j] - sums[k]));

return dp[m][n - 1];

}

};

Solution: Binary Search

Time complexity: O(log(sum(nums))*n)

Space complexity: O(1)

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

int splitArray(vector<int>& nums, int m) {

long l = *max_element(begin(nums), end(nums));

long r = accumulate(begin(nums), end(nums), 0L) + 1;

while (l < r) {

long limit = (r - l) / 2 + l;

if (min_groups(nums, limit) > m)

l = limit + 1;

else

r = limit;

}

return l;

}

private:

int min_groups(const vector<int>& nums, long limit) {

long sum = 0;

int groups = 1;

for (int num : nums) {

if (sum + num > limit) {

sum = num;

++groups;

} else {

sum += num;

}

return groups;

}

};

花花酱 LeetCode 845. Longest Mountain in Array

By zxi on June 3, 2018

Problem

题目大意：找出最长的山形子数组。

https://leetcode.com/problems/longest-mountain-in-array/description/

Let’s call any (contiguous) subarray B (of A) a mountain if the following properties hold:

B.length >= 3
There exists some 0 < i < B.length - 1 such that B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]

(Note that B could be any subarray of A, including the entire array A.)

Given an array A of integers, return the length of the longest mountain.

Return 0 if there is no mountain.

Example 1:

Input: [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.

Example 2:

Input: [2,2,2]
Output: 0
Explanation: There is no mountain.

Note:

0 <= A.length <= 10000
0 <= A[i] <= 10000

Solution: DP

Three passes

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int longestMountain(vector<int>& A) {

vector<int> inc(A.size());

vector<int> dec(A.size());

for (int i = 1; i < A.size(); ++i)

if (A[i] > A[i - 1]) inc[i] = inc[i - 1] + 1;

for (int i = A.size() - 2; i > 0; --i)

if (A[i] > A[i + 1]) dec[i] = dec[i + 1] + 1;

int ans = 0;

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

if (inc[i] && dec[i])

ans = max(ans, inc[i] + dec[i] + 1);

return ans >= 3 ? ans : 0;

}

};

One pass

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int longestMountain(vector<int>& A) {

int inc = 0;

int dec = 0;

int ans = 0;

for (int i = 1; i < A.size(); ++i) {

if (dec && A[i] > A[i - 1] || A[i] == A[i - 1])

dec = inc = 0;

inc += A[i] > A[i - 1];

dec += A[i] < A[i - 1];

if (inc && dec)

ans = max(ans, inc + dec + 1);

}

return ans >= 3 ? ans : 0;

}

};

花花酱 LeetCode 576. Out of Boundary Paths

By zxi on June 2, 2018

Problem

There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 10⁹ + 7.

Example 1:

Input:m = 2, n = 2, N = 2, i = 0, j = 0
Output: 6
Explanation:

Example 2:

Input:m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

Note:

Once you move the ball out of boundary, you cannot move it back.
The length and height of the grid is in range [1,50].
N is in range [0,50].

Solution

Time complexity: O(m*n + N * m * n)

Space complexity: O(N*m*n) -> O(m*n)

C++

// Author: Huahua

// Running time: 19 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

constexpr int kMod = 1000000007;

vector<vector<vector<int>>> dp(N + 1, vector<vector<int>>(m, vector<int>(n, 0)));

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

for (int s = 1; s <= N; ++s)

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

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

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

int tx = x + dirs[d];

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

if (tx < 0 || ty < 0 || tx >= n || ty >= m)

dp[s][y][x] += 1;

else

dp[s][y][x] = (dp[s][y][x] + dp[s - 1][ty][tx]) % kMod;

}

return dp[N][i][j];

}

};

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

constexpr int kMod = 1000000007;

vector<vector<int>> dp(m, vector<int>(n, 0));

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

for (int s = 1; s <= N; ++s) {

vector<vector<int>> tmp(m, vector<int>(n, 0));

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

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

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

int tx = x + dirs[d];

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

if (tx < 0 || ty < 0 || tx >= n || ty >= m)

tmp[y][x] += 1;

else

tmp[y][x] = (tmp[y][x] + dp[ty][tx]) % kMod;

}

dp.swap(tmp);

}

return dp[i][j];

}

};

Recursion with memorization

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

m_ = m;

n_ = n;

dp_ = vector<vector<vector<int>>>(N + 1, vector<vector<int>>(m, vector<int>(n, INT_MIN)));

return paths(N, j, i);

}

private:

vector<vector<vector<int>>> dp_;

int m_;

int n_;

// number of paths starts from out-of-boundary to (x, y) by moving at most N steps.

int paths(int N, int x, int y) {

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

static const int kMod = 1000000007;

// Out of boundary, one path.

if (x < 0 || x >= n_ || y < 0 || y >= m_) return 1;

// Can not move but still in the grid, no path.

if (N == 0) return 0;

// Already computed.

if (dp_[N][y][x] != INT_MIN) return dp_[N][y][x];

dp_[N][y][x] = 0;

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

int tx = x + dirs[d];

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

dp_[N][y][x] = (dp_[N][y][x] + paths(N - 1, tx, ty)) % kMod;

}

return dp_[N][y][x];

}

};

no loops

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

m_ = m;

n_ = n;

dp_ = vector<vector<vector<int>>>(N + 1, vector<vector<int>>(m, vector<int>(n, INT_MIN)));

return paths(N, j, i);

}

private:

vector<vector<vector<int>>> dp_;

int m_;

int n_;

// number of paths start from (x, y) and move at most N steps.

int paths(int N, int x, int y) {

static const int kMod = 1000000007;

if (x < 0 || x >= n_ || y < 0 || y >= m_) return 1;

if (N == 0) return 0;

if (dp_[N][y][x] != INT_MIN) return dp_[N][y][x];

long ans = 0;

ans += paths(N - 1, x + 1, y);

ans += paths(N - 1, x - 1, y);

ans += paths(N - 1, x, y + 1);

ans += paths(N - 1, x, y - 1);

ans %= kMod;

return dp_[N][y][x] = ans;

}

};

Posts tagged as “dp”

花花酱 LeetCode 871. Minimum Number of Refueling Stops

Problem

Solution1: DP

Solution2: Priority Queue

Related Problems

花花酱 LeetCode 542. 01 Matrix

Problem

Solution 1: DP

Solution 2: BFS

花花酱 LeetCode 410. Split Array Largest Sum

Problem

Solution: DP

Solution: Binary Search

花花酱 LeetCode 845. Longest Mountain in Array

Problem

Solution: DP

花花酱 LeetCode 576. Out of Boundary Paths

Problem

Solution

Related Problems