Posts tagged as “dp”

花花酱 LeetCode 198. House Robber

By zxi on December 5, 2017

Problem:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Idea:

Time complexity: O(n)

Space complexity: O(n) -> O(1)

Solution:

C++ / Recursion + Memorization

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 0 ms

class Solution {

public:

int rob(vector<int>& nums) {

const int n = nums.size();

m_ = vector<int>(n, -1);

return rob(nums, n - 1);

}

private:

int rob(const vector<int>& nums, int i) {

if (i < 0) return 0;

if (m_[i] >= 0) return m_[i];

return m_[i] = max(rob(nums, i - 2) + nums[i],

rob(nums, i - 1));

}

vector<int> m_;

};

C++ / DP

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int rob(vector<int>& nums) {

if (nums.empty()) return 0;

vector<int> dp(nums.size(), 0);

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

dp[i] = max((i > 1 ? dp[i - 2] : 0) + nums[i],

(i > 0 ? dp[i - 1] : 0));

return dp.back();

}

};

C++ / O(1) Space

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int rob(vector<int>& nums) {

if (nums.empty()) return 0;

int dp2 = 0;

int dp1 = 0;

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

int dp = max(dp2 + nums[i], dp1);

dp2 = dp1;

dp1 = dp;

}

return dp1;

}

};

花花酱 LeetCode 741. Cherry Pickup

By zxi on December 4, 2017

题目大意：给你樱桃田的地图（1: 樱桃, 0: 空, -1: 障碍物）。然你从左上角走到右下角（只能往右或往下），再从右下角走回左上角（只能往左或者往上）。问你最多能采到多少棵樱桃。

Problem:

In a N x N grid representing a field of cherries, each cell is one of three possible integers.

0 means the cell is empty, so you can pass through;
1 means the cell contains a cherry, that you can pick up and pass through;
-1 means the cell contains a thorn that blocks your way.

Your task is to collect maximum number of cherries possible by following the rules below:

Starting at the position (0, 0) and reaching (N-1, N-1) by moving right or down through valid path cells (cells with value 0 or 1);
After reaching (N-1, N-1), returning to (0, 0) by moving left or up through valid path cells;
When passing through a path cell containing a cherry, you pick it up and the cell becomes an empty cell (0);
If there is no valid path between (0, 0) and (N-1, N-1), then no cherries can be collected.

Example 1:

Input: grid =

[[0, 1, -1],

[1, 0, -1],

[1, 1, 1]]

Output: 5

Explanation:

The player started at (0, 0) and went down, down, right right to reach (2, 2).

4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].

Then, the player went left, up, up, left to return home, picking up one more cherry.

The total number of cherries picked up is 5, and this is the maximum possible.

Note:

grid is an N by N 2D array, with 1 <= N <= 50.
Each grid[i][j] is an integer in the set {-1, 0, 1}.
It is guaranteed that grid[0][0] and grid[N-1][N-1] are not -1.

Idea:

Key observation: (0,0) to (n-1, n-1) to (0, 0) is the same as (n-1,n-1) to (0,0) twice

Two people starting from (n-1, n-1) and go to (0, 0).
They move one step (left or up) at a time simultaneously. And pick up the cherry within the grid (if there is one).
if they ended up at the same grid with a cherry. Only one of them can pick up it.

Solution: DP / Recursion with memoization.

x1, y1, x2 to represent a state y2 can be computed: y2 = x1 + y1 – x2

dp(x1, y1, x2) computes the max cherries if start from {(x1, y1), (x2, y2)} to (0, 0), which is a recursive function.

Since two people move independently, there are 4 subproblems: (left, left), (left, up), (up, left), (left, up). Finally, we have:

dp(x1, y1, x2)= g[y1][x1] + g[y2][x2] + max{dp(x1-1,y1,x2-1), dp(x1,y1-1,x2-1), dp(x1-1,y1,x2), dp(x1,y1-1,x2)}

Time complexity: O(n^3)

Space complexity: O(n^3)

Solution: DP

Time complexity: O(n^3)

Space complexity: O(n^3)

C ++

// Author: Huahua

// Runtime: 32 ms

class Solution {

public:

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

const int n = grid.size();

grid_ = &grid;

m_ = vector<vector<vector<int>>>(

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

return max(0, dp(n - 1, n - 1, n - 1));

}

private:

// max cherries from (x1, y1) to (0, 0) + (x2, y2) to (0, 0)

int dp(int x1, int y1, int x2) {

const int y2 = x1 + y1 - x2;

if (x1 < 0 || y1 < 0 || x2 < 0 || y2 < 0) return -1;

if ((*grid_)[y1][x1] < 0 || (*grid_)[y2][x2] < 0) return -1;

if (x1 == 0 && y1 == 0) return (*grid_)[y1][x1];

if (m_[x1][y1][x2] != INT_MIN) return m_[x1][y1][x2];

int ans = max(max(dp(x1 - 1, y1, x2 - 1), dp(x1, y1 - 1, x2)),

max(dp(x1, y1 - 1, x2 - 1), dp(x1 - 1, y1, x2)));

if (ans < 0) return m_[x1][y1][x2] = -1;

ans += (*grid_)[y1][x1];

if (x1 != x2) ans += (*grid_)[y2][x2];

return m_[x1][y1][x2] = ans;

}

vector<vector<vector<int>>> m_;

vector<vector<int>>* grid_; // does not own the object

};

Java

// Author: Huahua

class Solution {

private int[][] grid;

private int[][][] memo;

public int cherryPickup(int[][] grid) {

this.grid = grid;

int m = grid.length;

int n = grid[0].length;

memo = new int[m][n][n];

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

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

Arrays.fill(memo[i][j], Integer.MIN_VALUE);

return Math.max(0, dp(n - 1, m - 1, n - 1));

}

private int dp(int x1, int y1, int x2) {

final int y2 = x1 + y1 - x2;

if (x1 < 0 || y1 < 0 || x2 < 0 || y2 < 0) return -1;

if (grid[y1][x1] < 0 || grid[y2][x2] < 0) return -1;

if (x1 == 0 && y1 == 0) return grid[y1][x1];

if (memo[y1][x1][x2] != Integer.MIN_VALUE) return memo[y1][x1][x2];

memo[y1][x1][x2] = Math.max(Math.max(dp(x1 - 1, y1, x2 - 1), dp(x1, y1 - 1, x2)),

Math.max(dp(x1, y1 - 1, x2 - 1), dp(x1 - 1, y1, x2)));

if (memo[y1][x1][x2] >= 0) {

memo[y1][x1][x2] += grid[y1][x1];

if (x1 != x2) memo[y1][x1][x2] += grid[y2][x2];

}

return memo[y1][x1][x2];

}

Related Problems:

花花酱 LeetCode 64. Minimum Path Sum

By zxi on November 27, 2017

Problem:

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example 1:

[[1,3,1],

[1,5,1],

[4,2,1]]

Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

Idea:

Solution 1:

C++ / Recursion with memoization

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

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

int m = grid.size();

if (m == 0) return 0;

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

s_ = vector<vector<int>>(m, vector<int>(n, 0));

return minPathSum(grid, n - 1, m - 1, n, m);

}

private:

long minPathSum(const vector<vector<int>>& grid,

int x, int y, int n, int m) {

if (x == 0 && y == 0) return grid[y][x];

if (x < 0 || y < 0) return INT_MAX;

if (s_[y][x] > 0) return s_[y][x];

int ans = grid[y][x] + min(minPathSum(grid, x - 1, y, n, m),

minPathSum(grid, x, y - 1, n, m));

return s_[y][x] = ans;

}

vector<vector<int>> s_;

};

Solution 2:

C++ / DP

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

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

int m = grid.size();

if (m == 0) return 0;

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

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

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

if (i == 0 && j == 0) continue;

if (i == 0)

grid[i][j] += grid[i][j - 1];

else if (j == 0)

grid[i][j] += grid[i - 1][j];

else

grid[i][j] += min(grid[i][j - 1], grid[i - 1][j]);

}

return grid[m - 1][n - 1];

}

};

Related Problems:

[解题报告] LeetCode 113. Path Sum II

花花酱 LeetCode 53. Maximum Subarray

By zxi on November 27, 2017

Problem:

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

Idea:

Solution:

C++

// Author: Huahua

// Runtime: 6 ms (better than 98.66%)

class Solution {

public:

int maxSubArray(vector<int>& nums) {

vector<int> f(nums.size());

f[0] = nums[0];

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

f[i] = max(f[i - 1] + nums[i], nums[i]);

return *std::max_element(f.begin(), f.end());

}

};

花花酱 LeetCode 312. Burst Balloons

By zxi on November 27, 2017

Problem:

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and rightthen becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

1 2	nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> [] coins = 315 + 358 + 138 + 181 = 167

Idea

Solution1:

C++ / Recursion with memoization

// Author: Huahua

// Runtime: 16 ms

class Solution {

public:

int maxCoins(vector<int>& nums) {

int n = nums.size();

nums.insert(nums.begin(), 1);

nums.push_back(1);

// c[i][j] = maxCoins(nums[i:j+1])

c_ = vector<vector<int>>(n+2, vector<int>(n+2, 0));

return maxCoins(nums, 1, n);

}

private:

int maxCoins(const vector<int>& nums, const int i, const int j) {

if(i>j) return 0;

if(c_[i][j]>0) return c_[i][j];

if(i==j) return nums[i-1]*nums[i]*nums[i+1];

int ans=0;

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

ans=max(ans, maxCoins(nums,i,k-1) + nums[i-1]*nums[k]*nums[j+1] + maxCoins(nums, k+1, j));

c_[i][j] = ans;

return c_[i][j];

}

vector<vector<int>> c_;

};

Java

// Author: Huahua

class Solution {

private int[][] memo;

private int[] vals;

public int maxCoins(int[] nums) {

final int n = nums.length;

vals = new int[n + 2];

for (int i = 0; i < n; ++i) vals[i + 1] = nums[i];

vals[0] = vals[n + 1] = 1;

memo = new int[n + 2][n + 2];

return maxCoins(1, n);

}

private int maxCoins(int i, int j) {

if (i > j) return 0;

if (i == j) return vals[i - 1] * vals[i] * vals[i + 1];

if (memo[i][j] > 0) return memo[i][j];

int ans = 0;

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

ans = Math.max(ans, maxCoins(i, k - 1) + vals[i - 1] * vals[k] * vals[j + 1] + maxCoins(k + 1, j));

memo[i][j] = ans;

return memo[i][j];

}

Solution2:

C++ / DP

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

int maxCoins(vector<int>& nums) {

int n = nums.size();

nums.insert(nums.begin(), 1);

nums.push_back(1);

// c[i][j] = maxCoins(nums[i:j+1])

vector<vector<int>> c(n+2, vector<int>(n+2, 0));

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

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

int j=i+l-1;

for(int k=i;k<=j;++k) {

c[i][j] = max(c[i][j], c[i][k-1] + nums[i-1]*nums[k]*nums[j+1] + c[k+1][j]);

}

return c[1][n];

}

};

Java / DP

Java

// Author: Huahua

class Solution {

public int maxCoins(int[] nums) {

final int n = nums.length;

int[] vals = new int[n + 2];

for (int i = 0; i < n; ++i) vals[i + 1] = nums[i];

vals[0] = vals[n + 1] = 1;

int[][] dp = new int[n + 2][n + 2];

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

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

int j = i + l - 1;

int best = 0;

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

best = Math.max(best,

dp[i][k - 1] + vals[i - 1] * vals[k] * vals[j + 1] + dp[k + 1][j]);

dp[i][j] = best;

}

return dp[1][n];

}