Posts published in “Dynamic Programming”

花花酱 LeetCode 377. Combination Sum IV

By zxi on December 15, 2017

Problem:

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]

target = 4

The possible combination ways are:

(1, 1, 1, 1)

(1, 1, 2)

(1, 2, 1)

(1, 3)

(2, 1, 1)

(2, 2)

(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.

题目大意：给你一些数和一个目标值，问使用这些数(每个数可以被使用任意次数)加起来等于目标值的组合（其实是不重复的排列）有多少种。

Idea:

Solution:

C++ / Recursion + Memorization

// Author: Huahua

// Runtime: 0 ms

class Solution {

public:

int combinationSum4(vector<int>& nums, int target) {

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

m_[0] = 1;

return dp(nums, target);

}

private:

int dp(const vector<int>& nums, int target) {

if (target < 0) return 0;

if (m_[target] != -1) return m_[target];

int ans = 0;

for (const int num : nums)

ans += dp(nums, target - num);

return m_[target] = ans;

}

vector<int> m_;

};

C++ / DP

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int combinationSum4(vector<int>& nums, int target) {

vector<int> dp(target + 1, 0); // dp[i] # of combinations sum up to i

dp[0] = 1;

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

for (const int num : nums)

if (i - num >= 0)

dp[i] += dp[i - num];

return dp[target];

}

};

Related Problems:

花花酱 LeetCode 740. Delete and Earn

By zxi on December 5, 2017

Problem:

Given an array nums of integers, you can perform operations on the array.

In each operation, you pick any nums[i] and delete it to earn nums[i] points. After, you must delete everyelement equal to nums[i] - 1 or nums[i] + 1.

You start with 0 points. Return the maximum number of points you can earn by applying such operations.

Example 1:

Input: nums = [3, 4, 2]

Output: 6

Explanation:

Delete 4 to earn 4 points, consequently 3 is also deleted.

Then, delete 2 to earn 2 points. 6 total points are earned.

Example 2:

Input: nums = [2, 2, 3, 3, 3, 4]

Output: 9

Explanation:

Delete 3 to earn 3 points, deleting both 2's and the 4.

Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.

9 total points are earned.

Note:

The length of nums is at most 20000.
Each element nums[i] is an integer in the range [1, 10000].

Idea:

Reduce the problem to House Robber Problem

Key observations: If we take nums[i]

We can safely take all of its copies.
We can’t take any of copies of nums[i – 1] and nums[i + 1]

This problem is reduced to 198 House Robber.

Houses[i] has all the copies of num whose value is i.

[3 4 2] -> [0 2 3 4], rob([0 2 3 4]) = 6

[2, 2, 3, 3, 3, 4] -> [0 2*2 3*3 4], rob([0 2*2 3*3 4]) = 9

Time complexity: O(n+r) reduction + O(r) solving rob = O(n + r)

Space complexity: O(r)

r = max(nums) – min(nums) + 1

Time complexity: O(n + r)

Space complexity: O(r)

Solution:

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

int deleteAndEarn(vector<int>& nums) {

if (nums.empty()) return 0;

const auto range = minmax_element(nums.begin(), nums.end());

const int l = *(range.first);

const int r = *(range.second);

vector<int> points(r - l + 1, 0);

for (const int num : nums)

points[num - l] += num;

return rob(points);

}

private:

// From LeetCode 198. House Robber

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

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;

}

};

Related Problem:

花花酱 LeetCode 198. House Robber

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