Posts published in “Dynamic Programming”

花花酱 LeetCode 174. Dungeon Game

By zxi on August 16, 2018

Problem

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0’s) or contain magic orbs that increase the knight’s health (positive integers).

In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Write a function to determine the knight’s minimum initial health so that he is able to rescue the princess.

For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

-2 (K)	-3	3
-5	-10	1
10	30	-5 (P)

Note:

The knight’s health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

Solution: DP

Time complexity: O(mn)

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

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int calculateMinimumHP(vector<vector<int>>& dungeon) {

const int m = dungeon.size();

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

// hp[y][x]: min hp required to reach bottom right (P).

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

hp[m][n - 1] = hp[m - 1][n] = 1;

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

for (int x = n - 1; x >= 0; --x)

hp[y][x] = max(1, min(hp[y + 1][x], hp[y][x + 1]) - dungeon[y][x]);

return hp[0][0];

}

};

O(n) space

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int calculateMinimumHP(vector<vector<int>>& dungeon) {

const int m = dungeon.size();

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

// hp[y][x]: min hp required to reach bottom right (P).

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

hp[n - 1] = 1;

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

for (int x = n - 1; x >= 0; --x)

hp[x] = max(1, min(hp[x], hp[x + 1]) - dungeon[y][x]);

return hp[0];

}

};

花花酱 LeetCode 887. Super Egg Drop

By zxi on August 13, 2018

Problem

You are given K eggs, and you have access to a building with N floors from 1 to N.

Each egg is identical in function, and if an egg breaks, you cannot drop it again.

You know that there exists a floor F with 0 <= F <= N such that any egg dropped at a floor higher than F will break, and any egg dropped at or below floor F will not break.

Each move, you may take an egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N).

Your goal is to know with certainty what the value of F is.

What is the minimum number of moves that you need to know with certainty what F is, regardless of the initial value of F?

Example 1:

Input: K = 1, N = 2
Output: 2
Explanation: 
Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
If it didn't break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.

Example 2:

Input: K = 2, N = 6
Output: 3

Example 3:

Input: K = 3, N = 14
Output: 4

Note:

1 <= K <= 100
1 <= N <= 10000

Solution 1: Recursion with Memorization (TLE)

dp[k][n] := min number of moves to test n floors with k eggs.

Base cases:

dp[0][n] = 0 # no eggs left.
dp[1][n] = n # one egg, need to test every floor.

Transition:

dp[k][n] = min(1 + max(dp[k][i – 1], dp[k – 1][n – i])) 1 <= i <= n

Time complexity: O(k*n^2)

Space complexity: O(k*n)

// Author: Huahua

// Running time: TLE (69/121 passed)

class Solution {

public:

int superEggDrop(int K, int N) {

m_ = vector<vector<int>>(K + 1, vector<int>(N + 1, INT_MIN));

return dp(K, N);

}

private:

// m[i][j] := min moves of i eggs and j floors

vector<vector<int>> m_;

int dp(int k, int n) {

if (k <= 0) return 0;

if (k == 1) return n;

if (n <= 1) return n;

if (m_[k][n] != INT_MIN) return m_[k][n];

int ans = INT_MAX;

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

ans = min(ans, 1 + max(dp(k - 1, i - 1), // broken at floor i, need to test i - 1 floors using k - 1 eggs.

dp(k, n - i))); // unbroken at floor i, need to test n - i floors using k eggs.

return m_[k][n] = ans;

}

};

Solution 2: Solution 1 + Binary Search

Time complexity: O(k*n*logn)

Space complexity: O(k*n)

C++

// Author: Huahua

// Running time: 40 ms

class Solution {

public:

int superEggDrop(int K, int N) {

m_ = vector<vector<int>>(K + 1, vector<int>(N + 1, INT_MIN));

return dp(K, N);

}

private:

// m[i][j] := min moves of i eggs and j floors

vector<vector<int>> m_;

int dp(int k, int n) {

if (k <= 0) return 0;

if (k == 1) return n;

if (n <= 1) return n;

if (m_[k][n] != INT_MIN) return m_[k][n];

// broken[i] = dp(k - 1, i - 1) is incresing with i.

// unbroken[i] = dp(k, n - i) is decresing with i.

// dp[k][n] = 1 + min(max(broken[i], unbroken[i])), 1 <= i <= n

// find the smallest i such that broken[i] >= unbroken[i],

// which minimizes max(broken[i], unbroken[i]).

int l = 1;

int r = n + 1;

while (l < r) {

int m = l + (r - l) / 2;

int broken = dp(k - 1, m - 1);

int unbroken = dp(k, n - m);

if (broken >= unbroken)

r = m;

else

l = m + 1;

}

return m_[k][n] = 1 + dp(k - 1, l - 1);

}

};

花花酱 LeetCode 97. Interleaving String

By zxi on August 5, 2018

Problem

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Solution: DP

Subproblems : whether s3[0:i+j] can be formed by interleaving s1[0:i] and s2[0:j].

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huhaua

// Running time: 0 ms

class Solution {

public:

bool isInterleave(string s1, string s2, string s3) {

int l1 = s1.length();

int l2 = s2.length();

int l3 = s3.length();

if (l1 + l2 != l3) return false;

// dp[i][j]: whehter s3[0:i+j] is a interleva of s1[0:i] and s2[0:j]

vector<vector<int>> dp(l1 + 1, vector<int>(l2 + 1));

dp[0][0] = true;

for (int i = 0; i <= l1; ++i)

for (int j = 0; j <= l2; ++j) {

if (i > 0) dp[i][j] |= dp[i - 1][j] && s1[i - 1] == s3[i + j - 1];

if (j > 0) dp[i][j] |= dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];

}

return dp[l1][l2];

}

};

Recursion + Memorization

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

bool isInterleave(string s1, string s2, string s3) {

m_ = vector<vector<int>>(s1.length() + 1, vector<int>(s2.length() + 1, INT_MIN));

return dp(s1, s1.length(), s2, s2.length(), s3, s3.length());

}

private:

// m_[i][j]: whehter s3[0:i+j] is a interleva of s1[0:i] and s2[0:j]

vector<vector<int>> m_;

bool dp(const string& s1, int l1, const string& s2, int l2, const string& s3, int l3) {

if (l1 + l2 != l3) return false;

if (l1 == 0 && l2 == 0) return true;

if (m_[l1][l2] != INT_MIN) return m_[l1][l2];

if (l1 > 0 & s3[l3 - 1] == s1[l1 - 1] && dp(s1, l1 - 1, s2, l2, s3, l3 - 1)

||l2 > 0 & s3[l3 - 1] == s2[l2 - 1] && dp(s1, l1, s2, l2 - 1, s3, l3 - 1))

m_[l1][l2] = 1;

else

m_[l1][l2] = 0;

return m_[l1][l2];

}

};

花花酱 LeetCode 546. Remove Boxes

By zxi on August 2, 2018

Problem

Given several boxes with different colors represented by different positive numbers.
You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (composed of k boxes, k >= 1), remove them and get k*k points.
Find the maximum points you can get.

Example 1:
Input:

1	[1, 3, 2, 2, 2, 3, 4, 3, 1]

Output:

Explanation:

[1, 3, 2, 2, 2, 3, 4, 3, 1] 
----> [1, 3, 3, 4, 3, 1] (3*3=9 points) 
----> [1, 3, 3, 3, 1] (1*1=1 points) 
----> [1, 1] (3*3=9 points) 
----> [] (2*2=4 points)

Note: The number of boxes n would not exceed 100.

Solution: Recursion + Memorization

Use dp[l][r][k] to denote the max score of subarray box[l] ~ box[r] with k boxes after box[r] that have the same color as box[r]

box[l], box[l+1], …, box[r], box[r+1], …, box[r+k]

e.g. “CDABACAAAAA”

dp[2][6][4] is the max score of [ABACA] followed by [AAAA]
dp[2][6][3] is the max score of [ABACA] followed by [AAA]

base case: l > r, empty array, return 0.

Transition:
dp[l][r][k] = max(dp[l][r-1][0] + (k + 1)*(k + 1),  # case 1
                  dp[l][i][k+1] + dp[i+1][r-1][0])  # case 2
# "ABACA|AAAA" 
# case 1: dp("ABAC") + score("AAAAA") drop j and the tail.
# case 2: box[i] == box[r], l <= i < r, try all break points
# max({dp("A|AAAAA") + dp("BAC")}, {dp("ABA|AAAAA") + dp("C")})

Time complexity: O(n^4)

Space complexity: O(n^3)

C++

// Author: Huahua

// Running time: 196 ms

class Solution {

public:

int removeBoxes(vector<int>& boxes) {

const int n = boxes.size();

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

return dfs(boxes, 0, n - 1, 0);

}

private:

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

int dfs(const vector<int>& boxes, int l, int r, int k) {

if (l > r) return 0;

if (m_[l][r][k] > 0) return m_[l][r][k];

m_[l][r][k] = dfs(boxes, l, r - 1, 0) + (k + 1) * (k + 1);

for (int i = l; i < r; ++i)

if (boxes[i] == boxes[r])

m_[l][r][k] = max(m_[l][r][k], dfs(boxes, l, i, k + 1) + dfs(boxes, i + 1, r - 1, 0));

return m_[l][r][k];

}

};

Optimized

// Author: Huahua

// Running time: 16 ms

int m[100][100][100];

class Solution {

public:

int removeBoxes(vector<int>& boxes) {

memset(m, 0, sizeof(m));

return dfs(boxes, 0, boxes.size() - 1, 0);

}

private:

int dfs(const vector<int>& boxes, int l, int r, int k) {

if (l > r) return 0;

if (m[l][r][k] > 0) return m[l][r][k];

int rr = r;

int kk = k;

while (l < r && boxes[r - 1] == boxes[r]) {--r; ++k;}

m[l][r][k] = dfs(boxes, l, r - 1, 0) + (k + 1) * (k + 1);

for (int i = l; i < r; ++i)

if (boxes[i] == boxes[r])

m[l][r][k] = max(m[l][r][k], dfs(boxes, l, i, k + 1) + dfs(boxes, i + 1, r - 1, 0));

for (int d = 1; d <= r - rr; ++d)

m[l][r + d][k - d] = m[l][r][k];

return m[l][r][k];

}

};

Use a HashTable to replace the 3D DP array since the DP array could be sparse when many consecutive boxes are the same color.

// Author: Huahua

// Running time: 8 ms (beats 100%)

class Solution {

public:

int removeBoxes(vector<int>& boxes) {

len_ = vector<int>(boxes.size());

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

if (boxes[i] == boxes[i - 1]) len_[i] = len_[i - 1] + 1;

return dfs(boxes, 0, boxes.size() - 1, 0);

}

private:

unordered_map<int, int> m_;

vector<int> len_;

int dfs(const vector<int>& boxes, int l, int r, int k) {

if (l > r) return 0;

k += len_[r];

r -= len_[r];

int key = (l * 100 + r) * 100 + k;

auto it = m_.find(key);

if (it != m_.end()) return it->second;

int& ans = m_[key];

ans = dfs(boxes, l, r - 1, 0) + (k + 1) * (k + 1);

for (int i = l; i < r; ++i)

if (boxes[i] == boxes[r])

ans = max(ans, dfs(boxes, l, i, k + 1) + dfs(boxes, i + 1, r - 1, 0));

return ans;

}

};

Problem

There are two types of soup: type A and type B. Initially we have N ml of each type of soup. There are four kinds of operations:

Serve 100 ml of soup A and 0 ml of soup B
Serve 75 ml of soup A and 25 ml of soup B
Serve 50 ml of soup A and 50 ml of soup B
Serve 25 ml of soup A and 75 ml of soup B

When we serve some soup, we give it to someone and we no longer have it. Each turn, we will choose from the four operations with equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as we can. We stop once we no longer have some quantity of both types of soup.

Note that we do not have the operation where all 100 ml’s of soup B are used first.

Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time.

Example:
Input: N = 50
Output: 0.625
Explanation: 
If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.

Notes:

0 <= N <= 10^9.
Answers within 10^-6 of the true value will be accepted as correct.

Solution 1: Recursion with Memorization

Time complexity: O(N^2) N ~ 5000 / 25 = 200

Space complexity: O(N^2)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

double soupServings(int N) {

if (N >= 5000) return 1.0;

return prob(N, N);

}

private:

unordered_map<int, unordered_map<int, double>> m_;

double prob(int A, int B) {

static int ops[][2] = {{100, 0}, {75, 25}, {50, 50}, {25, 75}};

if (A <= 0 && B <= 0) return 0.5;

if (A <= 0) return 1.0;

if (B <= 0) return 0.0;

if (m_.count(A) && m_[A].count(B)) return m_[A][B];

m_[A][B] = 0.0;

for (int i = 0; i < 4; ++i)

m_[A][B] += 0.25 * prob(A - ops[i][0], B - ops[i][1]);

return m_[A][B];

}

};

Posts published in “Dynamic Programming”

花花酱 LeetCode 174. Dungeon Game

Problem

Solution: DP

花花酱 LeetCode 887. Super Egg Drop

Problem

Solution 1: Recursion with Memorization (TLE)

Solution 2: Solution 1 + Binary Search

花花酱 LeetCode 97. Interleaving String

Problem

Solution: DP

花花酱 LeetCode 546. Remove Boxes

Problem

Solution: Recursion + Memorization

Related Problems

花花酱 LeetCode 808. Soup Servings

Problem

Solution 1: Recursion with Memorization