Posts tagged as “dynamic programming”

花花酱 LeetCode 1143. Longest Common Subsequence

By zxi on August 29, 2019

Given two strings text1 and text2, return the length of their longest common subsequence.

A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = "abcde", text2 = "ace" 
Output: 3  
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

1 <= text1.length <= 1000
1 <= text2.length <= 1000
The input strings consist of lowercase English characters only.

Solution: DP

Use dp[i][j] to represent the length of longest common sub-sequence of text1[0:i] and text2[0:j]
dp[i][j] = dp[i – 1][j – 1] + 1 if text1[i – 1] == text2[j – 1] else max(dp[i][j – 1], dp[i – 1][j])

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

C++

// Author: Huahua, 16ms 14.9 MB

class Solution {

public:

int longestCommonSubsequence(string text1, string text2) {

int m = text1.length();

int n = text2.length();

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

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

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

if (text1[i] == text2[j])

dp[i + 1][j + 1] = dp[i][j] + 1;

else

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

return dp[m][n];

}

};

C++/V2

// Author: Huahua, 8ms, 8.8MB

class Solution {

public:

int longestCommonSubsequence(string text1, string text2) {

int m = text1.length();

int n = text2.length();

vector<int> dp(n + 1);

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

int prev = 0; // dp[i][j]

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

int curr = dp[j + 1]; // dp[i][j + 1]

if (text1[i] == text2[j])

// dp[i + 1][j + 1] = dp[i][j] + 1

dp[j + 1] = prev + 1;

else

// dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j])

dp[j + 1] = max(curr, dp[j]);

prev = curr;

}

return dp[n]; // dp[m][n]

}

};

C++/V3

// Author: Huahua

class Solution {

public:

int longestCommonSubsequence(string text1, string text2) {

int m = text1.length();

int n = text2.length();

vector<int> dp1(n + 1);

vector<int> dp2(n + 1);

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

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

if (text1[i] == text2[j])

dp2[j + 1] = dp1[j] + 1;

else

dp2[j + 1] = max(dp1[j + 1], dp2[j]);

swap(dp1, dp2);

}

return dp1[n];

}

};

花花酱 LeetCode Weekly Contest 137

By zxi on May 18, 2019

1046. Last Stone Weight

Solution: Simulation (priority_queue)

Time complexity: O(nlogn)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int lastStoneWeight(vector<int>& stones) {

priority_queue<int> q;

for (int s : stones)

q.push(s);

while (q.size() > 1) {

int x = q.top(); q.pop();

int y = q.top(); q.pop();

if (x == y) continue;

q.push(abs(x - y));

}

return q.empty() ? 0 : q.top();

}

};

1047. Remove All Adjacent Duplicates In String

Solution: Stack / Deque

Time complexity: O(n)
Space complexity: O(n)

C++

class Solution {

public:

string removeDuplicates(string S) {

deque<char> s;

for (char c : S) {

if (!s.empty() && s.back() == c) s.pop_back();

else s.push_back(c);

}

return {begin(s), end(s)};

}

};

1048. Longest String Chain

Solution: DP

dp[i] := max length of chain of (A[0] ~ A[i-1])

dp[i] = max{dp[j] + 1} if A[j] is prederrsor of A[i], 1 <= j <i

Time complexity: O(n^2*l)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int longestStrChain(vector<string>& words) {

int n = words.size();

sort(begin(words), end(words), [](const string& a, const string& b) {

return a.length() < b.length();

});

vector<int> dp(n, 1);

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

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

if (valid(words[j], words[i]))

dp[i] = dp[j] + 1;

return *max_element(begin(dp), end(dp));

}

private:

bool valid(const string& a, const string& b) {

if (a.length() + 1 != b.length()) return false;

int count = 0;

for (int i = 0, j = 0; i < a.length() && j < b.length();) {

if (a[i] == b[j]) {

++i; ++j;

} else {

++count; ++j;

}

return count <= 1;

}

};

1049. Last Stone Weight II

Solution: DP / target sum

Time complexity: O(n * S) = O(n * 100)

Space complexity: O(S) = O(100)

C++

// Author: Huahua

class Solution {

public:

int lastStoneWeightII(vector<int>& stones) {

const int n = stones.size();

const int max_sum = accumulate(begin(stones), end(stones), 0);

unordered_set<int> sums;

sums.insert(stones[0]);

sums.insert(-stones[0]);

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

unordered_set<int> tmp;

for (int s : sums) {

tmp.insert(s + stones[i]);

tmp.insert(s - stones[i]);

}

swap(tmp, sums);

}

int ans = INT_MAX;

for (int s : sums)

ans = min(ans, abs(s));

return ans;

}

};

花花酱 LeetCode 1000. Minimum Cost to Merge Stones

By zxi on March 2, 2019

There are N piles of stones arranged in a row. The i-th pile has stones[i] stones.

A move consists of merging exactly K consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these K piles.

Find the minimum cost to merge all piles of stones into one pile. If it is impossible, return -1.

Example 1:

Input: stones = [3,2,4,1], K = 2
Output: 20
Explanation: 
We start with [3, 2, 4, 1].
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.

Example 2:

Input: stones = [3,2,4,1], K = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore.  So the task is impossible.

Example 3:

Input: stones = [3,5,1,2,6], K = 3
Output: 25
Explanation: 
We start with [3, 5, 1, 2, 6].
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.

Note:

1 <= stones.length <= 30
2 <= K <= 30
1 <= stones[i] <= 100

Solution: DP

dp[i][j][k] := min cost to merge subarray i ~ j into k piles
Init: dp[i][j][k] = 0 if i==j and k == 1 else inf
ans: dp[0][n-1][1]
transition:
1. dp[i][j][k] = min{dp[i][m][1] + dp[m+1][j][k-1]} for all i <= m < j
2. dp[i][j][1] = dp[i][j][K] + sum(A[i]~A[j])

Time complexity: O(n^3)
Space complexity: O(n^2*K)

C++

// Running time: 12 ms, 12.1 MB

class Solution {

public:

int mergeStones(vector<int>& stones, int K) {

const int n = stones.size();

if ((n - 1) % (K - 1)) return -1;

const int kInf = 1e9;

vector<int> sums(n + 1);

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

sums[i + 1] = sums[i] + stones[i];

// dp[i][j][k] := min cost to merge subarray i~j into k piles.

vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(K + 1, kInf)));

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

dp[i][i][1] = 0;

for (int l = 2; l <= n; ++l) // subproblem length

for (int i = 0; i <= n - l; ++i) { // start

int j = i + l - 1; // end

for (int k = 2; k <= K; ++k) // piles

for (int m = i; m < j; m += K - 1) // split point

dp[i][j][k] = min(dp[i][j][k], dp[i][m][1] + dp[m + 1][j][k - 1]);

dp[i][j][1] = dp[i][j][K] + sums[j + 1] - sums[i];

}

return dp[0][n - 1][1];

}

};

C++/top down

// Author: Huahua, running time: 12 ms

class Solution {

public:

int mergeStones(vector<int>& stones, int K) {

const int n = stones.size();

if ((n - 1) % (K - 1)) return -1;

const int kInf = 1e9;

vector<int> sums(n + 1);

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

sums[i + 1] = sums[i] + stones[i];

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

std::function<int(int, int, int)> dp = [&stones, &sums, &cache, kInf, n, K, &dp](int i, int j, int k) {

if ((j - i + 1 - k) % (K - 1)) return kInf;

if (i == j) return k == 1 ? 0 : kInf;

if (cache[i][j][k] != INT_MAX) return cache[i][j][k];

if (k == 1)

return cache[i][j][k] = dp(i, j, K) + sums[j + 1] - sums[i];

int ans = kInf;

for (int m = i; m < j; m += K - 1) {

int l = dp(i, m, 1);

if (l >= ans) continue;

int r = dp(m + 1, j, k - 1);

if (r >= ans) continue;

ans = min(ans, l + r);

}

return cache[i][j][k] = ans;

};

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

}

};

Solution 2: DP

dp[l][i] := min cost to merge [i, i + l) into as less piles as possible. Number of merges will be (l-1) / (K – 1) and
Transition: dp[l][i] = min(dp[m][i] + dp[l – m][i + m]) for 1 <= m < l
if ((l – 1) % (K – 1) == 0) [i, i + l) can be merged into 1 pile, dp[l][i] += sum(A[i:i+l])

Time complexity: O(n^3 / k)
Space complexity: O(n^2)

C++

// Author: Huahua 4ms, 9.6 MB

class Solution {

public:

int mergeStones(vector<int>& stones, int K) {

const int n = stones.size();

if ((n - 1) % (K - 1)) return -1;

vector<int> sums(n + 1);

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

sums[i + 1] = sums[i] + stones[i];

const int kInf = 1e9;

// dp[i][j] := min cost to merge subarray A[i] ~ A[j] into (j-i)%(K-1) + 1 piles

vector<vector<int>> dp(n, vector<int>(n, kInf));

for (int i = 0; i < n; ++i) dp[i][i] = 0;

for (int l = 2; l <= n; ++l) // subproblem length

for (int i = 0; i <= n - l; ++i) { // start

int j = i + l - 1;

for (int m = i; m < j ; m += K - 1) // split point

dp[i][j] = min(dp[i][j], dp[i][m] + dp[m + 1][j]);

// If the current length can be merged into 1 pile

// The cost is independent of the merge orders.

if ((l - 1) % (K - 1) == 0)

dp[i][j] += sums[j + 1] - sums[i];

}

return dp[0][n - 1];

}

};

C++/Top-Down

// Author: Huahua, 4 ms

class Solution {

public:

int mergeStones(vector<int>& stones, int K) {

const int n = stones.size();

if ((n - 1) % (K - 1)) return -1;

const int kInf = 1e9;

vector<vector<int>> cache(n, vector<int>(n, kInf));

vector<int> sums(n + 1);

for (int i = 0; i < stones.size(); ++i) sums[i + 1] = sums[i] + stones[i];

std::function<int(int, int)> dp;

dp = [&stones, &sums, &cache, &dp, K, kInf](int i, int j) {

const int l = j - i + 1;

if (l < K) return 0;

if (cache[i][j] != kInf) return cache[i][j];

int ans = kInf;

for (int m = i; m < j ; m += K - 1) // split point

ans = min(ans, dp(i, m) + dp(m + 1, j));

if ((l - 1) % (K - 1) == 0)

ans += sums[j + 1] - sums[i];

return cache[i][j] = ans;

};

return dp(0, n - 1);

}

};

花花酱 Leetcode 140. Word Break II

By zxi on September 2, 2017

Problem

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution

Time complexity: O(2^n)

Space complexity: O(2^n)

C++

// Author: Huahua

class Solution {

public:

vector<string> wordBreak(string s, vector<string>& wordDict) {

unordered_set<string> dict(wordDict.cbegin(), wordDict.cend());

return wordBreak(s, dict);

}

private:

// >> append({"cats and", "cat sand"}, "dog");

// {"cats and dog", "cat sand dog"}

vector<string> append(const vector<string>& prefixes, const string& word) {

vector<string> results;

for(const auto& prefix : prefixes)

results.push_back(prefix + " " + word);

return results;

}

const vector<string>& wordBreak(string s, unordered_set<string>& dict) {

// Already in memory, return directly

if(mem_.count(s)) return mem_[s];

// Answer for s

vector<string> ans;

// s in dict, add it to the answer array

if(dict.count(s))

ans.push_back(s);

for(int j=1;j<s.length();++j) {

// Check whether right part is a word

const string& right = s.substr(j);

if (!dict.count(right)) continue;

// Get the ans for left part

const string& left = s.substr(0, j);

const vector<string> left_ans =

append(wordBreak(left, dict), right);

// Notice, can not use mem_ here,

// since we haven't got the ans for s yet.

ans.insert(ans.end(), left_ans.begin(), left_ans.end());

}

// memorize and return

mem_[s].swap(ans);

return mem_[s];

}

private:

unordered_map<string, vector<string>> mem_;

};

Python3

"""

Author: Huahua

Running time: 56 ms

"""

class Solution:

def wordBreak(self, s, wordDict):

words = set(wordDict)

mem = {}

def wordBreak(s):

if s in mem: return mem[s]

ans = []

if s in words: ans.append(s)

for i in range(1, len(s)):

right = s[i:]

if right not in words: continue

ans += [w + " " + right for w in wordBreak(s[0:i])]

mem[s] = ans

return mem[s]

return wordBreak(s)

Problem

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Idea:

Time complexity O(n^2)

Space complexity O(n^2)

Solutions:

C++

// Author: Huahua

class Solution {

public:

bool wordBreak(string s, vector<string>& wordDict) {

// Create a hashset of words for fast query.

unordered_set<string> dict(wordDict.cbegin(), wordDict.cend());

// Query the answer of the original string.

return wordBreak(s, dict);

}

bool wordBreak(const string& s, const unordered_set<string>& dict) {

// In memory, directly return.

if(mem_.count(s)) return mem_[s];

// Whole string is a word, memorize and return.

if(dict.count(s)) return mem_[s]=true;

// Try every break point.

for(int j=1;j<s.length();j++) {

const string left = s.substr(0,j);

const string right = s.substr(j);

// Find the solution for s.

if(dict.count(right) && wordBreak(left, dict))

return mem_[s]=true;

}

// No solution for s, memorize and return.

return mem_[s]=false;

}

private:

unordered_map<string, bool> mem_;

};

C++ V2 without using dict. Updated: 1/9/2018

// Author: Huahua

// Running time: 9 ms

class Solution {

public:

bool wordBreak(string s, vector<string>& wordDict) {

// Mark evert word as breakable.

for (const string& word : wordDict)

mem_.emplace(word, true);

// Query the answer of the original string.

return wordBreak(s);

}

bool wordBreak(const string& s) {

// In memory, directly return.

if (mem_.count(s)) return mem_[s];

// Try every break point.

for (int j = 1; j < s.length(); j++) {

auto it = mem_.find(s.substr(j));

// Find the solution for s.

if (it != mem_.end() && it->second && wordBreak(s.substr(0, j)))

return mem_[s] = true;

}

// No solution for s, memorize and return.

return mem_[s] = false;

}

private:

unordered_map<string, bool> mem_;

};

Java

// Author: Huahua

// Runtime: 13 ms

class Solution {

public boolean wordBreak(String s, List<String> wordDict) {

Set<String> dict = new HashSet<String>(wordDict);

Map<String, Boolean> mem = new HashMap<String, Boolean>();

return wordBreak(s, mem, dict);

}

private boolean wordBreak(String s,

Map<String, Boolean> mem,

Set<String> dict) {

if (mem.containsKey(s)) return mem.get(s);

if (dict.contains(s)) {

mem.put(s, true);

return true;

}

for (int i = 1; i < s.length(); ++i) {

if (dict.contains(s.substring(i)) && wordBreak(s.substring(0, i), mem, dict)) {

mem.put(s, true);

return true;

}

mem.put(s, false);

return false;

}

Python

"""

Author: Huahua

Runtime: 42 ms

"""

class Solution(object):

def wordBreak(self, s, wordDict):

def canBreak(s, m, wordDict):

if s in m: return m[s]

if s in wordDict:

m[s] = True

return True

for i in range(1, len(s)):

r = s[i:]

if r in wordDict and canBreak(s[0:i], m, wordDict):

m[s] = True

return True

m[s] = False

return False

return canBreak(s, {}, set(wordDict))

Posts tagged as “dynamic programming”

花花酱 LeetCode 1143. Longest Common Subsequence

C++

C++/V2

C++/V3

花花酱 LeetCode Weekly Contest 137

1046. Last Stone Weight

C++

1047. Remove All Adjacent Duplicates In String

C++

1048. Longest String Chain

C++

1049. Last Stone Weight II

C++

花花酱 LeetCode 1000. Minimum Cost to Merge Stones

Solution: DP

C++

C++/top down

Solution 2: DP

C++

C++/Top-Down

花花酱 Leetcode 140. Word Break II

Problem

Solution

Related Problems

花花酱 Leetcode 139. Word Break

Problem

Idea:

Solutions: