Posts published in “Dynamic Programming”

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];

}

花花酱 LeetCode 730. Count Different Palindromic Subsequences

By zxi on November 20, 2017

Problem:

Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i.

Example 1:

Input: 
S = 'bccb'
Output: 6
Explanation: 
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice.

Example 2:

Input: 
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
Output: 104860361
Explanation: 
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.

Note:

The length of S will be in the range [1, 1000].
Each character S[i] will be in the set {'a', 'b', 'c', 'd'}.

Idea:

Solution 1: Recursion with memoization

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

C++

// Author: Huahua

// Runtime: 72 ms

class Solution {

public:

int countPalindromicSubsequences(const string& S) {

const int n = S.length();

m_ = vector<vector<int>>(n + 1, vector<int>(n + 1, 0));

return count(S, 0, S.length() - 1);

}

private:

static constexpr long kMod = 1000000007;

long count(const string& S, int s, int e) {

if (s > e) return 0;

if (s == e) return 1;

if (m_[s][e] > 0) return m_[s][e];

long ans = 0;

if (S[s] == S[e]) {

int l = s + 1;

int r = e - 1;

while (l <= r && S[l] != S[s]) ++l;

while (l <= r && S[r] != S[s]) --r;

if (l > r)

ans = count(S, s + 1, e - 1) * 2 + 2;

else if (l == r)

ans = count(S, s + 1, e - 1) * 2 + 1;

else

ans = count(S, s + 1, e - 1) * 2

- count(S, l + 1, r - 1);

} else {

ans = count(S, s, e - 1)

+ count(S, s + 1, e)

- count(S, s + 1, e - 1);

}

return m_[s][e] = (ans + kMod) % kMod;

}

vector<vector<int>> m_;

};

Python

"""

Author: Huahua

Runtime: 3582 ms

"""

class Solution:

def countPalindromicSubsequences(self, S):

def count(S, i, j):

if i > j: return 0

if i == j: return 1

if self.m_[i][j]: return self.m_[i][j]

if S[i] == S[j]:

ans = count(S, i + 1, j - 1) * 2

l = i + 1

r = j - 1

while l <= r and S[l] != S[i]: l += 1

while l <= r and S[r] != S[i]: r -= 1

if l > r: ans += 2

elif l == r: ans += 1

else: ans -= count(S, l + 1, r - 1)

else:

ans = count(S, i + 1, j) + count(S, i, j - 1) - count(S, i + 1, j - 1)

self.m_[i][j] = ans % 1000000007

return self.m_[i][j]

n = len(S)

self.m_ = [[None for _ in range(n)] for _ in range(n)]

return count(S, 0, n - 1)

Java

// Author: Huahua

// Runtime: 107 ms

class Solution {

private int[][] m_;

private static final int kMod = 1000000007;

public int countPalindromicSubsequences(String S) {

int n = S.length();

m_ = new int[n][n];

return count(S.toCharArray(), 0, n - 1);

}

private int count(char[] s, int i, int j) {

if (i > j) return 0;

if (i == j) return 1;

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

long ans = 0;

if (s[i] == s[j]) {

ans += count(s, i + 1, j - 1) * 2;

int l = i + 1;

int r = j - 1;

while (l <= r && s[l] != s[i]) ++l;

while (l <= r && s[r] != s[i]) --r;

if (l > r) ans += 2;

else if (l == r) ans += 1;

else ans -= count(s, l + 1, r - 1);

} else {

ans = count(s, i, j - 1)

+ count(s, i + 1, j)

- count(s, i + 1, j - 1);

}

m_[i][j] = (int)((ans + kMod) % kMod);

return m_[i][j];

}

Solution 2: DP

C++

// Author: Huahua

// Runtime: 79 ms

class Solution {

public:

int countPalindromicSubsequences(const string& S) {

int n = S.length();

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

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

dp[i][i] = 1;

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

for (int i = 0; i < n - len; ++i) {

const int j = i + len;

if (S[i] == S[j]) {

dp[i][j] = dp[i + 1][j - 1] * 2;

int l = i + 1;

int r = j - 1;

while (l <= r && S[l] != S[i]) ++l;

while (l <= r && S[r] != S[i]) --r;

if (l == r) dp[i][j] += 1;

else if (l > r) dp[i][j] += 2;

else dp[i][j] -= dp[l + 1][r - 1];

} else {

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

}

dp[i][j] = (dp[i][j] + kMod) % kMod;

}

return dp[0][n - 1];

}

private:

static constexpr long kMod = 1000000007;

};

Pyhton

"""

Author: Huahua

Runtime: 3639 ms

"""

class Solution:

def countPalindromicSubsequences(self, S):

n = len(S)

if n == 0: return 0

dp = [[0 for _ in range(n)] for _ in range(n)]

for i in range(n): dp[i][i] = 1

for size in range(2, n + 1):

for i in range(n - size + 1):

j = i + size - 1

if S[i] == S[j]:

dp[i][j] = dp[i + 1][j - 1] * 2

l = i + 1

r = j - 1

while l <= r and S[l] != S[i]: l += 1

while l <= r and S[r] != S[i]: r -= 1

if l > r: dp[i][j] += 2

elif l == r: dp[i][j] += 1

else: dp[i][j] -= dp[l + 1][r - 1]

else:

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

dp[i][j] %= 1000000007

return dp[0][n - 1]

Related Problems:

[解题报告] LeetCode 664. Strange Printer

花花酱 LeetCode 639. Decode Ways II

By zxi on November 18, 2017

Problem:

A message containing letters from A-Z is being encoded to numbers using the following mapping way:

'A' -> 1

'B' -> 2

...

'Z' -> 26

Beyond that, now the encoded string can also contain the character ‘*’, which can be treated as one of the numbers from 1 to 9.

Given the encoded message containing digits and the character ‘*’, return the total number of ways to decode it.

Also, since the answer may be very large, you should return the output mod 10⁹ + 7.

Example 1:

Input: "*"

Output: 9

Explanation: The encoded message can be decoded to the string:

"A", "B", "C", "D", "E", "F", "G", "H", "I".

Example 2:

1 2	Input: "1*" Output: 9 + 9 = 18

Note:

The length of the input string will fit in range [1, 10⁵].
The input string will only contain the character ‘*’ and digits ‘0’ – ‘9’.

Idea:

Time complexity: O(n)

Space complexity: O(1)

Solution:

C++

// Author: Huahua

// Runtime: 58 ms

class Solution {

public:

int numDecodings(string s) {

if (s.empty()) return 0;

// dp[-1] dp[0]

long dp[2] = {1, ways(s[0])};

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

long dp_i = ways(s[i]) * dp[1] + ways(s[i - 1], s[i]) * dp[0];

dp_i %= kMod;

dp[0] = dp[1];

dp[1] = dp_i;

}

return dp[1];

}

private:

static constexpr int kMod = 1000000007;

int ways(char c) {

if (c == '0') return 0;

if (c == '*') return 9;

return 1;

}

int ways(char c1, char c2) {

if (c1 == '*' && c2 == '*')

return 15;

if (c1 == '*') {

return (c2 >= '0' && c2 <= '6') ? 2 : 1;

} else if (c2 == '*') {

switch (c1) {

case '1': return 9;

case '2': return 6;

default: return 0;

}

} else {

int prefix = (c1 - '0') * 10 + (c2 - '0');

return prefix >= 10 && prefix <= 26;

}

};

Related Problems:

[解题报告] LeetCode 91. Decode Ways

花花酱 LeetCode 321. Create Maximum Number

By zxi on November 14, 2017

Problem:

Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the k digits. You should try to optimize your time and space complexity.

Example 1:

nums1 = [3, 4, 6, 5]
nums2 = [9, 1, 2, 5, 8, 3]
k = 5
return [9, 8, 6, 5, 3]

Example 2:

nums1 = [6, 7]
nums2 = [6, 0, 4]
k = 5
return [6, 7, 6, 0, 4]

Example 3:

nums1 = [3, 9]
nums2 = [8, 9]
k = 3
return [9, 8, 9]

题目大意：给你两个数字数组和k，返回从两个数组中选取k个数字能够组成的最大值。

Idea: Greedy + DP

Solution:

Time complexity: O(k * (n1+n2)^2)

Space complexity: O(n1+n2)

C++

// Author: Huahua

// Runtime: 26 ms

class Solution {

public:

vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {

vector<int> ans;

const int n1 = nums1.size();

const int n2 = nums2.size();

for (int i = max(0, k - n2); i <= min(k, n1); ++i)

ans = max(ans, maxNumber(maxNumber(nums1, i),

maxNumber(nums2, k - i)));

return ans;

}

private:

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

vector<int> ans(k);

int j = 0;

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

while (j > 0 && nums[i] > ans[j - 1]

&& nums.size() - i > k - j) --j;

if (j < k) ans[j++] = nums[i];

}

return ans;

}

vector<int> maxNumber(const vector<int>& nums1, const vector<int>& nums2) {

vector<int> ans(nums1.size() + nums2.size());

auto s1 = nums1.cbegin();

auto e1 = nums1.cend();

auto s2 = nums2.cbegin();

auto e2 = nums2.cend();

int index = 0;

while (s1 != e1 || s2 != e2)

ans[index++] =

lexicographical_compare(s1, e1, s2, e2) ? *s2++ : *s1++;

return ans;

}

};

Java

// Author: Huahua

// Runtime: 18 ms

class Solution {

public int[] maxNumber(int[] nums1, int[] nums2, int k) {

int[] best = new int[0];

for (int i = Math.max(0, k - nums2.length);

i <= Math.min(k, nums1.length); ++i)

best = max(best, 0,

maxNumber(maxNumber(nums1, i),

maxNumber(nums2, k - i)), 0);

return best;

}

private int[] maxNumber(int[] nums, int k) {

int[] ans = new int[k];

int j = 0;

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

while (j > 0 && nums[i] > ans[j - 1]

&& nums.length - i > k - j) --j;

if (j < k)

ans[j++] = nums[i];

}

return ans;

}

private int[] maxNumber(int[] nums1, int[] nums2) {

int[] ans = new int[nums1.length + nums2.length];

int s1 = 0;

int s2 = 0;

int index = 0;

while (s1 != nums1.length || s2 != nums2.length)

ans[index++] = max(nums1, s1, nums2, s2) == nums1 ?

nums1[s1++] : nums2[s2++];

return ans;

}

private int[] max(int[] nums1, int s1, int[] nums2, int s2) {

for (int i = s1; i < nums1.length; ++i) {

int j = s2 + i - s1;

if (j >= nums2.length) return nums1;

if (nums1[i] < nums2[j]) return nums2;

if (nums1[i] > nums2[j]) return nums1;

}

return nums2;

}