花花酱 LeetCode 657. Judge Route Circle

By zxi on November 27, 2017

Problem:

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place.

The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L (Left), U (Up) and D (down). The output should be true or false representing whether the robot makes a circle.

Example 1:

1 2	Input: "UD" Output: true

Example 2:

1 2	Input: "LL" Output: false

Idea:

Simulation

Solution:

C++

// Author: Huahua

// Runtime: 19 ms

class Solution {

public:

bool judgeCircle(string moves) {

int cx{0};

int cy{0};

for (const char move : moves) {

switch(move) {

case 'U':

cy--;

break;

case 'D':

cy++;

break;

case 'R':

cx++;

break;

case 'L':

cx--;

break;

}

return cx==0 && cy==0;

}

};

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 1. Two Sum

By zxi on November 23, 2017

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Idea:

Brute force / Hashtable

Solution1:

Brute force / C++

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Runtime: 166 ms

class Solution {

public:

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

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

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

int sum = nums[i] + nums[j];

if (sum == target) {

return {i, j};

}

return {};

}

};

Solution 2:

Hashtable / C++

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

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

unordered_map<int, int> indies;

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

indies[nums[i]] = i;

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

int left = target - nums[i];

if (indies.count(left) && indies[left] != i) {

return {i, indies[left]};

}

return {};

}

};

Playlists by difficulties

By zxi on November 22, 2017

I have created YouTube playlists by difficulties.

Easy link

Medium link

Hard link

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

Huahua's Tech Road

花花酱 LeetCode 657. Judge Route Circle

花花酱 LeetCode 312. Burst Balloons

Java

Java

花花酱 LeetCode 1. Two Sum

Playlists by difficulties

花花酱 LeetCode 730. Count Different Palindromic Subsequences

Solution 1: Recursion with memoization

C++

Python

Java

Solution 2: DP

C++

Pyhton