Posts published in “Dynamic Programming”

花花酱 LeetCode 724. Find Pivot Index

By zxi on November 13, 2017

Problem:

Given an array of integers nums, write a method that returns the “pivot” index of this array.

We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index.

If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.

Example 1:

Input:

nums = [1, 7, 3, 6, 5, 6]

Output: 3

Explanation:

The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.

Also, 3 is the first index where this occurs.

Example 2:

Input:

nums = [1, 2, 3]

Output: -1

Explanation:

There is no index that satisfies the conditions in the problem statement.

Note:

The length of nums will be in the range [0, 10000].
Each element nums[i] will be an integer in the range [-1000, 1000].

Idea:

Solution:

C++

// Author: Huahua

// Runtime: 43 ms

class Solution {

public:

int pivotIndex(vector<int>& nums) {

const int sum = accumulate(nums.begin(), nums.end(), 0);

int l = 0;

int r = sum;

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

r -= nums[i];

if (l == r) return i;

l += nums[i];

}

return -1;

}

};

Java

// Author: Huahua

// Runtime: 133 ms

class Solution {

public int pivotIndex(int[] nums) {

int sum = IntStream.of(nums).sum();

int l = 0;

int r = sum;

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

r -= nums[i];

if (l == r) return i;

l += nums[i];

}

return -1;

}

Python

"""

Author: Huahua

Runtime: 112 ms

"""

class Solution:

def pivotIndex(self, nums):

l, r = 0, sum(nums)

for i in range(len(nums)):

r -= nums[i]

if l == r: return i

l += nums[i]

return -1

花花酱 LeetCode 91. Decode Ways

By zxi on November 9, 2017

题目大意：

给你一个加密的数字字符串，问你一共有多少种不同的解密方式。

Problem:

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

‘A’ -> 1

‘B’ -> 2

…

‘Z’ -> 26

Given an encoded message containing digits, determine the total number of ways to decode it.

For example,
Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).

The number of ways decoding "12" is 2.

Idea:

Dynamic Programming

Solution:

C++

Time complexity: O(n^2)

Space complexity: O(n^2)

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

int numDecodings(string s) {

if(s.length() == 0) return 0;

m_ways[""] = 1;

return ways(s);

}

private:

int ways(const string& s) {

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

if (s[0] == '0') return 0;

if (s.length() == 1) return 1;

int w = ways(s.substr(1));

const int prefix = stoi(s.substr(0, 2));

if (prefix <= 26)

w += ways(s.substr(2));

m_ways[s] = w;

return w;

}

unordered_map<string, int> m_ways;

};

C++

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int numDecodings(string s) {

if(s.length() == 0) return 0;

return ways(s, 0, s.length() - 1);

}

private:

int ways(const string& s, int l, int r) {

if (m_ways.count(l)) return m_ways[l];

if (s[l] == '0') return 0;

if (l >= r) return 1; // Single digit or empty.

int w = ways(s, l + 1, r);

const int prefix = (s[l] - '0') * 10 + (s[l + 1] - '0');

if (prefix <= 26)

w += ways(s, l + 2, r);

m_ways[l] = w;

return w;

}

// Use l as key.

unordered_map<int, int> m_ways;

};

C++

Time complexity: O(n)

Space complexity: O(1)

class Solution {

public:

int numDecodings(string s) {

if (s.empty() || s[0] == '0') return 0;

if (s.length() == 1) return 1;

const int n = s.length();

int w1 = 1;

int w2 = 1;

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

int w = 0;

if (!isValid(s[i]) && !isValid(s[i - 1], s[i])) return 0;

if (isValid(s[i])) w = w1;

if (isValid(s[i - 1], s[i])) w += w2;

w2 = w1;

w1 = w;

}

return w1;

}

private:

bool isValid(const char c) { return c != '0'; }

bool isValid(const char c1, const char c2) {

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

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

}

};

Related Problems:

[解题报告] LeetCode 639. Decode Ways II – 花花酱

花花酱 LeetCode 72. Edit Distance

By zxi on October 15, 2017

Problem:

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

Idea:

Dynamic Programming

Solution:

Recursive

// Author: Huahua

// Runtime: 13 ms

class Solution {

public:

int minDistance(string word1, string word2) {

int l1 = word1.length();

int l2 = word2.length();

d_ = vector<vector<int>>(l1 + 1, vector<int>(l2 + 1, -1));

return minDistance(word1, word2, l1, l2);

}

private:

vector<vector<int>> d_;

// minDistance from word1[0:l1-1] to word2[0:l2-1]

int minDistance(const string& word1, const string& word2, int l1, int l2) {

if (l1 == 0) return l2;

if (l2 == 0) return l1;

if (d_[l1][l2] >= 0) return d_[l1][l2];

int ans;

if (word1[l1 - 1] == word2[l2 - 1])

ans = minDistance(word1, word2, l1 - 1, l2 - 1);

else

ans = min(minDistance(word1, word2, l1 - 1, l2 - 1),

min(minDistance(word1, word2, l1 - 1, l2),

minDistance(word1, word2, l1, l2 - 1))) + 1;

return d_[l1][l2] = ans;

}

};

Iterative

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

int minDistance(string word1, string word2) {

int l1 = word1.length();

int l2 = word2.length();

// d[i][j] := minDistance(word1[0:i - 1], word2[0:j - 1]);

vector<vector<int>> d(l1 + 1, vector<int>(l2 + 1, 0));

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

d[i][0] = i;

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

d[0][j] = j;

for (int i = 1; i <= l1; ++i)

for (int j = 1; j <= l2; ++j) {

int c = (word1[i - 1] == word2[j - 1]) ? 0 : 1;

d[i][j] = min(d[i - 1][j - 1] + c,

min(d[i][j - 1], d[i - 1][j]) + 1);

}

return d[l1][l2];

}

};

花花酱 688. Knight Probability in Chessboard

By zxi on October 1, 2017

https://leetcode.com/problems/knight-probability-in-chessboard/description/

Problem:

On an NxN chessboard, a knight starts at the r-th row and c-th column and attempts to make exactly Kmoves. The rows and columns are 0 indexed, so the top-left square is (0, 0), and the bottom-right square is (N-1, N-1).

A chess knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.

The knight continues moving until it has made exactly K moves or has moved off the chessboard. Return the probability that the knight remains on the board after it has stopped moving.

Example:

Input: 3, 2, 0, 0

Output: 0.0625

Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board.

From each of those positions, there are also two moves that will keep the knight on

the board. The total probability the knight stays on the board is 0.0625.

Note:

N will be between 1 and 25.
K will be between 0 and 100.
The knight always initially starts on the board.

Idea:

Dynamic programming

Count the ways to reach (x, y) after k moves from start.

Time Complexity: O(k*n^2)

Space Complexity: O(n^2)

Solution:

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

double knightProbability(int N, int K, int r, int c) {

vector<vector<double>> dp0(N, vector<double>(N, 0.0));

dp0[r][c] = 1.0;

int dirs[8][2] = {{1, 2}, {-1, -2}, {1, -2}, {-1, 2},

{2, 1}, {-2, -1}, {2, -1}, {-2, 1}};

for (int k = 0; k < K; ++k) {

vector<vector<double>> dp1(N, vector<double>(N, 0.0));

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

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

for (int m = 0; m < 8; ++m) {

int x = j + dirs[m][0];

int y = i + dirs[m][1];

if (x < 0 || y < 0 || x >= N || y >= N) continue;

dp1[i][j] += dp0[y][x];

}

std::swap(dp0, dp1);

}

double total = 0;

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

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

total += dp0[i][j];

return total / pow(8, K);

}

};

Related problems:

花花酱 LeetCode 678. Valid Parenthesis String

By zxi on September 30, 2017

Problem:

Given a string containing only three types of characters: ‘(‘, ‘)’ and ‘*’, write a function to check whether this string is valid. We define the validity of a string by these rules:

Any left parenthesis '(' must have a corresponding right parenthesis ')'.
Any right parenthesis ')' must have a corresponding left parenthesis '('.
Left parenthesis '(' must go before the corresponding right parenthesis ')'.
'*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string.
An empty string is also valid.

Example 1:

1 2	Input: "()" Output: True

Example 2:

1 2	Input: "(*)" Output: True

Example 3:

1 2	Input: "(*))" Output: True

Note:

The string size will be in the range [1, 100].

Idea:

Dynamic Programming / Counting

Solution 1:

C++ / DP / Top-down O(n^3)

// Author: Huahua

// Runtime: 12 ms

class Solution {

public:

bool checkValidString(const string& s) {

int l = s.length();

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

return isValid(s, 0, l - 1);

}

private:

vector<vector<int>> m_;

bool isValid(const string& s, int i, int j) {

if (i > j) return true;

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

if (i == j) return m_[i][j] = (s[i] == '*');

if ((s[i] == '(' || s[i] == '*')

&&(s[j] == ')' || s[j] == '*')

&& isValid(s, i + 1, j - 1))

return m_[i][j] = 1;

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

if (isValid(s, i, p) && isValid(s, p + 1, j))

return m_[i][j] = 1;

return m_[i][j] = 0;

}

};

C++ / DP/ Bottom-up O(n^3)

// Author: Huahua

// Runtime: 12 ms

class Solution {

public:

bool checkValidString(const string& s) {

int l = s.length();

if (l == 0) return true;

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

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

if (s[i] == '*') dp[i][i] = 1;

for (int len = 2; len <= l; ++len)

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

int j = i + len - 1;

// (...), *...), (...*, *...*

if ((s[i] == '(' || s[i] == '*')

&&(s[j] == ')' || s[j] == '*'))

if (len == 2 || dp[i + 1][j - 1]) {

dp[i][j] = 1;

continue;

}

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

if (dp[i][k] && dp[k + 1][j]) {

dp[i][j] = 1;

break;

}

return dp[0][l - 1];

}

};

C++ / Counting O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

bool checkValidString(string s) {

int min_op = 0;

int max_op = 0;

for (char c : s) {

if (c == '(') ++min_op; else --min_op;

if (c != ')') ++max_op; else --max_op;

if (max_op < 0) return false;

min_op = max(0, min_op);

}

return min_op == 0;

}

};

Java / DP / Bottom-up O(n^3)

class Solution {

public boolean checkValidString(String s) {

int l = s.length();

if (l == 0) return true;

boolean[][] dp = new boolean[l][l];

char[] ss = s.toCharArray();

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

if (ss[i] == '*') dp[i][i] = true;

for (int len = 2; len <= l; ++len)

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

int j = i + len - 1;

if ((ss[i] == '*' || ss[i] == '(')

&&(ss[j] == '*' || ss[j] == ')')) {

if (len == 2 || dp[i + 1][j - 1]) {

dp[i][j] = true;

continue;

}

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

if (dp[i][k] && dp[k + 1][j]) {

dp[i][j] = true;

break;

}

return dp[0][l - 1];

}