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花花酱 LeetCode Weekly Contest 130 (1017, 1018, 1019, 1020)

By zxi on March 31, 2019

1017. Convert to Base -2

Solution: Math / Simulation

Time complexity: O(logn)
Space complexity: O(logn)

C++

// Author: Huahua

class Solution {

public:

string baseNeg2(int N) {

if (N == 0) return "0";

vector<char> ans;

while (N) {

ans.push_back((N & 1) + '0');

N = -(N >> 1);

}

return {rbegin(ans), rend(ans)};

}

};

base K

// Author: Huahua

class Solution {

public:

string baseNeg2(int N) {

if (N == 0) return "0";

return baseNegK(N, -2);

}

private:

string baseNegK(int N, int K) {

vector<char> ans;

while (N) {

int r = N % K;

N /= K;

if (r < 0) {

r += -K;

N += 1;

}

ans.push_back(r + '0');

}

return {rbegin(ans), rend(ans)};

}

};

1018. Binary Prefix Divisible By 5

Solution: Math

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

vector<bool> prefixesDivBy5(vector<int>& A) {

int num = 0;

vector<bool> ans(A.size());

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

num = ((num << 1) | A[i]) % 5;

ans[i] = num == 0;

}

return ans;

}

};

1019. Next Greater Node In Linked List

Solution: Reverse + Monotonic Stack

Process in reverse order and keep a monotonically increasing stack, pop all the elements that are smaller than the current one, then the top of the stack (if exists) will be the next greater node.

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

C++

// Author: Huahua

class Solution {

public:

vector<int> nextLargerNodes(ListNode* head) {

vector<int> nums;

while (head) {

nums.push_back(head->val);

head = head->next;

}

stack<int> s;

vector<int> ans(nums.size());

for (int i = nums.size() - 1; i >= 0; --i) {

while (!s.empty() && s.top() <= nums[i]) s.pop();

ans[i] = s.empty() ? 0 : s.top();

s.push(nums[i]);

}

return ans;

}

};

1020. Number of Enclaves

Solution: DFS / Connected Components

Time complexity: O(mn)
Space complexity: O(mn)

C++

// Author: Huahua

class Solution {

public:

int numEnclaves(vector<vector<int>>& A) {

int ans = 0;

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

for (int j = 0; j < A[0].size(); ++j) {

int count = 0;

if (!dfs(A, j, i, count))

ans += count;

}

return ans;

}

private:

static constexpr int dirs[5]{0, -1, 0, 1, 0};

bool dfs(vector<vector<int>>& A, int x, int y, int& count) {

if (x < 0 || x == A[0].size() || y < 0 || y == A.size()) return true;

if (A[y][x] == 0) return false;

++count;

A[y][x] = 0;

bool valid = false;

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

valid |= dfs(A, x + dirs[i], y + dirs[i + 1], count);

return valid;

}

};

花花酱 LeetCode 1003. Check If Word Is Valid After Substitutions

By zxi on March 3, 2019

We are given that the string "abc" is valid.

From any valid string V, we may split V into two pieces X and Y such that X + Y (X concatenated with Y) is equal to V. (X or Y may be empty.) Then, X + "abc" + Y is also valid.

If for example S = "abc", then examples of valid strings are: "abc", "aabcbc", "abcabc", "abcabcababcc". Examples of invalid strings are: "abccba", "ab", "cababc", "bac".

Return true if and only if the given string S is valid.

Example 1:

Input: "aabcbc"
Output: true
Explanation: 
We start with the valid string "abc".
Then we can insert another "abc" between "a" and "bc", resulting in "a" + "abc" + "bc" which is "aabcbc".

Example 2:

Input: "abcabcababcc"
Output: true
Explanation: 
"abcabcabc" is valid after consecutive insertings of "abc".
Then we can insert "abc" before the last letter, resulting in "abcabcab" + "abc" + "c" which is "abcabcababcc".

Example 3:

Input: "abccba"
Output: false

Example 4:

Input: "cababc"
Output: false

Note:

1 <= S.length <= 20000
S[i] is 'a', 'b', or 'c'

Solution: Stack

If current char can be appended to the stack do so, if the top of stack is “abc” pop, otherwise push the current char to the stack. Check whether the stack is empty after all chars were processed.

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

C++

// Author: Huahua, running time: 28 ms, 11.9 MB

class Solution {

public:

bool isValid(string S) {

stack<string> st;

for (char ch : S) {

if (st.empty()) {

st.push("");

}

string& s = st.top();

char exp = 'a' + s.length();

if (ch == exp) {

s += ch;

if (s == "abc") st.pop();

} else if (ch != 'a'){

return false;

} else {

st.push("a");

}

return st.empty();

}

};

花花酱 LeetCode 964. Least Operators to Express Number

By zxi on December 28, 2018

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, *, or /). For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.

When writing such an expression, we adhere to the following conventions:

The division operator (/) returns rational numbers.
There are no parentheses placed anywhere.
We use the usual order of operations: multiplication and division happens before addition and subtraction.
It’s not allowed to use the unary negation operator (-). For example, “x - x” is a valid expression as it only uses subtraction, but “-x + x” is not because it uses negation.

We would like to write an expression with the least number of operators such that the expression equals the given target. Return the least number of operators used.

Example 1:

Input: x = 3, target = 19 
Output: 5 
Explanation: 3 * 3 + 3 * 3 + 3 / 3.  The expression contains 5 operations.

Example 2:

Input: x = 5, target = 501 
Output: 8 
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.  The expression contains 8 operations.

Example 3:

Input: x = 100, target = 100000000 
Output: 3 
Explanation: 100 * 100 * 100 * 100.  The expression contains 3 operations.

Note:

2 <= x <= 100
1 <= target <= 2 * 10^8

Solution: Dijkstra

Find the shortest path from target to 0 using ops.

Time complexity: O(nlogn)
Space complexity: O(nlogn)
n = x * log(t) / log(x)

C++/set

class Solution {

public:

int leastOpsExpressTarget(int x, int target) {

set<pair<int, int>> q; // # x -> number

unordered_set<int> s;

q.emplace(0, target);

while (!q.empty()) {

const auto it = begin(q);

const int c = it->first;

const int t = it->second;

q.erase(it);

if (t == 0) return c - 1;

if (s.count(t)) continue;

s.insert(t);

int n = log(t) / log(x);

int l = t - pow(x, n);

q.emplace(c + (n == 0 ? 2 : n), l);

int r = pow(x, n + 1) - t;

q.emplace(c + n + 1, r);

}

return -1;

}

};

C++/heap

class Solution {

public:

int leastOpsExpressTarget(int x, int target) {

priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;

unordered_set<int> s;

q.emplace(0, target);

while (!q.empty()) {

const int c = q.top().first;

const int t = q.top().second;

q.pop();

if (t == 0) return c - 1;

if (s.count(t)) continue;

s.insert(t);

int n = log(t) / log(x);

int l = t - pow(x, n);

q.emplace(c + (n == 0 ? 2 : n), l);

int r = pow(x, n + 1) - t;

q.emplace(c + n + 1, r);

}

return -1;

}

};

Solution 2: DFS + Memoization

Time complexity: O(x * log(t)/log(x))
Space complexity: O(x * log(t)/log(x))

C++

// Author: Huahua, running time: 8 ms

class Solution {

public:

int leastOpsExpressTarget(int x, int target) {

return dp(x, target);

}

private:

unordered_map<int, int> m_;

int dp(int x, int t) {

if (t == 0) return 0;

if (t < x) return min(2 * t - 1, 2 * (x - t));

if (m_.count(t)) return m_.at(t);

int k = log(t) / log(x);

long p = pow(x, k);

if (t == p) return m_[t] = k - 1;

int ans = dp(x, t - p) + k; // t - p < t

long left = p * x - t;

if (left < t) // only rely on smaller sub-problems

ans = min(ans, dp(x, left) + k + 1);

return m_[t] = ans;

}

};

花花酱 LeetCode 957. Prison Cells After N Days

By zxi on December 20, 2018

There are 8 prison cells in a row, and each cell is either occupied or vacant.

Each day, whether the cell is occupied or vacant changes according to the following rules:

If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
Otherwise, it becomes vacant.

(Note that because the prison is a row, the first and the last cells in the row can’t have two adjacent neighbors.)

We describe the current state of the prison in the following way: cells[i] == 1 if the i-th cell is occupied, else cells[i] == 0.

Given the initial state of the prison, return the state of the prison after N days (and N such changes described above.)

Example 1:

Input: cells = [0,1,0,1,1,0,0,1], N = 7
Output: [0,0,1,1,0,0,0,0]
Explanation: The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]

Example 2:

Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]

Note:

cells.length == 8
cells[i] is in {0, 1}
1 <= N <= 10^9

Solution: Simulation

Simulate the process, since there must be loops, record the last day when a state occurred.

Time complexity: O(2^8)
Space complexity: O(2^8)

C++

// Author: Huahua, running time: 4 ms

class Solution {

public:

vector<int> prisonAfterNDays(vector<int>& cells, int N) {

vector<int> seen(1 << 8, -1);

auto getKey = [] (const vector<int>& cells) {

int key = 0;

for (int i = 0; i < 8; ++i)

key |= cells[i] << i;

return key;

};

while (N > 0) {

const int key = getKey(cells);

if (seen[key] > 0)

N %= seen[key] - N;

seen[key] = N--;

if (N < 0) break;

vector<int> c(8);

for (int i = 1; i < 7; ++i)

c[i] = cells[i - 1] == cells[i + 1];

cells.swap(c);

}

return cells;

}

};

花花酱 LeetCode 922. Sort Array By Parity II

By zxi on October 14, 2018

Problem

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even.

You may return any answer array that satisfies this condition.

Example 1:

Input: [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Note:

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

Solution 1: Brute Force

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<int> sortArrayByParityII(vector<int>& A) {

vector<int> evens;

vector<int> odds;

for (int a : A)

if (a % 2 == 0)

evens.push_back(a);

else

odds.push_back(a);

auto it1 = begin(evens);

auto it2 = begin(odds);

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

A[i] = *it1++;

swap(it1, it2);

}

return A;

}

};