Posts tagged as “array”

花花酱 LeetCode 496. Next Greater Element I

By zxi on March 5, 2018

题目大意：给你一个组数A里面每个元素都不相同。再给你一个数组B，元素是A的子集，问对于B中的每个元素，在A数组中相同元素之后第一个比它的元素是多少。

Problem:

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1‘s elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].

Output: [-1,3,-1]

Explanation:

For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.

For number 1 in the first array, the next greater number for it in the second array is 3.

For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].

Output: [3,-1]

Explanation:

For number 2 in the first array, the next greater number for it in the second array is 3.

For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:

All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Running time: 13 ms

class Solution {

public:

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

vector<int> ans;

for (int num : findNums) {

ans.push_back(-1);

int index = std::find(nums.begin(), nums.end(), num) - nums.begin();

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

if (nums[i] > num) {

ans.back() = nums[i];

break;

}

return ans;

}

};

Solution 2: HashTable + Brute Force

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 15 ms

class Solution {

public:

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

unordered_map<int, int> indices;

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

indices[nums[i]] = i;

vector<int> ans;

for (int num : findNums) {

ans.push_back(-1);

for (int i = indices[num]; i < nums.size(); ++i)

if (nums[i] > num) {

ans.back() = nums[i];

break;

}

return ans;

}

};

Solution 3: Stack + HashTable

Using a stack to store the nums whose next greater isn’t found yet.

// Author: Huahua

// Running time: 11 ms

class Solution {

public:

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

stack<int> s;

unordered_map<int, int> next;

for (int num : nums) {

while (!s.empty() && num > s.top()) {

next[s.top()] = num;

s.pop();

}

s.push(num);

}

vector<int> ans;

for (int num : findNums)

ans.push_back(next.count(num) ? next[num] : -1);

return ans;

}

};

花花酱 LeetCode 11. Container With Most Water

By zxi on December 3, 2017

Problem:

Given n non-negative integers a₁, a₂, …, a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

Examples:

input: [1 3 2 4]
output: 6
explanation: use 3, 4, we have the following, which contains (4th-2nd) * min(3, 4) = 2 * 3 = 6 unit of water.

|~~|

Idea:

Two pointers

Time complexity: O(n)

Space complexity: O(1)

Solution:

C++ two pointers

// Author: Huahua

// Runtime: 19 ms

class Solution {

public:

int maxArea(const vector<int>& height) {

int ans = 0;

int l = 0;

int r = height.size() - 1;

while (l < r) {

int h = min(height[l], height[r]);

ans = max(ans, h * (r - l));

if (height[l] < height[r])

++l;

else

--r;

}

return ans;

}

};

Java

// Author: Huahua

// Runtime: 13 ms

class Solution {

public int maxArea(int[] height) {

int l = 0;

int r = height.length - 1;

int ans = 0;

while (l < r) {

int h = Math.min(height[l], height[r]);

ans = Math.max(ans, h * (r - l));

if (height[l] < height[r])

++l;

else

--r;

}

return ans;

}

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 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];

}

花花酱 LeetCode 683. K Empty Slots

By zxi on September 29, 2017

题目大意：有n个花盆，第i天，第flowers[i]个花盆的花会开。问是否存在一天，两朵花之间有k个空花盆。

Problem:

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.

Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.

For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.

Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.

If there isn’t such day, output -1.

Example 1:

Input: 
flowers: [1,3,2]
k: 1
Output: 2
Explanation: In the second day, the first and the third flower have become blooming.

Example 2:

Input: 
flowers: [1,2,3]
k: 1
Output: -1

Note:

The given array will be in the range [1, 20000].

Idea:

BST/Buckets

Solution 2: BST

C++

// Author: Huahua

// Runtime: 228 ms

class Solution {

public:

int kEmptySlots(vector<int>& flowers, int k) {

int n = flowers.size();

if (n == 0 || k >= n) return -1;

set<int> xs;

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

int x = flowers[i];

auto r = xs.insert(x).first;

auto l = r;

if (++r != xs.end() && *r == x + k + 1) return i + 1;

if (l != xs.begin() && *(--l) == x - k - 1) return i + 1;

}

return -1;

}

};

Solution 3: Buckets

C++

// Author: Huahua

// Runtime: 196 ms (better than 94%)

class Solution {

public:

int kEmptySlots(vector<int>& flowers, int k) {

int n = flowers.size();

if (n == 0 || k >= n) return -1;

++k;

int bs = (n + k - 1) / k;

vector<int> lows(bs, INT_MAX);

vector<int> highs(bs, INT_MIN);

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

int x = flowers[i];

int p = x / k;

if (x < lows[p]) {

lows[p] = x;

if (p > 0 && highs[p - 1] == x - k) return i + 1;

}

if (x > highs[p]) {

highs[p] = x;

if (p < bs - 1 && lows[p + 1] == x + k) return i + 1;

}

return -1;

}

};

Solution 1: TLE with latest test cases

C++

// Author: Huahua

// Runtime: 192 ms (better than 97.85%)

class Solution {

public:

int kEmptySlots(vector<int>& flowers, int k) {

int n = flowers.size();

if (n == 0 || k >= n) return -1;

std::unique_ptr<char[]> f(new char[n+1]);

memset(f.get(), 0, (n + 1)*sizeof(char));

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

if (IsValid(flowers[i], k, n, f.get()))

return i + 1;

return -1;

}

private:

bool IsValid(int x, int k, int n, char* f) {

f[x] = 1;

if (x + k + 1 <= n && f[x + k + 1]) {

bool valid = true;

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

if (f[x + i]) {

valid = false;

break;

}

if (valid) return true;

}

if (x - k - 1 > 0 && f[x - k - 1]) {

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

if (f[x - i]) return false;

return true;

}

return false;

}

};

Java

// Author: Huahua

// Runtime: 17 ms

class Solution {

public int kEmptySlots(int[] flowers, int k) {

int n = flowers.length;

if (n == 0 || k >= n) return -1;

int[] f = new int[n + 1];

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

if (IsValid(flowers[i], k, n, f))

return i + 1;

return -1;

}

private boolean IsValid(int x, int k, int n, int[] f) {

f[x] = 1;

if (x + k + 1 <= n && f[x + k + 1] == 1) {

boolean valid = true;

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

if (f[x + i] == 1) {

valid = false;

break;

}

if (valid) return true;

}

if (x - k - 1 > 0 && f[x - k - 1] == 1) {

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

if (f[x - i] == 1) return false;

return true;

}

return false;

}

Python

"""

Author: Huahua

Runtime: 398 ms

"""

class Solution:

def kEmptySlots(self, flowers, k):

n = len(flowers)

f = [0] * (n + 1)

i = 0

def isValid(x, k, n, f):

f[x] = 1

if x + k + 1 <= n and f[x + k + 1] == 1:

valid = True

for i in range(k):

if f[x + i + 1] == 1:

valid = False

break

if valid: return True

if x - k - 1 > 0 and f[x - k - 1] == 1:

for i in range(k):

if f[x - i - 1] == 1:

return False

return True

return False

for x in flowers:

i += 1

if isValid(x, k, n, f): return i

return -1