花花酱 LeetCode 238. Product of Array Except Self

By zxi on July 16, 2018

Problem:

Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input:  [1,2,3,4] Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

Solution: DP

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Running time: 36 ms

class Solution {

public:

vector<int> productExceptSelf(vector<int>& nums) {

const int n = nums.size();

vector<int> l(n, 1); // l[i] = prod(nums[0] ~ nums[i - 1])

vector<int> r(n, 1); // l[i] = prod(nums[i + 1] ~ nums[n - 1])

vector<int> ans(n);

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

l[i] = l[i - 1] * nums[i - 1];

for (int i = n - 2; i >= 0; --i)

r[i] = r[i + 1] * nums[i + 1];

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

ans[i] = l[i] * r[i];

return ans;

}

};

// Author: Huahua

// Running time: 32 ms

class Solution {

public:

vector<int> productExceptSelf(vector<int>& nums) {

const int n = nums.size();

int l = 1; // l = prod(nums[0] ~ nums[i - 1])

int r = 1; // r = prod(nums[i + 1] ~ nums[n - 1])

vector<int> ans(n, 1);

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

ans[i] *= l;

ans[n - i - 1] *= r;

l *= nums[i];

r *= nums[n - i - 1];

}

return ans;

}

};

花花酱 LeetCode 739. Daily Temperatures

By zxi on July 16, 2018

Problem

Given a list of daily temperatures, produce a list that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.

For example, given the list temperatures = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].

Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].

Solution: Stack

Use a stack to track indices of future warmer days. From top to bottom: recent to far away.

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Running time: 148 ms

class Solution {

public:

vector<int> dailyTemperatures(vector<int>& temperatures) {

const int n = temperatures.size();

stack<int> s;

vector<int> ans(n);

for (int i = n - 1; i >= 0; --i) {

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

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

s.push(i);

}

return ans;

}

};

花花酱 LeetCode 643. Maximum Average Subarray I

By zxi on July 16, 2018

Problem

题目大意：找出k长度的子数组的平均值的最大值。

https://leetcode.com/problems/maximum-average-subarray-i/description/

Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.

Example 1:

Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75

Note:

1 <= k <= n <= 30,000.
Elements of the given array will be in the range [-10,000, 10,000].

Solution: Sliding Window

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 124 ms

class Solution {

public:

double findMaxAverage(vector<int>& nums, int k) {

const int n = nums.size();

int sum = 0;

int ans = INT_MIN;

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

if (i >= k) sum -= nums[i - k];

sum += nums[i];

if (i + 1 >= k) ans = max(ans, sum);

}

return static_cast<double>(ans) / k;

}

};

Problem

A car travels from a starting position to a destination which is target miles east of the starting position.

Along the way, there are gas stations. Each station[i] represents a gas station that is station[i][0] miles east of the starting position, and has station[i][1] liters of gas.

The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses 1 liter of gas per 1 mile that it drives.

When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.

What is the least number of refueling stops the car must make in order to reach its destination? If it cannot reach the destination, return -1.

Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. If the car reaches the destination with 0 fuel left, it is still considered to have arrived.

Example 1:

Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.

Example 2:

Input: target = 100, startFuel = 1, stations = [[10,100]]
Output: -1
Explanation: We can't reach the target (or even the first gas station).

Example 3:

Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]]
Output: 2
Explanation: 
We start with 10 liters of fuel.
We drive to position 10, expending 10 liters of fuel.  We refuel from 0 liters to 60 liters of gas.
Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
and refuel from 10 liters to 50 liters of gas.  We then drive to and reach the target.
We made 2 refueling stops along the way, so we return 2.

Note:

1 <= target, startFuel, stations[i][1] <= 10^9
0 <= stations.length <= 500
0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target

Solution1: DP

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

const int n = stations.size();

// dp[i]: max distance to go with i stops.

vector<long> dp(n + 1, startFuel);

// For each station

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

// for j stops, start from high to low to avoid reusing station i

for (int j = i + 1; j >= 1; --j)

if (dp[j - 1] >= stations[i][0])

// if we can reach station i with j - 1 stops,

// we can reach dp[j - 1] + station[i][1] with j stops.

dp[j] = max(dp[j], dp[j - 1] + stations[i][1]);

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

if (dp[i] >= target) return i;

return -1;

}

};

Solution2: Priority Queue

Time complexity: O(nlogn)

Space complexity: O(n)

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

int cur = startFuel;

int stops = 0;

int i = 0;

priority_queue<int> q; // gas (high to low) of reachable stations.

while (true) {

if (cur >= target) return stops;

while (i < stations.size() && stations[i][0] <= cur)

q.push(stations[i++][1]);

if (q.empty()) break;

cur += q.top(); q.pop();

++stops;

}

return -1;

}

};

V2: Iterator

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

int cur = startFuel;

int stops = 0;

priority_queue<int> q; // gas (high to low) of reachable stations.

auto it = begin(stations);

while (true) {

if (cur >= target) return stops;

while (it != end(stations) && it->at(0) <= cur)

q.push(it++->at(1));

if (q.empty()) break;

cur += q.top(); q.pop();

++stops;

}

return -1;

}

};

Problem

Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].

Return any permutation of A that maximizes its advantage with respect to B.

Example 1:

Input: A = [2,7,11,15], B = [1,10,4,11]
Output: [2,11,7,15]

Example 2:

Input: A = [12,24,8,32], B = [13,25,32,11]
Output: [24,32,8,12]

Note:

1 <= A.length = B.length <= 10000
0 <= A[i] <= 10^9
0 <= B[i] <= 10^9

Solution: Greedy 田忌赛马

Use the smallest unused number A[j] in A such that A[j] > B[i], if not possible, use the smallest number in A.

Time complexity: O(nlogn)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 124 ms

class Solution {

public:

vector<int> advantageCount(vector<int>& A, vector<int>& B) {

multiset<int> s(begin(A), end(A));

vector<int> ans;

for (int b : B) {

auto it = s.upper_bound(b);

if (it == s.end()) it = s.begin();

ans.push_back(*it);

s.erase(it);

}

return ans;

}

};

花花酱 LeetCode 238. Product of Array Except Self

Problem:

Solution: DP

花花酱 LeetCode 739. Daily Temperatures

Problem

Solution: Stack

花花酱 LeetCode 643. Maximum Average Subarray I

Problem

Solution: Sliding Window

Related Problems

花花酱 LeetCode 871. Minimum Number of Refueling Stops

Problem

Solution1: DP

Solution2: Priority Queue

Related Problems

花花酱 LeetCode 870. Advantage Shuffle

Problem

Solution: Greedy 田忌赛马

Huahua's Tech Road

花花酱 LeetCode 238. Product of Array Except Self

Problem:

Solution: DP

花花酱 LeetCode 739. Daily Temperatures

Problem

Solution: Stack

花花酱 LeetCode 643. Maximum Average Subarray I

Problem

Solution: Sliding Window

Related Problems

花花酱 LeetCode 871. Minimum Number of Refueling Stops

Problem

Solution1: DP

Solution2: Priority Queue

Related Problems

花花酱 LeetCode 870. Advantage Shuffle

Problem

Solution: Greedy 田忌赛马