Posts published in July 2018

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 869. Reordered Power of 2

By zxi on July 14, 2018

Problem

Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this in a way such that the resulting number is a power of 2.

Example 1:

Input: 1
Output: true

Example 2:

Input: 10
Output: false

Example 3:

Input: 16
Output: true

Example 4:

Input: 24
Output: false

Example 5:

1 2	<strong>Input: </strong><span id="example-input-5-1">46</span> <strong>Output: </strong><span id="example-output-5">true</span>

Note:

1 <= N <= 10^9

Solution: HashTable

Compare the counter of digit string with that of all power of 2s.

e.g. 64 -> {4: 1, 6: 1} == 46 {4:1, 6: 1}

Time complexity: O(1)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool reorderedPowerOf2(int N) {

auto m = countMap(N);

for (int i = 0; i < 31; ++i)

if (m == countMap(1 << i)) return true;

return false;

}

private:

map<int, int> countMap(int n) {

map<int, int> m;

while (n) {

++m[n % 10];

n /= 10;

}

return m;

}

};

花花酱 LeetCode 287. Find the Duplicate Number

By zxi on July 14, 2018

Problem

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

Note:

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n²).
There is only one duplicate number in the array, but it could be repeated more than once.

Solution1: Binary Search

Time complexity: O(nlogn)

Space complexity: O(1)

Find the smallest m such that len(nums <= m) > m, which means m is the duplicate number.

In the sorted form [1, 2, …, m1, m2, m + 1, …, n]

There are m+1 numbers <= m

// Author: Huahua

// Running time: 8 ms

class Solution {

public:

int findDuplicate(vector<int>& nums) {

int l = 1;

int r = nums.size();

while (l < r) {

int m = (r - l) / 2 + l;

int count = 0; // len(nums <= m)

for (int num : nums)

if (num <= m) ++count;

if (count <= m)

l = m + 1;

else

r = m;

}

return l;

}

};

Solution: Linked list cycle

Convert the problem to find the entry point of the cycle in a linked list.

Take the number in the array as the index of next node.

[1,3,4,2,2]

0->1->3->2->4->2 cycle: 2->4->2

[3,1,3,4,2]

0->3->4->2->3->4->2 cycle 3->4->2->3

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int findDuplicate(vector<int>& nums) {

int slow = 0;

int fast = 0;

while (true) {

slow = nums[slow];

fast = nums[nums[fast]];

if (slow == fast) break;

}

fast = 0;

while (fast != slow) {

slow = nums[slow];

fast = nums[fast];

}

return slow;

}

};

Posts published in July 2018

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

花花酱 LeetCode 869. Reordered Power of 2

Problem

Solution: HashTable

花花酱 LeetCode 287. Find the Duplicate Number

Problem

Solution1: Binary Search

Solution: Linked list cycle