Posts tagged as “sort”

花花酱 LeetCode 1402. Reducing Dishes

By zxi on April 4, 2020

A chef has collected data on the satisfaction level of his n dishes. Chef can cook any dish in 1 unit of time.

Like-time coefficient of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. time[i]*satisfaction[i]

Return the maximum sum of Like-time coefficient that the chef can obtain after dishes preparation.

Dishes can be prepared in any order and the chef can discard some dishes to get this maximum value.

Example 1:

Input: satisfaction = [-1,-8,0,5,-9]
Output: 14
Explanation: After Removing the second and last dish, the maximum total Like-time coefficient will be equal to (-1*1 + 0*2 + 5*3 = 14). Each dish is prepared in one unit of time.

Example 2:

Input: satisfaction = [4,3,2]
Output: 20
Explanation: Dishes can be prepared in any order, (2*1 + 3*2 + 4*3 = 20)

Example 3:

Input: satisfaction = [-1,-4,-5]
Output: 0
Explanation: People don't like the dishes. No dish is prepared.

Example 4:

Input: satisfaction = [-2,5,-1,0,3,-3]
Output: 35

Constraints:

n == satisfaction.length
1 <= n <= 500
-10^3 <= satisfaction[i] <= 10^3

Solution 1: Sort + Brute Force

Time complexity: O(nlogn + n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int maxSatisfaction(vector<int>& s) {

const int n = s.size();

sort(begin(s), end(s));

int ans = 0;

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

int v = 0;

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

v += s[j] * (j - i + 1);

ans = max(ans, v);

}

return ans;

}

};

Solution 2: Sort + prefix sum

Sort in reverse order, accumulate prefix sum until prefix sum <= 0.

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

[9, 8, 5, 2, 1, -1]
sum = 9 * 4 + 8 * 3 + 2 * 3 + 1 * 2 + -1 * 1
<=>
sum += 9
sum += (9 + 8 = 17)
sum += (17 + 2 = 19)
sum += (19 + 1 = 20)
sum += (20 – 1 = 19)

C++

// Author: Huahua

class Solution {

public:

int maxSatisfaction(vector<int>& s) {

const int n = s.size();

sort(rbegin(s), rend(s));

int ans = 0;

int prefix = 0;

for (int v : s) {

prefix += v;

if (prefix <= 0) break;

ans += prefix;

}

return ans;

}

};

花花酱 LeetCode 1385. Find the Distance Value Between Two Arrays

By zxi on March 21, 2020

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays.

The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d.

Example 1:

Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
Output: 2
Explanation: 
For arr1[0]=4 we have: 
|4-10|=6 > d=2 
|4-9|=5 > d=2 
|4-1|=3 > d=2 
|4-8|=4 > d=2 
For arr1[1]=5 we have: 
|5-10|=5 > d=2 
|5-9|=4 > d=2 
|5-1|=4 > d=2 
|5-8|=3 > d=2
For arr1[2]=8 we have:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2

Example 2:

Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
Output: 2

Example 3:

Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
Output: 1

Constraints:

1 <= arr1.length, arr2.length <= 500
-10^3 <= arr1[i], arr2[j] <= 10^3
0 <= d <= 100

Solution 1: All pairs

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

C++

// Author: Huahua

class Solution {

public:

int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {

int ans = 0;

for (int x : arr1) {

bool flag = true;

for (int y : arr2)

if (abs(x - y) <= d) {

flag = false;

break;

}

ans += flag;

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:

return sum(all(abs(x - y) > d for y in arr2) for x in arr1)

Solution 2: Two Pointers

Sort arr1 in ascending order and sort arr2 in descending order.
Time complexity: O(mlogm + nlogn + m + n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {

sort(begin(arr1), end(arr1));

sort(rbegin(arr2), rend(arr2));

int ans = 0;

for (int a : arr1) {

while (arr2.size() && arr2.back() < a - d)

arr2.pop_back();

ans += arr2.empty() || arr2.back() > a + d;

}

return ans;

}

};

Solution 3: Binary Search

Sort arr2 in ascending order. and do two binary searches for each element to determine the range of [a-d, a+d], if that range is empty we increase the counter

Time complexity: O(mlogm + nlogm)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {

sort(begin(arr2), end(arr2));

int ans = 0;

for (int a : arr1) {

auto it1 = lower_bound(begin(arr2), end(arr2), a - d);

auto it2 = upper_bound(begin(arr2), end(arr2), a + d);

if (it1 == it2) ++ans;

}

return ans;

}

};

花花酱 LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

By zxi on March 2, 2020

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

Constraints:

2 <= nums.length <= 500
0 <= nums[i] <= 100

Solution 1: Brute Force

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

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

const int n = nums.size();

vector<int> ans(n);

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

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

if (i != j && nums[j] < nums[i]) ++ans[i];

return ans;

}

};

Solution 2: Sort + Binary Search

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

C++

// Author: Huahua

class Solution {

public:

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

const int n = nums.size();

vector<int> s(nums);

sort(begin(s), end(s));

vector<int> ans(n);

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

ans[i] = distance(begin(s), lower_bound(begin(s), end(s), nums[i]));

return ans;

}

};

Solution 3: Cumulative frequency

Time complexity: O(n)
Space complexity: O(101)

C++

// Author: Huahua

class Solution {

public:

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

vector<int> f(101);

for (int x : nums) ++f[x];

partial_sum(begin(f), end(f), begin(f));

for (int& x : nums)

x = x == 0 ? 0 : f[x - 1];

return nums;

}

};

花花酱 LeetCode 1356. Sort Integers by The Number of 1 Bits

By zxi on February 22, 2020

Given an integer array arr. You have to sort the integers in the array in ascending order by the number of 1’s in their binary representation and in case of two or more integers have the same number of 1’s you have to sort them in ascending order.

Return the sorted array.

Example 1:

Input: arr = [0,1,2,3,4,5,6,7,8]
Output: [0,1,2,4,8,3,5,6,7]
Explantion: [0] is the only integer with 0 bits.
[1,2,4,8] all have 1 bit.
[3,5,6] have 2 bits.
[7] has 3 bits.
The sorted array by bits is [0,1,2,4,8,3,5,6,7]

Example 2:

Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
Output: [1,2,4,8,16,32,64,128,256,512,1024]
Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.

Example 3:

Input: arr = [10000,10000]
Output: [10000,10000]

Example 4:

Input: arr = [2,3,5,7,11,13,17,19]
Output: [2,3,5,17,7,11,13,19]

Example 5:

Input: arr = [10,100,1000,10000]
Output: [10,100,10000,1000]

Constraints:

1 <= arr.length <= 500
0 <= arr[i] <= 10^4

Solution: Sorting

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

C++

// Author: Huahua

class Solution {

public:

vector<int> sortByBits(vector<int>& arr) {

sort(begin(arr), end(arr), [](const int a, const int b){

int key_a = __builtin_popcount(a);

int key_b = __builtin_popcount(b);

if (key_a != key_b) return key_a < key_b;

return a < b;

});

return arr;

}

};

Python3

# Author: Huahua

class Solution:

def sortByBits(self, arr: List[int]) -> List[int]:

return sorted(arr, key=lambda x: (bin(x).count('1') << 16) + x)

花花酱 LeetCode 853. Car Fleet

By zxi on April 28, 2019

N cars are going to the same destination along a one lane road. The destination is target miles away.

Each car i has a constant speed speed[i] (in miles per hour), and initial position position[i] miles towards the target along the road.

A car can never pass another car ahead of it, but it can catch up to it, and drive bumper to bumper at the same speed.

The distance between these two cars is ignored – they are assumed to have the same position.

A car fleet is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.

If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.

How many car fleets will arrive at the destination?

Example 1:

Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]
Output: 3
Explanation:
The cars starting at 10 and 8 become a fleet, meeting each other at 12.
The car starting at 0 doesn't catch up to any other car, so it is a fleet by itself.
The cars starting at 5 and 3 become a fleet, meeting each other at 6.
Note that no other cars meet these fleets before the destination, so the answer is 3.

Note:

0 <= N <= 10 ^ 4
0 < target <= 10 ^ 6
0 < speed[i] <= 10 ^ 6
0 <= position[i] < target
All initial positions are different.

Solution: Greedy

Compute the time when each car can reach target.
Sort cars by position DESC

Answer will be number slow cars in the time array.

Ex 1: 
target = 12
p = [10,8,0,5,3] 
v = [2,4,1,1,3]


p     t
10    1  <- slow -- ^
 8    1             |
 5    7  <- slow -- ^
 3    3             |
 0   12  <- slow -- ^

Ex 2
target = 10
p = [5, 4, 3, 2, 1]
v = [1, 2, 3, 4, 5]

p     t
5     5  <- slow -- ^
4     3             |
3     2.33          |
2     2             |
1     1.8           |

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

C++

// Author: Huahua, running time: 52 ms / 10.9 MB

class Solution {

public:

int carFleet(int target, vector<int>& position, vector<int>& speed) {

vector<pair<int, float>> cars(position.size());

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

cars[i] = {position[i], static_cast<float>(target - position[i]) / speed[i]};

sort(rbegin(cars), rend(cars));

int ans{0};

float max_t{0};

for (const auto& [p, t] : cars)

if (t > max_t) {

max_t = t;

++ans;

}

return ans;

}

};

Python3

class Solution:

def carFleet(self, target: int, position: List[int], speed: List[int]) -> int:

cars = sorted([(p, (target - p) / s) for p, s in zip(position, speed)],

key=lambda x: -x[0])

ans, max_t = 0, 0

for p, t in cars:

if t > max_t:

ans += 1

max_t = t

return ans