Posts published in “Array”

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 1395. Count Number of Teams

By zxi on March 28, 2020

There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
A team is valid if: (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1).

Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.

Example 3:

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

Constraints:

n == rating.length
1 <= n <= 200
1 <= rating[i] <= 10^5

Solution 1: Brute Force

Time complexity: O(n^3)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int numTeams(vector<int>& rating) {

int n = rating.size();

int ans = 0;

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

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

for (int k = j + 1; k < n; ++k)

if (rating[i] < rating[j] && rating[j] < rating[k]

|| rating[i] > rating[j] && rating[j] > rating[k])

++ans;

return ans;

}

};

Solution 2: Math

For each soldier j, count how many soldiers on his left has smaller ratings as left[j], count how many soldiers his right side has larger ratings as right[j]

ans = sum(left[j] * right[j] + (j – left[j]) * (n – j – 1 * right[j])

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

C++

// Author: Huahua, 4 ms

class Solution {

public:

int numTeams(vector<int>& rating) {

int n = rating.size();

int ans = 0;

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

int l = 0;

int r = 0;

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

if (rating[i] < rating[j]) ++l;

for (int k = j + 1; k < n; ++k)

if (rating[j] < rating[k]) ++r;

ans += (l * r) + (j - l) * (n - j - 1 - r);

}

return ans;

}

};

花花酱 LeetCode 1380. Lucky Numbers in a Matrix

By zxi on March 15, 2020

Given a m * n matrix of distinct numbers, return all lucky numbers in the matrix in any order.

A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.

Example 1:

Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
Output: [15]
Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column

Example 2:

Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
Output: [12]
Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.

Example 3:

Input: matrix = [[7,8],[1,2]]
Output: [7]

Constraints:

m == mat.length
n == mat[i].length
1 <= n, m <= 50
1 <= matrix[i][j] <= 10^5.
All elements in the matrix are distinct.

Solution: Pre-processing

Two pass. First pass, record the min val of each row, and max val of each column.
Second pass, identify lucky numbers.

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

C++

// Author: Huahua

class Solution {

public:

vector<int> luckyNumbers (vector<vector<int>>& matrix) {

const int m = matrix.size();

const int n = matrix[0].size();

vector<int> rows(m, INT_MAX);

vector<int> cols(n, INT_MIN);

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

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

rows[i] = min(rows[i], matrix[i][j]);

cols[j] = max(cols[j], matrix[i][j]);

}

vector<int> ans;

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

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

if (matrix[i][j] == rows[i] &&

matrix[i][j] == cols[j])

ans.push_back(matrix[i][j]);

return ans;

}

};

花花酱 LeetCode 240. Search a 2D Matrix II

By zxi on March 13, 2020

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Solution 1: Two Pointers

Start from first row + last column, if the current value is larger than target, –column; if smaller then ++row.

e.g.
1. r = 0, c = 4, v = 15, 15 > 5 => –c
2. r = 0, c = 3, v = 11, 11 > 5 => –c
3. r = 0, c = 2, v = 7, 7 > 5 => –c
4. r = 0, c = 1, v = 4, 4 < 5 => ++r
5. r = 1, c = 1, v = 5, 5 = 5, found it!

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

C++

// Author: Huahua

class Solution {

public:

bool searchMatrix(vector<vector<int>>& matrix, int target) {

if (matrix.empty() || matrix[0].empty()) return false;

const int m = matrix.size();

const int n = matrix[0].size();

int r = 0;

int c = matrix[0].size() - 1;

while (r < m && c >= 0) {

if (matrix[r][c] == target) return true;

else if (matrix[r][c] > target) --c;

else ++r;

}

return false;

}

};

花花酱 LeetCode 229. Majority Element II

By zxi on March 13, 2020

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Note: The algorithm should run in linear time and in O(1) space.

Example 1:

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

Example 2:

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

Solution: Boyer–Moore Voting Algorithm

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

C++

// Author: Huahua

class Solution {

public:

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

int n1 = 0;

int c1 = 0;

int n2 = 1;

int c2 = 0;

for (int num : nums) {

if (num == n1) {

++c1;

} else if (num == n2) {

++c2;

} else if (c1 == 0) {

n1 = num;

c1 = 1;

} else if (c2 == 0) {

n2 = num;

c2 = 1;

} else {

--c1;

--c2;

}

c1 = c2 = 0;

for (int num : nums) {

if (num == n1) ++c1;

else if (num == n2) ++c2;

}

const int c = nums.size() / 3;

vector<int> ans;

if (c1 > c) ans.push_back(n1);

if (c2 > c) ans.push_back(n2);

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def majorityElement(self, nums: List[int]) -> List[int]:

n1, c1, n2, c2 = 0, 0, 1, 0

for num in nums:

if num == n1: c1 += 1

elif num == n2: c2 += 1

elif c1 == 0: n1, c1 = num, 1

elif c2 == 0: n2, c2 = num, 1

else: c1, c2 = c1 - 1, c2 -1

c1, c2 = 0, 0

for num in nums:

if num == n1: c1 += 1

elif num == n2: c2 += 1

ans = []

if c1 > len(nums) // 3: ans.append(n1)

if c2 > len(nums) // 3: ans.append(n2)

return ans

Posts published in “Array”

花花酱 LeetCode 1402. Reducing Dishes

Solution 1: Sort + Brute Force

C++

Solution 2: Sort + prefix sum

C++

花花酱 LeetCode 1395. Count Number of Teams

Solution 1: Brute Force

C++

Solution 2: Math

C++

花花酱 LeetCode 1380. Lucky Numbers in a Matrix

Solution: Pre-processing

C++

花花酱 LeetCode 240. Search a 2D Matrix II

Solution 1: Two Pointers

C++

花花酱 LeetCode 229. Majority Element II

Solution: Boyer–Moore Voting Algorithm

C++

Python3

Related Problem