Posts published in “Algorithms”

花花酱 LeetCode 1582. Special Positions in a Binary Matrix

By zxi on September 12, 2020

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

Example 1:

Input: mat = [[1,0,0],
              [0,0,1],
              [1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:

Input: mat = [[1,0,0],
              [0,1,0],
              [0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions.

Example 3:

Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:

Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

Constraints:

rows == mat.length
cols == mat[i].length
1 <= rows, cols <= 100
mat[i][j] is 0 or 1.

Solution: Sum for each row and column

Brute force:
Time complexity: O(R*C*(R+C))
Space complexity: O(1)

We can pre-compute the sums for each row and each column, ans = sum(mat[r][c] == 1 and rsum[r] == 1 and csum[c] == 1)

Time complexity: O(R*C)
Space complexity: O(R+C)

C++

class Solution {

public:

int numSpecial(vector<vector<int>>& mat) {

const int rows = mat.size();

const int cols = mat[0].size();

vector<int> rs(rows);

vector<int> cs(cols);

for (int r = 0; r < rows; ++r)

for (int c = 0; c < cols; ++c) {

rs[r] += mat[r][c];

cs[c] += mat[r][c];

}

int ans = 0;

for (int r = 0; r < rows; ++r)

for (int c = 0; c < cols; ++c)

ans += mat[r][c] && rs[r] == 1 && cs[c] == 1;

return ans;

}

};

花花酱 LeetCode 1572. Matrix Diagonal Sum

By zxi on September 5, 2020

Given a square matrix mat, return the sum of the matrix diagonals.

Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagonal.

Example 1:

Input: mat = [[1,2,3],
              [4,5,6],
              [7,8,9]]
Output: 25
Explanation: Diagonals sum: 1 + 5 + 9 + 3 + 7 = 25
Notice that element mat[1][1] = 5 is counted only once.

Example 2:

Input: mat = [[1,1,1,1],
              [1,1,1,1],
              [1,1,1,1],
              [1,1,1,1]]
Output: 8

Example 3:

Input: mat = [[5]]
Output: 5

Constraints:

n == mat.length == mat[i].length
1 <= n <= 100
1 <= mat[i][j] <= 100

Solution: Brute Force

Note: if n is odd, be careful not to double count the center one.

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

C++

class Solution {

public:

int diagonalSum(vector<vector<int>>& mat) {

const int n = mat.size();

int ans = 0;

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

ans += mat[i][i] + mat[i][n - i - 1];

if (n & 1) ans -= mat[n / 2][n / 2];

return ans;

}

};

花花酱 LeetCode 1566. Detect Pattern of Length M Repeated K or More Times

By zxi on August 29, 2020

Given an array of positive integers arr, find a pattern of length m that is repeated k or more times.

A pattern is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times consecutively without overlapping. A pattern is defined by its length and the number of repetitions.

Return true if there exists a pattern of length m that is repeated k or more times, otherwise return false.

Example 1:

Input: arr = [1,2,4,4,4,4], m = 1, k = 3
Output: true
Explanation: The pattern (4) of length 1 is repeated 4 consecutive times. Notice that pattern can be repeated k or more times but not less.

Example 2:

Input: arr = [1,2,1,2,1,1,1,3], m = 2, k = 2
Output: true
Explanation: The pattern (1,2) of length 2 is repeated 2 consecutive times. Another valid pattern (2,1) is also repeated 2 times.

Example 3:

Input: arr = [1,2,1,2,1,3], m = 2, k = 3
Output: false
Explanation: The pattern (1,2) is of length 2 but is repeated only 2 times. There is no pattern of length 2 that is repeated 3 or more times.

Example 4:

Input: arr = [1,2,3,1,2], m = 2, k = 2
Output: false
Explanation: Notice that the pattern (1,2) exists twice but not consecutively, so it doesn't count.

Example 5:

Input: arr = [2,2,2,2], m = 2, k = 3
Output: false
Explanation: The only pattern of length 2 is (2,2) however it's repeated only twice. Notice that we do not count overlapping repetitions.

Constraints:

2 <= arr.length <= 100
1 <= arr[i] <= 100
1 <= m <= 100
2 <= k <= 100

Solution 1: Brute Force

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

C++

class Solution {

public:

bool containsPattern(vector<int>& arr, int m, int k) {

const int n = arr.size();

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

bool valid = true;

for (int j = 1; j < k && valid; ++j)

for (int p = 0; p < m && valid; ++p)

if (arr[i + j * m + p] != arr[i + p])

valid = false;

if (valid) return true;

}

return false;

}

};

Solution 2: Shift and count

Since we need k consecutive subarrays, we can compare arr[i] with arr[i + m], if they are the same, increase the counter, otherwise reset the counter. If the counter reaches (k – 1) * m, it means we found k consecutive subarrays of length m.

ex1: arr = [1,2,4,4,4,4], m = 1, k = 3
i arr[i], arr[i + m] counter
0 1. 2. 0
0 2. 4. 0
0 4. 4. 1
0 4. 4. 2. <– found

ex2: arr = [1,2,1,2,1,1,1,3], m = 2, k = 2
i arr[i], arr[i + m] counter
0 1. 1. 1
0 2. 2. 2 <– found

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

C++

class Solution {

public:

bool containsPattern(vector<int>& arr, int m, int k) {

const int n = arr.size();

int count = 0;

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

if (arr[i] == arr[i + m]) {

if (++count == (k - 1) * m) return true;

} else {

count = 0;

}

return false;

}

};

花花酱 LeetCode 1552. Magnetic Force Between Two Balls

By zxi on August 15, 2020

In universe Earth C-137, Rick discovered a special form of magnetic force between two balls if they are put in his new invented basket. Rick has n empty baskets, the i^th basket is at position[i], Morty has m balls and needs to distribute the balls into the baskets such that the minimum magnetic force between any two balls is maximum.

Rick stated that magnetic force between two different balls at positions x and y is |x - y|.

Given the integer array position and the integer m. Return the required force.

Example 1:

Input: position = [1,2,3,4,7], m = 3
Output: 3
Explanation: Distributing the 3 balls into baskets 1, 4 and 7 will make the magnetic force between ball pairs [3, 3, 6]. The minimum magnetic force is 3. We cannot achieve a larger minimum magnetic force than 3.

Example 2:

Input: position = [5,4,3,2,1,1000000000], m = 2
Output: 999999999
Explanation: We can use baskets 1 and 1000000000.

Constraints:

n == position.length
2 <= n <= 10^5
1 <= position[i] <= 10^9
All integers in position are distinct.
2 <= m <= position.length

Solution: Binary Search

Find the max distance that we can put m balls.

Time complexity: O(n*log(distance))
Space complexity: O(1)

C++

class Solution {

public:

int maxDistance(vector<int>& position, int m) {

const int n = position.size();

sort(begin(position), end(position));

auto count = [&](int d) -> int {

int last = position[0];

int ans = 1;

for (int x : position) {

if (x - last >= d) {

last = x;

++ans;

}

return ans;

};

int l = 0;

int t = (position.back() - position.front()) + 1;

int r = t;

while (l < r) {

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

if (count(t - mid) >= m) // find min of l => max of t - l

r = mid;

else

l = mid + 1;

}

return t - l;

}

};

Java

class Solution {

public int maxDistance(int[] position, int m) {

Arrays.sort(position);

final int n = position.length;

int l = 0;

int r = position[n - 1] - position[0] + 1;

int t = r;

while (l < r) {

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

if (getCount(position, t - mid) >= m)

r = mid;

else

l = mid + 1;

}

return t - l;

}

private int getCount(int[] position, int d) {

int count = 1;

int last = position[0];

for (int x : position) {

if (x - last >= d) {

++count;

last = x;

}

return count;

}

Python3

class Solution:

def maxDistance(self, position: List[int], m: int) -> int:

position.sort()

def getCount(d: int) -> int:

last, count = position[0], 1

for x in position:

if x - last >= d:

last = x

count += 1

return count

l, r = 0, position[-1] - position[0] + 1

t = r

while l < r:

mid = l + (r - l) // 2

if getCount(t - mid) >= m:

r = mid

else:

l = mid + 1

return t - l

花花酱 LeetCode 1550. Three Consecutive Odds

By zxi on August 15, 2020

Given an integer array arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false.

Example 1:

Input: arr = [2,6,4,1]
Output: false
Explanation: There are no three consecutive odds.

Example 2:

Input: arr = [1,2,34,3,4,5,7,23,12]
Output: true
Explanation: [5,7,23] are three consecutive odds.

Constraints:

1 <= arr.length <= 1000
1 <= arr[i] <= 1000

Solution: Counting

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

C++

class Solution {

public:

bool threeConsecutiveOdds(vector<int>& arr) {

int count = 0;

for (int x : arr) {

if (x & 1) {

if (++count == 3) return true;

} else {

count = 0;

}

return false;

}

};