Today I have done Count Negative Numbers in a Sorted Matrix and leetcode's May LeetCoding Challenge with cpp.

Count Negative Numbers in a Sorted Matrix

Description

Given a m x n matrix grid which is sorted in non-increasing order both row-wise and column-wise, return the number of negative numbers in grid.

Example 1:

Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
Output: 8
Explanation: There are 8 negatives number in the matrix.

Example 2:

Input: grid = [[3,2],[1,0]]
Output: 0

Example 3:

Input: grid = [[1,-1],[-1,-1]]
Output: 3

Example 4:

Input: grid = [[-1]]
Output: 1

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• -100 <= grid[i][j] <= 100

Follow up: Could you find an O(n + m) solution?

Solution

class Solution {
public:
  int countNegatives(vector<vector<int>>& grid) {
    int answer = 0;
    for(auto &row : grid) {
      for(auto i : row) {
        answer += i < 0;
      }
    }
    return answer;
  }
};

because it's sorted, so we can solve it in $O(n+m)$ time

class Solution {
public:
  int countNegatives(vector<vector<int>>& grid) {
    int rows = grid.size();
    int cols = grid.front().size();
    int row = rows;
    int col = 0;
    int answer = 0;
    while(row-- && col < cols) {
      while(col < cols && grid[row][col] >= 0) {
        col += 1;
      }
      answer += cols - col;
    }
    return answer;
  }
};

May LeetCoding Challenge 24

Description

To Lower Case

Given a string s, return the string after replacing every uppercase letter with the same lowercase letter.

Example 1:

Input: s = "Hello"
Output: "hello"

Example 2:

Input: s = "here"
Output: "here"

Example 3:

Input: s = "LOVELY"
Output: "lovely"

Constraints:

• 1 <= s.length <= 100
• s consists of printable ASCII characters.

Solution

class Solution {
public:
  string toLowerCase(string s) {
    for(auto &c : s) c = tolower(c);
    return s;
  }
};