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

Count Negative Numbers in a Sorted Matrix

Description

Given a m * 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

Solution

nothing to say

class Solution {
 public:
  int countNegatives(vector<vector<int>>& grid) {
    int l = grid.size(), ll = grid[0].size(), cnt = 0;
    for (int i = 0; i < l; ++i) {
      for (int j = 0; j < ll; ++j) {
        cnt += grid[i][j] < 0;
      }
    }
    return cnt;
  }
};

October LeetCoding Challenge 15

Description

Rotate Array

Given an array, rotate the array to the right by k steps, where k is non-negative.

Follow up:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

• 1 <= nums.length <= 2 * 104
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

Solution

nothing to say

class Solution {
 public:
  void rotate(vector<int>& nums, int k) {
    vector<int> ans;
    int offset = (nums.size() - (k % nums.size())) % nums.size();
    auto it = nums.begin();
    advance(it, offset);
    for (auto i = it; i < nums.end(); ++i) {
      ans.push_back(*i);
    }
    for (auto i = nums.begin(); i < it; ++i) {
      ans.push_back(*i);
    }
    nums = ans;
  }
};