Today I have done Rotate Image and leetcode's February LeetCoding Challenge with cpp.

Rotate Image

Description

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

Example 3:

Input: matrix = [[1]]
Output: [[1]]

Example 4:

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

Constraints:

• matrix.length == n
• matrix[i].length == n
• 1 <= n <= 20
• -1000 <= matrix[i][j] <= 1000

Solution

check comment for for details

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int low = n / 2;
        int high = n - low;
        for(int j = 0; j < high; ++j) {
            for(int i = 0; i+j < n-1-j; ++i) {
                swap(matrix[j][j+i], matrix[j+i][n-1-j]);
                swap(matrix[j][j+i], matrix[n-1-i-j][j]);
                swap(matrix[n-1-j][n-1-i-j], matrix[n-1-i-j][j]);
            }
        }
    }
};

// [0, 0], [0, n-1], [n-1, n-1], [n-1, 0] -> [n-1, 0], [0, 0], [0, n-1], [n-1, n-1]
// [0, 1], [1, n-1], [n-1, n-2], [n-2, 0] -> [n-2, 0], [0, 1], [1, n-1], [n-1, n-2]
// ...
// [0, i], [i, n-1], [n-1, n-1-i], [n-1-i, 0] -> ...
// ...
// [0, n-2], [n-2, n-1], [n-1, 1], [1, 0] -> ...

// [1, 1], [1, n-2], [n-2, n-2], [n-2, 1] -> ...
// [1, 2], [2, n-2], [n-2, n-3], [n-3, 1] -> ...
// ...
// [1, 1+i], [1+i, n-2], [n-2, n-2-i], [n-2-i, 1] -> ...

// [j, j+i], [j+i, n-1-j], [n-1-j, n-1-i-j], [n-1-i-j, j] -> ...
    

February LeetCoding Challenge 12

Description

Number of Steps to Reduce a Number to Zero

Given a non-negative integer num, return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.

Example 1:

Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7. 
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3. 
Step 4) 3 is odd; subtract 1 and obtain 2. 
Step 5) 2 is even; divide by 2 and obtain 1. 
Step 6) 1 is odd; subtract 1 and obtain 0.

Example 2:

Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4. 
Step 2) 4 is even; divide by 2 and obtain 2. 
Step 3) 2 is even; divide by 2 and obtain 1. 
Step 4) 1 is odd; subtract 1 and obtain 0.

Example 3:

Input: num = 123
Output: 12

Constraints:

• 0 <= num <= 10^6

Solution

class Solution {
public:
    int numberOfSteps (int num) {
        return num ? log2(num) + __builtin_popcount(num) : 0;
    }
};