2022-04-11 Daily-Challenge
Today I have done leetcode's April LeetCoding Challenge with cpp
.
April LeetCoding Challenge 11
Description
Shift 2D Grid
Given a 2D grid
of size m x n
and an integer k
. You need to shift the grid
k
times.
In one shift operation:
- Element at
grid[i][j]
moves togrid[i][j + 1]
. - Element at
grid[i][n - 1]
moves togrid[i + 1][0]
. - Element at
grid[m - 1][n - 1]
moves togrid[0][0]
.
Return the 2D grid after applying shift operation k
times.
Example 1:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output: [[9,1,2],[3,4,5],[6,7,8]]
Example 2:
Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
Example 3:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
Output: [[1,2,3],[4,5,6],[7,8,9]]
Constraints:
m == grid.length
n == grid[i].length
1 <= m <= 50
1 <= n <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100
Solution
auto speedup = [](){
cin.tie(nullptr);
cout.tie(nullptr);
ios::sync_with_stdio(false);
return 0;
}();
class Solution {
public:
vector<vector<int>> shiftGrid(vector<vector<int>>& grid, int k) {
int rows = grid.size();
int cols = grid.front().size();
k %= (rows * cols);
if(!k) return move(grid);
vector<int> temp;
temp.reserve(k);
for(int row = 0; temp.size() != k && row < rows; ++row) {
for(int col = 0; temp.size() != k && col < cols; ++col) {
temp.push_back(grid[row][col]);
}
}
for(int row = rows - 1; row >= 0; --row) {
for(int col = cols - 1; row * cols + col >= k && col >= 0; --col) {
cout << row << ' '<< col << endl;
int targetPos = (row * cols + col + k) % (rows * cols);
grid[targetPos / cols][targetPos % cols] = grid[row][col];
}
}
for(int i = 0; i < k; ++i) {
int targetPos = (i + k) % (rows * cols);
grid[targetPos / cols][targetPos % cols] = temp[i];
}
return move(grid);
}
};
// Accepted
// 107/107 cases passed (46 ms)
// Your runtime beats 47.01 % of cpp submissions
// Your memory usage beats 100 % of cpp submissions (13.6 MB)