Leetcode 2850: Minimum Moves to Spread Stones Over Grid
You are given a 3x3 grid representing stones placed in each cell. In one move, you can move a stone from its current cell to an adjacent cell. The goal is to place one stone in each cell, minimizing the number of moves.
Problem
Approach
Steps
Complexity
Input: The input consists of a 3x3 grid where each cell contains an integer representing the number of stones in that cell.
Example: grid = [[1, 1, 0], [1, 1, 1], [1, 2, 1]]
Constraints:
• grid.length == grid[i].length == 3
• 0 <= grid[i][j] <= 9
• Sum of all grid[i][j] values is equal to 9.
Output: Return the minimum number of moves required to place exactly one stone in each cell of the grid.
Example: Output: 3
Constraints:
• The output is an integer indicating the minimum number of moves.
Goal: Calculate the minimum number of moves to place exactly one stone in each cell.
Steps:
• Identify cells that are empty (i.e., containing zero stones).
• Determine the shortest move distance to transfer stones from cells containing more than one stone to empty cells.
• Repeat until all cells contain exactly one stone.
Goal: Constraints on the grid values and dimensions.
Steps:
• grid.length == grid[i].length == 3
• 0 <= grid[i][j] <= 9
• Sum of grid[i][j] is exactly 9.
Assumptions:
• The grid is always 3x3 and contains exactly 9 stones.
• Stones can be moved to any adjacent cell.
• Input: grid = [[1, 1, 0], [1, 1, 1], [1, 2, 1]]
• Explanation: Starting with stones placed as shown in the grid, it takes 3 moves to move the stones around so that each cell contains exactly one stone.
• Input: grid = [[1, 3, 0], [1, 0, 0], [1, 0, 3]]
• Explanation: In this case, it takes 4 moves to rearrange the stones and place exactly one in each cell.
Approach: The approach involves moving stones from cells with excess stones to empty cells while minimizing the number of moves required to distribute the stones evenly.
Observations:
• We need to consider the cells that already contain stones and find the nearest empty cells to move stones into.
• A greedy approach can be employed to always move stones to the closest empty cell, minimizing the distance traveled.
Steps:
• Iterate through the grid and identify cells with no stones.
• For each empty cell, calculate the shortest distance to cells with more than one stone.
• Move a stone to the empty cell, update the grid, and repeat the process until every cell contains exactly one stone.
Empty Inputs:
• There are no empty grid inputs since the grid is always 3x3.
Large Inputs:
• The algorithm should efficiently handle the grid size, which is always fixed at 3x3.
Special Values:
• When a grid has already one stone in each cell, no moves are needed.
Constraints:
• The solution needs to be efficient enough to handle up to 9 stones in the grid.
int minimumMoves(vector<vector<int>>& grid) {
// Base Case
int t = 0;
for (int i = 0; i < 3; ++i)
for (int j = 0; j < 3; ++j)
if (grid[i][j] == 0)
t++;
if (t == 0)
return 0;
int ans = INT_MAX;
for (int i = 0; i < 3; ++i)
{
for (int j = 0; j < 3; ++j)
{
if (grid[i][j] == 0)
{
for (int ni = 0; ni < 3; ++ni)
{
for (int nj = 0; nj < 3; ++nj)
{
int d = abs(ni - i) + abs(nj - j);
if (grid[ni][nj] > 1)
{
grid[ni][nj]--;
grid[i][j]++;
ans = min(ans, d + minimumMoves(grid));
grid[ni][nj]++;
grid[i][j]--;
}
}
}
}
}
}
return ans;
}
1 : Function Definition
int minimumMoves(vector<vector<int>>& grid) {
This line defines the function 'minimumMoves' which takes a 2D vector 'grid' as input, representing the current state of the grid.
2 : Variable Declaration
int t = 0;
This initializes a variable 't' to zero, which will be used to count the number of zeroes in the grid.
3 : Loop
for (int i = 0; i < 3; ++i)
This is the outer loop that iterates over the rows of the grid.
4 : Loop
for (int j = 0; j < 3; ++j)
This inner loop iterates over the columns of the grid.
5 : Condition
if (grid[i][j] == 0)
This condition checks if the current cell in the grid contains zero.
6 : Action
t++;
If a zero is found, increment the variable 't' to track the number of zeroes.
7 : Condition
if (t == 0)
This condition checks if there are no zeroes in the grid, in which case no moves are needed.
8 : Return Statement
return 0;
If there are no zeroes in the grid, the function returns 0, indicating no moves are required.
9 : Variable Declaration
int ans = INT_MAX;
This initializes a variable 'ans' to the maximum possible integer value, which will hold the minimum number of moves.
10 : Loop
for (int i = 0; i < 3; ++i)
This loop iterates over the rows of the grid again for checking possible moves.
11 : Loop
for (int j = 0; j < 3; ++j)
Inner loop to check each column within the current row.
12 : Condition
if (grid[i][j] == 0)
This condition checks again for the presence of a zero in the current grid cell.
13 : Loop
for (int ni = 0; ni < 3; ++ni)
This loop iterates over the rows to check for possible movements.
14 : Loop
for (int nj = 0; nj < 3; ++nj)
Inner loop iterating over the columns to check each grid element for possible moves.
15 : Block Start
{
Start of the block where the distance calculation and move checking occurs.
16 : Distance Calculation
int d = abs(ni - i) + abs(nj - j);
This calculates the Manhattan distance between the current cell and the target cell.
17 : Condition
if (grid[ni][nj] > 1)
This condition checks if the current target cell is eligible for a move (greater than 1).
18 : Action
grid[ni][nj]--;
Decrement the value of the target cell.
19 : Action
grid[i][j]++;
Increment the value of the current zero cell.
20 : Recursive Call
ans = min(ans, d + minimumMoves(grid));
Make a recursive call to calculate the minimum number of moves after the current swap.
21 : Action
grid[ni][nj]++;
Restore the value of the target cell after the recursive call.
22 : Action
grid[i][j]--;
Restore the value of the current zero cell after the recursive call.
23 : Return Statement
return ans;
Return the minimum number of moves found.
Best Case: O(1)
Average Case: O(1)
Worst Case: O(1)
Description: The time complexity is O(1) because the grid size is fixed and the solution involves a fixed number of operations for a 3x3 grid.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we only use a fixed amount of space to store the grid and perform calculations.
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