Leetcode 931: Minimum Falling Path Sum
Given a square matrix of integers, your task is to find the minimum sum of any falling path through the matrix. A falling path starts at any element in the first row and chooses the next element from the row directly below it, which can be either directly below, diagonally left, or diagonally right.
Problem
Approach
Steps
Complexity
Input: You are given an n x n matrix, where each element represents a value. You need to find the falling path with the minimum sum.
Example: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Constraints:
• n == matrix.length == matrix[i].length
• 1 <= n <= 100
• -100 <= matrix[i][j] <= 100
Output: Return the minimum sum of a falling path.
Example: Output: 13
Constraints:
• The matrix will always be a square.
Goal: To find the falling path with the minimum sum, update the matrix values row by row, adding the smallest possible sum from the previous row to the current row.
Steps:
• 1. Start from the second row and update each element by adding the minimum of the three possible elements from the previous row (directly above, diagonally left, diagonally right).
• 2. Once the matrix is updated, the minimum sum will be the smallest value in the last row.
Goal: The input matrix will always be a square with integer values.
Steps:
• 1 <= n <= 100
• -100 <= matrix[i][j] <= 100
Assumptions:
• The matrix is square, and all elements are integers.
• Input: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
• Explanation: To get the minimum falling path sum, start with any element in the first row, and for each subsequent row, pick the minimum value from the adjacent elements in the previous row. For this example, the minimum sum is 13, which is the path [1, 5, 7].
• Input: Input: matrix = [[-19,57],[-40,-5]]
• Explanation: For this example, the minimum falling path sum is -59, which follows the path [-19, -40].
Approach: We solve the problem using dynamic programming by updating the matrix to store the minimum path sum at each position. This allows us to compute the minimum sum in a bottom-up manner.
Observations:
• The problem can be solved efficiently using dynamic programming, where each element in the matrix is updated to represent the minimum falling path sum up to that point.
• By using dynamic programming, we avoid recalculating the same subproblems multiple times, making the solution more efficient.
Steps:
• Step 1: Traverse the matrix starting from the second row.
• Step 2: For each element in the current row, add the minimum of the three possible elements from the previous row (directly above, diagonally left, diagonally right).
• Step 3: Once the matrix is updated, the result will be the minimum value in the last row.
Empty Inputs:
• The matrix will not be empty, as per the constraints.
Large Inputs:
• The algorithm should handle matrices up to 100x100 efficiently.
Special Values:
• The algorithm should work for matrices with negative numbers and ensure the minimum path sum is correctly calculated.
Constraints:
• The matrix is always square, which simplifies the implementation.
int minFallingPathSum(vector<vector<int>>& mtx) {
int m = mtx.size(), n = mtx[0].size();
for(int i = 1; i < m; i++)
for(int j = 0; j < n; j++) {
int l = max(j - 1, 0);
int r = min(j + 1, n - 1);
mtx[i][j] += min(mtx[i- 1][l], min(mtx[i - 1][j], mtx[i - 1][r]));
}
int res = mtx[m - 1][0];
for(int j = 0; j < n; j++)
res = min(res, mtx[m - 1][j]);
return res;
}
1 : Function Definition
int minFallingPathSum(vector<vector<int>>& mtx) {
Define the function 'minFallingPathSum', which takes a matrix 'mtx' and returns the minimum sum of a falling path.
2 : Matrix Size
int m = mtx.size(), n = mtx[0].size();
Store the number of rows 'm' and columns 'n' in the matrix 'mtx'.
3 : Outer Loop
for(int i = 1; i < m; i++)
Loop through each row starting from the second row (index 1), as the first row does not require modification.
4 : Inner Loop
for(int j = 0; j < n; j++) {
Loop through each column 'j' in the current row 'i'.
5 : Left Bound Calculation
int l = max(j - 1, 0);
Calculate the index 'l' of the left cell. If 'j - 1' is less than 0, set 'l' to 0 to prevent out-of-bounds access.
6 : Initialize Result
int res = mtx[m - 1][0];
Initialize 'res' to the value of the first element in the last row of the matrix, which will be used to track the minimum sum.
7 : Final Row Loop
for(int j = 0; j < n; j++)
Loop through each element in the last row of the matrix to find the minimum value.
8 : Update Minimum Result
res = min(res, mtx[m - 1][j]);
Update 'res' to the minimum value between the current 'res' and the current element in the last row of the matrix.
9 : Return Result
return res;
Return the minimum falling path sum from the top to the bottom of the matrix.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is O(n^2), where n is the number of rows (or columns) in the matrix, as we traverse each element once and update it based on the previous row.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1), as we update the matrix in place and do not use any extra space.
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