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Minimum Path Sum(LeetCode)

Problem Statement​

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Examples​

Example 1:

image

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

Constraints​

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 200
  • 0 <= grid[i][j] <= 200

Solution​

The problem is to find the minimum path sum from the top-left corner to the bottom-right corner of a grid. You can only move either down or right at any point in time. Each cell in the grid contains a non-negative integer which represents the cost to traverse that cell.

Approach​

Algorithm​

  1. Initialization:
  • Create a 1D array cur of size m (the number of rows in the grid) to store the minimum path sum for the current column.
  • Initialize cur[0] to grid[0][0] since the starting point is the cost of the top-left cell.
  • Fill the first column of cur by accumulating the costs of the cells in the first column of the grid.
  1. Dynamic Programming:
  • Iterate through each column starting from the second column.
  • For each column, update the first cell by adding the cost of the current cell to the value of the first cell in cur from the previous column.
  • For the remaining cells in the column, update each cell by taking the minimum of the value from the left cell and the value from the top cell (stored in cur) and adding the cost of the current cell.
  1. Result:
  • After iterating through the grid, cur[m-1] will contain the minimum path sum to reach the bottom-right corner of the grid.

Implementation​

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<int> cur(m, grid[0][0]);
        
        // Initialize the first column
        for (int i = 1; i < m; i++)
            cur[i] = cur[i - 1] + grid[i][0]; 
        
        // Fill the rest of the grid
        for (int j = 1; j < n; j++) {
            cur[0] += grid[0][j]; // Update the top cell of the current column
            for (int i = 1; i < m; i++)
                cur[i] = min(cur[i - 1], cur[i]) + grid[i][j];
        }
        return cur[m - 1];
    }
};

Complexity Analysis​

  • Time complexity: O(M∗N)O(M*N) - where m is the number of rows and n is the number of columns. Each cell in the grid is processed once.
  • Space complexity: O(M)O(M) - where m is the number of rows. We use a single 1D array of size m to store the minimum path sums for the current column.

Conclusion​

This dynamic programming solution efficiently computes the minimum path sum in a grid by maintaining only a 1D array to store the current state of the path sums, which significantly optimizes space usage.