Minimum Path Sum
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Move from the top-left cell of a grid of non-negative numbers to the bottom-right cell, stepping only down or right. Of all such routes, return the smallest possible sum of the cells you walk through.
Example 1
- Input
- grid = [[2,4,3],[5,1,6],[7,8,2]]
- Output
- 15
- Why
- Path: 2 -> 4 -> 1 -> 6 -> 2 = 15. Alternative 2->4->3->6->2 = 17 or 2->5->1->6->2 = 16.
Example 2
- Input
- grid = [[3,1,4],[2,5,6],[7,8,9]]
- Output
- 23
- Why
- Path: 3->1->4->6->9 = 23.
Example 3
- Input
- grid = [[1]]
- Output
- 1
- Why
- Single cell grid, the path sum is just that cell.
Constraints
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 200
- 0 <= grid[i][j] <= 100
How to think about it
Updated 2026-09-29Every cell (r, c) can only be entered from the cell directly above or the cell directly to the left. The cheapest cost to enter (r, c) is grid[r][c] plus the minimum of those two predecessors. No backtracking is needed because path steps are strictly monotonic.
Approaches, worst first
Recursive path search
time O(2^(m+n)) · space O(m + n)
From (0, 0), branch right and down, accumulating cell values to find the minimum leaf. Exponentially explores overlapping paths.
2D dynamic programming grid
time O(m * n) · space O(m * n)
Create an m x n matrix dp where dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]). Initialize top-left cell and fill borders as running prefix sums.
1D rolling rowWrite this one
time O(m * n) · space O(n)
Use a single row array of size n. Initialize with row 0's prefix sums. For subsequent rows, update dp[0] += grid[i][0] and dp[j] = grid[i][j] + min(dp[j], dp[j-1]).
Where people lose marks · 3
- Uninitialized border transitions: cell (0, j) can only come from (0, j-1) and cell (i, 0) can only come from (i-1, 0). Blindly taking min with uninitialized cells injects false zeroes.
- Single cell grid (1x1): ensure loop boundaries do not skip the initial grid[0][0] value.
- Overwriting the input grid directly when pure functional immutability is required by the caller.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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