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‎posts/algos/leetcode-62-unique-paths.txt‎

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LeetCode 62, "Unique Paths," is a classic problem that often serves as an excellent introduction to dynamic programming. It challenges us to find the number of unique paths a robot can take to reach the bottom-right corner of a `m x n` grid, starting from the top-left corner. The robot can only move either down or right at any point in time.
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## Problem Description
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Imagine a robot positioned at the top-left cell (0,0) of a grid with `m` rows and `n` columns. The robot's goal is to reach the bottom-right cell (`m-1`, `n-1`). The only allowed moves are one step down or one step right. We need to calculate the total number of distinct paths the robot can take to reach its destination.
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Let's visualize a simple `3 x 7` grid:
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```
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S . . . . . .
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. . . . . . .
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. . . . . . F
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```
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Where `S` is the start and `F` is the finish.
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## Dynamic Programming Approach
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This problem has optimal substructure and overlapping subproblems, making it a perfect candidate for dynamic programming.
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Consider a cell `(i, j)` in the grid. To reach this cell, the robot must have come either from the cell directly above it (`i-1, j`) by moving down, or from the cell directly to its left (`i, j-1`) by moving right.
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Therefore, the number of unique paths to reach `(i, j)` is the sum of unique paths to reach `(i-1, j)` and unique paths to reach `(i, j-1)`.
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Let `dp[i][j]` represent the number of unique paths to reach cell `(i, j)`.
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The recurrence relation is:
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`dp[i][j] = dp[i-1][j] + dp[i][j-1]`
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**Base Cases:**
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* For any cell in the first row (`i=0`), there's only one way to reach it: by moving right repeatedly from `(0,0)`. So, `dp[0][j] = 1`.
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* For any cell in the first column (`j=0`), there's only one way to reach it: by moving down repeatedly from `(0,0)`. So, `dp[i][0] = 1`.
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* The starting cell `(0,0)` has `dp[0][0] = 1` path (it's already there).
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We can build a 2D array (or even optimize space to a 1D array) to store these path counts.
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## Go Solution
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Here's an implementation of the dynamic programming approach in Go:
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```go
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func uniquePaths(m int, n int) int {
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// Create a 2D DP array initialized with 1s for the base cases
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dp := make([][]int, m)
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for i := range dp {
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dp[i] = make([]int, n)
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}
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// Initialize the first row and first column with 1s
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// since there's only one way to reach any cell in the first row/column
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// (by only moving right or only moving down respectively).
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for i := 0; i < m; i++ {
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dp[i][0] = 1
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}
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for j := 0; j < n; j++ {
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dp[0][j] = 1
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}
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// Fill the DP table
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for i := 1; i < m; i++ {
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for j := 1; j < n; j++ {
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dp[i][j] = dp[i-1][j] + dp[i][j-1]
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}
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}
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// The result is the value at the bottom-right corner
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return dp[m-1][n-1]
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}
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```
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### Combinatorial Approach
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Alternatively, this problem can be solved using a combinatorial approach. To reach the bottom-right corner of an `m x n` grid, the robot must make exactly `m-1` 'down' moves and `n-1` 'right' moves. The total number of moves will be `(m-1) + (n-1)`.
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The problem then reduces to finding the number of ways to arrange these `m-1` down moves and `n-1` right moves. This is a classic combinatorial problem: choosing `m-1` positions for the 'down' moves (or `n-1` positions for the 'right' moves) out of a total of `(m-1) + (n-1)` moves.
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The formula for combinations is C(N, K) = N! / (K! * (N-K)!), where N is the total number of steps and K is the number of 'down' (or 'right') moves.
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### Go Solution (Combinatorial)
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```go
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func uniquePathsCombinatorial(m int, n int) int {
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downMoves := m - 1
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rightMoves := n - 1
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totalSteps := downMoves + rightMoves
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// Choose the smaller of downMoves or rightMoves for k to minimize calculations
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k := downMoves
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if rightMoves < downMoves {
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k = rightMoves
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}
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var comb float64 = 1.0
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// Formula: C(N, K) = (N/1) * ((N-1)/2) * ... * ((N-k+1)/k)
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// This avoids large factorial calculations by performing multiplications and divisions iteratively.
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for i := 1; i <= k; i++ {
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comb = comb * float64(totalSteps - i + 1) / float64(i)
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}
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return int(comb)
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}
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```
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## Conclusion
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The "Unique Paths" problem demonstrates the power of dynamic programming in breaking down a complex problem into simpler, overlapping subproblems. By carefully defining our state and recurrence relation, we can build up the solution efficiently. This particular problem also has a combinatorial solution using binomial coefficients, but the dynamic programming approach is often more intuitive for beginners to DP.

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"title": "LeetCode 62: Unique Paths - A Dynamic Programming Approach",
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"description": "Exploring LeetCode 62, Unique Paths, and solving it efficiently using dynamic programming in Go.",
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