Unique Paths in Grid DP in DSA Go - Time & Space Complexity

Time Complexity: Unique Paths in Grid DP

O(m * n)

Understanding Time Complexity

We want to know how the time needed to find unique paths in a grid grows as the grid size increases.

How does the number of steps change when the grid gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func uniquePaths(m int, n int) int {
    dp := make([][]int, m)
    for i := 0; i < m; i++ {
        dp[i] = make([]int, n)
        dp[i][0] = 1
    }
    for j := 0; j < n; j++ {
        dp[0][j] = 1
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
        }
    }
    return dp[m-1][n-1]
}
    

This code calculates the number of unique paths from the top-left to the bottom-right of an m by n grid using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops filling the dp table.
• How many times: The inner loop runs (m-1) * (n-1) times, covering almost all cells except the first row and column.

How Execution Grows With Input

As the grid size grows, the number of operations grows roughly with the number of cells in the grid.

Input Size (m x n)	Approx. Operations
10 x 10	81 (9*9)
100 x 100	9,801 (99*99)
1000 x 1000	998,001 (999*999)

Pattern observation: Operations grow roughly as the product of m and n, so doubling both roughly quadruples the work.

Final Time Complexity

Time Complexity: O(m * n)

This means the time needed grows proportionally to the number of cells in the grid.

Common Mistake

[X] Wrong: "The time grows exponentially because there are many paths."

[OK] Correct: The dynamic programming approach calculates each cell once, so it avoids repeated work and runs in polynomial time, not exponential.

Interview Connect

Understanding how nested loops affect time helps you explain and optimize grid problems clearly in interviews.

Self-Check

"What if we used recursion without memoization? How would the time complexity change?"

Practice

1. What does the problem "Unique Paths in Grid DP" ask you to find?
Unique Paths means:

easy

A. The shortest path length from top-left to bottom-right in a grid

B. The number of ways to move from top-left to bottom-right in a grid using only right and down moves

C. The maximum sum path from top-left to bottom-right in a grid

D. The number of obstacles in the grid

Solution

1. Step 1: Understand the problem statement

   The problem asks for counting all possible paths from the top-left corner to the bottom-right corner of a grid, moving only right or down.

2. Step 2: Identify the meaning of "Unique Paths"

   Unique paths means counting distinct ways, not shortest or maximum sum paths.

3. Final Answer:

   The number of ways to move from top-left to bottom-right in a grid using only right and down moves -> Option B

4. Quick Check:

   Unique Paths = Count of right/down moves paths [OK]

Hint: Unique Paths count all right/down routes from start to end [OK]

Common Mistakes:

• Confusing unique paths with shortest path length
• Thinking about obstacles when none are given
• Mixing path count with path sum

2. Which of the following Go code snippets correctly initializes a 2D slice dp of size m x n with zeros for Unique Paths DP?

easy

A. dp := [][]int{}
dp = append(dp, make([]int, m))
dp = append(dp, make([]int, n))

B. dp := make([]int, m*n)
for i := 0; i < m*n; i++ {
dp[i] = 0
}

C. var dp [m][n]int
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
dp[i][j] = 1
}
}

D. dp := make([][]int, m)
for i := range dp {
dp[i] = make([]int, n)
}

Solution

1. Step 1: Understand Go slice initialization for 2D arrays

   In Go, a 2D slice is created by first making the outer slice, then initializing each inner slice.

2. Step 2: Check each option

   dp := make([][]int, m) for i := range dp { dp[i] = make([]int, n) } correctly creates a slice of length m, then each inner slice of length n, all initialized to zero by default.

3. Final Answer:

   dp := make([][]int, m) for i := range dp { dp[i] = make([]int, n) } -> Option D

4. Quick Check:

   2D slice init with nested make loops = dp := make([][]int, m) for i := range dp { dp[i] = make([]int, n) } [OK]

Hint: Use nested make to create 2D slices in Go [OK]

Common Mistakes:

• Using one-dimensional slice for 2D data
• Initializing inner slices with wrong length
• Appending slices incorrectly

3. What is the output of the following Go code for Unique Paths in a 3x3 grid?

func uniquePaths(m, n int) int {
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    for i := 0; i < m; i++ {
        dp[i][0] = 1
    }
    for j := 0; j < n; j++ {
        dp[0][j] = 1
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
        }
    }
    return dp[m-1][n-1]
}

fmt.Println(uniquePaths(3, 3))

medium

A. 3

B. 9

C. 6

D. 1

Solution

1. Step 1: Initialize dp with 1s on first row and column

   First row and first column are set to 1 because only one way to reach those cells (all right or all down).

2. Step 2: Fill dp table using dp[i][j] = dp[i-1][j] + dp[i][j-1]

   Calculate paths for each cell by summing paths from top and left cells. For 3x3 grid, dp[2][2] = 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Unique paths in 3x3 grid = 6 [OK]

Hint: Use dp[i][j] = dp[i-1][j] + dp[i][j-1] to fill grid [OK]

Common Mistakes:

• Not initializing first row and column to 1
• Using wrong indices in dp update
• Returning wrong dp cell

4. The following Go code for Unique Paths has a bug. What is the error?

func uniquePaths(m, n int) int {
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    for i := 0; i < m; i++ {
        dp[i][0] = 1
    }
    for j := 0; j < n; j++ {
        dp[0][j] = 1
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            dp[i][j] = dp[i-1][j] * dp[i][j-1]
        }
    }
    return dp[m-1][n-1]
}

medium

A. The dp update uses multiplication instead of addition

B. The dp slice is not initialized properly

C. The loops for initializing first row and column are swapped

D. The function returns dp[0][0] instead of dp[m-1][n-1]

Solution

1. Step 1: Check dp update formula

   The code uses dp[i][j] = dp[i-1][j] * dp[i][j-1], which multiplies paths instead of adding them.

2. Step 2: Correct formula for Unique Paths

   The correct formula is dp[i][j] = dp[i-1][j] + dp[i][j-1] to sum the number of ways from top and left cells.

3. Final Answer:

   The dp update uses multiplication instead of addition -> Option A

4. Quick Check:

   Use addition, not multiplication in dp update [OK]

Hint: Sum paths from top and left, don't multiply [OK]

Common Mistakes:

• Using * instead of + in dp update
• Forgetting to initialize first row/column
• Returning wrong dp cell

5. Given a 4x4 grid with some obstacles represented by 1s, how would you modify the Unique Paths DP approach to count paths avoiding obstacles?
Obstacles grid:

0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

hard

A. Set dp[i][j] = 0 if obstacle at (i,j), else dp[i][j] = dp[i-1][j] + dp[i][j-1]

B. Ignore obstacles and use normal dp formula

C. Set dp[i][j] = 1 if obstacle at (i,j), else 0

D. Use multiplication of dp[i-1][j] and dp[i][j-1] for obstacles

Solution

1. Step 1: Identify obstacle handling in DP

   When a cell has an obstacle, no path can go through it, so dp[i][j] must be 0.

2. Step 2: Modify dp update to skip obstacles

   For cells without obstacles, dp[i][j] = dp[i-1][j] + dp[i][j-1]. For obstacles, dp[i][j] = 0.

3. Final Answer:

   Set dp[i][j] = 0 if obstacle at (i,j), else dp[i][j] = dp[i-1][j] + dp[i][j-1] -> Option A

4. Quick Check:

   Obstacles block paths, set dp to zero there [OK]

Hint: Zero dp at obstacles, sum paths otherwise [OK]

Common Mistakes:

• Ignoring obstacles in dp calculation
• Setting dp to 1 at obstacles
• Multiplying dp values instead of adding