Unique Paths in Grid DP in DSA C - 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 changes as the grid size grows.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int uniquePaths(int m, int n) {
    int dp[m][n];
    for (int i = 0; i < m; i++) dp[i][0] = 1;
    for (int j = 0; j < n; j++) dp[0][j] = 1;
    for (int i = 1; i < m; i++) {
        for (int 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: The nested loops that fill 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: The operations grow roughly by the product of m and n, so doubling both dimensions roughly quadruples the work.

Final Time Complexity

Time Complexity: O(m * n)

This means the time needed grows proportionally to the total 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 avoids repeated work by storing results, so it only visits each cell once, making the time grow linearly with the grid size.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency by avoiding repeated calculations, a key skill in many coding challenges.

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?

easy

A. The maximum sum path from the top-left to the bottom-right corner of a grid.

B. The shortest path length from the top-left to the bottom-right corner of a grid.

C. The number of obstacles in the grid.

D. The number of different ways to move from the top-left to the bottom-right corner of a grid using only right and down moves.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible paths from the start (top-left) to the end (bottom-right) moving only right or down.

2. Step 2: Clarify movement constraints

   Only moves allowed are right or down, no diagonal or backward moves.

3. Final Answer:

   The number of different ways to move from the top-left to the bottom-right corner of a grid using only right and down moves. -> Option D

4. Quick Check:

   Unique paths = count of right/down moves combinations [OK]

Hint: Count paths by moving only right or down [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Confusing path count with shortest path
• Counting obstacles instead of paths

2. Which of the following is the correct way to initialize the DP array for a grid of size m x n in C?

easy

A. int dp[m][n] = {0};

B. int dp[m][n]; dp = 0;

C. int dp[m][n]; memset(dp, 0, sizeof(dp));

D. int dp[m][n] = NULL;

Solution

1. Step 1: Understand array initialization in C

   Arrays cannot be assigned directly after declaration; memset is used to set all bytes to zero.

2. Step 2: Check each option

   int dp[m][n] = {0}; tries to initialize with {0} which works only for static arrays, not variable length arrays. int dp[m][n]; memset(dp, 0, sizeof(dp)); uses memset correctly. int dp[m][n]; dp = 0; tries invalid assignment. int dp[m][n] = NULL; assigns NULL which is invalid for arrays.

3. Final Answer:

   int dp[m][n]; memset(dp, 0, sizeof(dp)); -> Option C

4. Quick Check:

   Use memset to zero initialize arrays in C [OK]

Hint: Use memset to zero initialize arrays in C [OK]

Common Mistakes:

• Trying to assign 0 directly to array
• Using NULL for arrays
• Assuming {0} works for variable length arrays

3. What is the output of the following code snippet for m=3, n=3?

int uniquePaths(int m, int n) {
    int dp[m][n];
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < m; i++) dp[i][0] = 1;
    for (int j = 0; j < n; j++) dp[0][j] = 1;
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = dp[i-1][j] + dp[i][j-1];
        }
    }
    return dp[m-1][n-1];
}
printf("%d", uniquePaths(3, 3));

medium

A. 3

B. 6

C. 9

D. 1

Solution

1. Step 1: Initialize first row and column

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

2. Step 2: Fill dp table using sum of top and left cells

   For each cell dp[i][j], value = dp[i-1][j] + dp[i][j-1]. For 3x3 grid, dp[2][2] = 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Unique paths in 3x3 grid = 6 [OK]

Hint: Sum paths from top and left cells to fill DP [OK]

Common Mistakes:

• Not initializing first row and column to 1
• Adding instead of summing dp[i-1][j] and dp[i][j-1]
• Returning wrong dp cell

4. Find the error in this code snippet for unique paths calculation:

int uniquePaths(int m, int n) {
    int dp[m][n];
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i <= m; i++) dp[i][0] = 1;
    for (int j = 0; j <= n; j++) dp[0][j] = 1;
    for (int i = 1; i < m; i++) {
        for (int 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 loops initializing dp go out of bounds causing runtime error.

B. The dp array is not zero initialized.

C. The return statement returns wrong cell.

D. The nested loops start from 0 instead of 1.

Solution

1. Step 1: Check loop bounds for initialization

   Loops use <= m and <= n which causes index to go out of bounds since valid indices are 0 to m-1 and 0 to n-1.

2. Step 2: Identify correct loop bounds

   Loops should use < m and < n to avoid accessing invalid indices.

3. Final Answer:

   The loops initializing dp go out of bounds causing runtime error. -> Option A

4. Quick Check:

   Array indices must be within 0 to size-1 [OK]

Hint: Use < not <= for array indices in loops [OK]

Common Mistakes:

• Using <= instead of < in loops
• Ignoring array bounds causing crashes
• Incorrect initialization of dp array

5. Given a 4x4 grid with some blocked cells (obstacles) represented by 1, how would you modify the unique paths DP solution to count paths avoiding 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 cell is blocked; else dp[i][j] = dp[i-1][j] + dp[i][j-1].

B. Ignore obstacles and calculate paths normally.

C. Set dp[i][j] = 1 for blocked cells and 0 otherwise.

D. Only count paths that move diagonally to avoid obstacles.

Solution

1. Step 1: Mark blocked cells in dp as zero

   If a cell is blocked (value 1), no paths can go through it, so dp[i][j] = 0.

2. Step 2: Calculate dp for free cells

   For free cells (value 0), dp[i][j] = dp[i-1][j] + dp[i][j-1], summing paths from top and left.

3. Final Answer:

   Set dp[i][j] = 0 if cell is blocked; else dp[i][j] = dp[i-1][j] + dp[i][j-1]. -> Option A

4. Quick Check:

   Blocked cells have zero paths [OK]

Hint: Zero dp for obstacles, sum paths for free cells [OK]

Common Mistakes:

• Ignoring obstacles in path count
• Assigning 1 to blocked cells
• Using diagonal moves to avoid obstacles