Minimum Path Sum in Grid in DSA C++ - Time & Space Complexity

Time Complexity: Minimum Path Sum in Grid

O(m x n)

Understanding Time Complexity

We want to know how the time needed to find the minimum path sum 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 minPathSum(vector<vector<int>>& grid) {
    int m = grid.size();
    int n = grid[0].size();
    vector<vector<int>> dp(m, vector<int>(n));
    dp[0][0] = grid[0][0];
    for (int i = 1; i < m; i++)
        dp[i][0] = dp[i-1][0] + grid[i][0];
    for (int j = 1; j < n; j++)
        dp[0][j] = dp[0][j-1] + grid[0][j];
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
        }
    }
    return dp[m-1][n-1];
}
    

This code finds the smallest sum of numbers on a path from the top-left to the bottom-right of a grid, moving only down or right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops filling the dp table for each cell.
• How many times: The inner loop runs for each cell except the first row and column, so roughly m x n times.

How Execution Grows With Input

As the grid size grows, the number of cells to process grows with the product of rows and columns.

Input Size (m x n)	Approx. Operations
10 x 10	100
100 x 100	10,000
1000 x 1000	1,000,000

Pattern observation: The operations grow roughly by the square of the grid size, meaning doubling rows and columns quadruples work.

Final Time Complexity

Time Complexity: O(m x n)

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

Common Mistake

[X] Wrong: "Since we only move right or down, the path count is exponential, so the time is exponential too."

[OK] Correct: The code uses dynamic programming to store results for each cell once, avoiding repeated work and keeping time linear in the number of cells.

Interview Connect

Understanding this time complexity shows you can analyze grid-based dynamic programming efficiently, a common skill in coding challenges and interviews.

Self-Check

"What if we allowed moving diagonally as well? How would the time complexity change?"

Practice

1. What does the "Minimum Path Sum in Grid" problem ask you to find?

easy

A. The largest sum of numbers along any path in the grid.

B. The shortest path length from top-left to bottom-right.

C. The number of paths from top-left to bottom-right.

D. The smallest sum of numbers along a path from top-left to bottom-right, moving only right or down.

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum sum of values collected when moving from the top-left corner to the bottom-right corner of a grid, moving only right or down.

2. Step 2: Clarify movement constraints

   Only moves to the right or down are allowed, so the path is restricted and the sum depends on chosen cells.

3. Final Answer:

   The smallest sum of numbers along a path from top-left to bottom-right, moving only right or down. -> Option D

4. Quick Check:

   Minimum path sum = smallest sum path [OK]

Hint: Focus on sum along right/down moves only [OK]

Common Mistakes:

• Thinking diagonal moves are allowed
• Looking for maximum sum instead of minimum
• Counting number of paths instead of sum

2. Which of the following C++ code snippets correctly initializes a 2D vector named grid with 3 rows and 4 columns filled with zeros?

easy

A. std::vector grid(3, std::vector<int>(4, 0));

B. std::vector<int> grid(3, std::vector<int>(4, 0));

C. std::vector grid(4, std::vector<int>(3, 0));

D. std::vector grid(3, 4, 0);

Solution

1. Step 1: Understand 2D vector initialization syntax

   To create a 2D vector with 3 rows and 4 columns filled with zeros, we use vector of vectors: outer vector size 3, inner vector size 4, each element 0.

2. Step 2: Check each option

   std::vector grid(3, std::vector<int>(4, 0)); correctly uses std::vector grid(3, std::vector<int>(4, 0)); which means 3 rows, each row is a vector of 4 zeros.

3. Final Answer:

   std::vector grid(3, std::vector<int>(4, 0)); -> Option A

4. Quick Check:

   Correct 2D vector init = std::vector grid(3, std::vector<int>(4, 0)); [OK]

Hint: Use vector of vectors with inner vector size and default value [OK]

Common Mistakes:

• Using wrong vector type for 2D
• Swapping rows and columns sizes
• Trying to pass three arguments to vector constructor

3. What is the output of the following C++ code that calculates the minimum path sum in a 2x2 grid?

std::vector grid = {{1, 3}, {1, 5}};
int m = grid.size();
int n = grid[0].size();
for (int i = 1; i < m; i++) grid[i][0] += grid[i-1][0];
for (int j = 1; j < n; j++) grid[0][j] += grid[0][j-1];
for (int i = 1; i < m; i++) {
  for (int j = 1; j < n; j++) {
    grid[i][j] += std::min(grid[i-1][j], grid[i][j-1]);
  }
}
std::cout << grid[m-1][n-1] << std::endl;

medium

A. 7

B. 8

C. 9

D. 10

Solution

1. Step 1: Calculate first column cumulative sums

   grid[1][0] = grid[1][0] + grid[0][0] = 1 + 1 = 2

2. Step 2: Calculate first row cumulative sums

   grid[0][1] = grid[0][1] + grid[0][0] = 3 + 1 = 4

3. Step 3: Calculate inner cell minimum path sums

   grid[1][1] = grid[1][1] + min(grid[0][1], grid[1][0]) = 5 + min(4, 2) = 5 + 2 = 7

4. Final Answer:

   7 -> Option A

5. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Update first row and column cumulatively, then inner cells [OK]

Common Mistakes:

• Not updating first row and column before inner cells
• Adding max instead of min
• Mixing indices i and j

4. The following code attempts to compute the minimum path sum in a grid but has a bug. What is the error?

std::vector grid = {{1, 2, 3}, {4, 5, 6}};
int m = grid.size();
int n = grid[0].size();
for (int i = 1; i < m; i++) grid[i][0] += grid[i-1][0];
for (int j = 1; j < n; j++) grid[0][j] += grid[0][j-1];
for (int i = 1; i < m; i++) {
  for (int j = 1; j < n; j++) {
    grid[i][j] += std::max(grid[i-1][j], grid[i][j-1]);
  }
}
std::cout << grid[m-1][n-1] << std::endl;

medium

A. Wrong loop ranges for i and j

B. Not initializing first row and column

C. Using std::max instead of std::min in inner loop

D. Incorrect grid initialization syntax

Solution

1. Step 1: Check the inner loop operation

   The code uses std::max to update grid[i][j], but the problem requires minimum path sum, so std::min should be used.

2. Step 2: Verify other parts

   First row and column are correctly initialized, loops have correct ranges, and grid initialization is valid.

3. Final Answer:

   Using std::max instead of std::min in inner loop -> Option C

4. Quick Check:

   Use min for minimum path sum [OK]

Hint: Use min, not max, when updating path sums [OK]

Common Mistakes:

• Confusing max and min functions
• Skipping first row/column initialization
• Off-by-one errors in loops

5. Given the grid below, what is the minimum path sum from top-left to bottom-right?

grid = {
  {1, 3, 1, 2},
  {1, 5, 1, 1},
  {4, 2, 1, 3}
};

You can only move right or down. Calculate the minimum sum.

hard

A. 8

B. 10

C. 9

D. 11

Solution

1. Step 1: Compute first row and first column cumulative sums

   First row: [1, 4, 5, 7] (1, 1+3=4, 4+1=5, 5+2=7) First column: [1, 2, 6] (1, 1+1=2, 2+4=6)

2. Step 2: Calculate inner cells minimum path sums

   Row 1, col 1: 5 + min(4,2) = 5 + 2 = 7 Row 1, col 2: 1 + min(5,7) = 1 + 5 = 6 Row 1, col 3: 1 + min(7,6) = 1 + 6 = 7 Row 2, col 1: 2 + min(7,6) = 2 + 6 = 8 Row 2, col 2: 1 + min(6,8) = 1 + 6 = 7 Row 2, col 3: 3 + min(7,7) = 3 + 7 = 10

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Minimum path sum = 10 [OK]

Hint: Fill first row/column, then inner cells with min sums [OK]

Common Mistakes:

• Adding max instead of min
• Incorrect cumulative sums for first row/column
• Misindexing grid cells