Minimum Path Sum in Grid in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Minimum Path Sum Calculation

What is the output of the following Java code that calculates the minimum path sum in a 3x3 grid?

DSA Java

public class Solution {
    public static int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][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] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
            }
        }
        return dp[m-1][n-1];
    }

    public static void main(String[] args) {
        int[][] grid = {
            {1, 3, 1},
            {1, 5, 1},
            {4, 2, 1}
        };
        System.out.println(minPathSum(grid));
    }
}

A 7

B 8

C 9

D 10

💡 Hint

Trace the path from top-left to bottom-right choosing the minimum sum at each step.

The minimum path sum is calculated by dynamic programming. The path is 1 -> 3 -> 1 -> 1 -> 1 with sum 7.

🧠 Conceptual

intermediate

Understanding Movement Constraints in Minimum Path Sum

In the Minimum Path Sum problem, which movements are allowed when moving from the top-left corner to the bottom-right corner of the grid?

A Only diagonal moves are allowed.

B Moves can be in any direction including diagonals.

C Moves can be up, down, left, or right.

D Only moves down or right are allowed.

💡 Hint

Think about the problem constraints that simplify the path choices.

The problem restricts movement to only down or right to ensure a unique path and simplify calculation.

🔧 Debug

advanced

Identify the Error in Minimum Path Sum Code

What error will the following Java code produce when run on a non-empty grid?

DSA Java

public class Solution {
    public static int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][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] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
            }
        }
        return dp[m-1][n-1];
    }
}

A NullPointerException

B ArrayIndexOutOfBoundsException

C No error, returns correct result

D ArithmeticException

💡 Hint

Check the loop conditions for accessing array indices.

The loops use <= which causes index to go out of bounds since arrays are 0-indexed.

🚀 Application

advanced

Minimum Path Sum with Obstacles

Given a grid where some cells are blocked (represented by -1), how would you modify the minimum path sum algorithm to avoid blocked cells?

A Treat blocked cells as having infinite cost and skip them in calculations.

B Ignore blocked cells and calculate minimum path normally.

C Replace blocked cells with zero and proceed.

D Only consider paths that move diagonally to avoid blocked cells.

💡 Hint

Think about how to represent unreachable cells in the dynamic programming table.

Blocked cells should be treated as unreachable by assigning a very high cost so paths avoid them.

❓ Predict Output

expert

Output of Minimum Path Sum with Large Grid

What is the output of the following Java code that calculates the minimum path sum for a 4x4 grid?

DSA Java

public class Solution {
    public static int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] dp = new int[m][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] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
            }
        }
        return dp[m-1][n-1];
    }

    public static void main(String[] args) {
        int[][] grid = {
            {1, 2, 5, 2},
            {3, 2, 1, 4},
            {4, 3, 2, 1},
            {5, 6, 1, 1}
        };
        System.out.println(minPathSum(grid));
    }
}

A 12

B 13

C 14

D 15

💡 Hint

Calculate the minimum sum path step by step using the dp table.

The minimum path sum is 14 following the path with the least cumulative cost.

Practice

1. What does the "Minimum Path Sum in Grid" problem ask you to find?
Given a grid of numbers, find the path from top-left to bottom-right with the smallest sum.

easy

A. The path with the smallest sum of numbers from top-left to bottom-right

B. The path with the largest sum of numbers from top-left to bottom-right

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

D. The maximum number in the grid

Solution

1. Step 1: Understand the problem goal

   The problem asks to find a path from the top-left cell to the bottom-right cell of a grid such that the sum of all numbers on the path is as small as possible.

2. Step 2: Clarify path movement

   Movement is only allowed down or right at each step, so the path is constrained.

3. Final Answer:

   The path with the smallest sum of numbers from top-left to bottom-right -> Option A

4. Quick Check:

   Minimum path sum means smallest total sum [OK]

Hint: Look for smallest sum path moving only right or down [OK]

Common Mistakes:

• Thinking the path can move diagonally or up
• Looking for maximum sum instead of minimum
• Counting number of paths instead of sum

2. Which of the following Java code snippets correctly initializes a 2D array named grid with 3 rows and 3 columns filled with zeros?

easy

A. int grid[][] = new int[3];

B. int[][] grid = new int[3][3];

C. int[] grid = new int[3][3];

D. int[][] grid = new int[3];

Solution

1. Step 1: Understand Java 2D array syntax

   In Java, a 2D array is declared as int[][] name = new int[rows][columns];. This creates a grid with specified rows and columns.

2. Step 2: Check each option

   int[][] grid = new int[3][3]; correctly declares a 3x3 int 2D array. Options A, C, and D have incorrect syntax or dimensions.

3. Final Answer:

   int[][] grid = new int[3][3]; -> Option B

4. Quick Check:

   Correct 2D array syntax = int[][] grid = new int[3][3]; [OK]

Hint: Use int[][] name = new int[rows][cols] for 2D arrays [OK]

Common Mistakes:

• Declaring 2D arrays with wrong brackets
• Initializing only one dimension
• Confusing 1D and 2D array syntax

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

int[][] grid = {{1, 3}, {1, 5}};
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][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] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
  }
}
System.out.println(dp[m-1][n-1]);

medium

A. 7

B. 8

C. 9

D. 10

Solution

1. Step 1: Initialize dp array with first row and column sums

   dp[0][0] = 1
   dp[1][0] = dp[0][0] + grid[1][0] = 1 + 1 = 2
   dp[0][1] = dp[0][0] + grid[0][1] = 1 + 3 = 4

2. Step 2: Calculate dp[1][1]

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

3. Final Answer:

   7 -> Option A

4. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Fill dp row and column first, then use min of top and left [OK]

Common Mistakes:

• Adding max instead of min
• Not initializing first row and column correctly
• Mixing indices in loops

4. Identify the error in the following Java code snippet that tries to compute the minimum path sum in a grid:

int[][] grid = {{1,2,3},{4,5,6}};
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][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] = Math.max(dp[i-1][j], dp[i][j-1]) + grid[i][j];
  }
}

medium

A. Grid dimensions are wrong

B. Not initializing dp[0][0]

C. Loop ranges are incorrect

D. Using Math.max instead of Math.min in dp calculation

Solution

1. Step 1: Check dp initialization

   dp[0][0] is correctly initialized to grid[0][0]. Loops for first row and column are correct.

2. Step 2: Identify the error in dp update

   The inner loop uses Math.max instead of Math.min, which finds the maximum path sum, not minimum.

3. Final Answer:

   Using Math.max instead of Math.min in dp calculation -> Option D

4. Quick Check:

   Minimum path sum needs Math.min, not Math.max [OK]

Hint: Use Math.min to find minimum path sum, not Math.max [OK]

Common Mistakes:

• Using max instead of min in dp update
• Forgetting to initialize dp[0][0]
• Incorrect loop boundaries

5. Given a 3x3 grid:

int[][] grid = {
  {1, 2, 5},
  {3, 2, 1},
  {4, 1, 1}
};

What is the minimum path sum from top-left to bottom-right moving only right or down?

hard

A. 9

B. 8

C. 7

D. 10

Solution

1. Step 1: Initialize dp first row and column sums

   dp[0][0] = 1
   dp[0][1] = 1 + 2 = 3
   dp[0][2] = 3 + 5 = 8
   dp[1][0] = 1 + 3 = 4
   dp[2][0] = 4 + 4 = 8

2. Step 2: Calculate dp values for inner cells

   dp[1][1] = min(dp[0][1], dp[1][0]) + 2 = min(3,4) + 2 = 5
   dp[1][2] = min(dp[0][2], dp[1][1]) + 1 = min(8,5) + 1 = 6
   dp[2][1] = min(dp[1][1], dp[2][0]) + 1 = min(5,8) + 1 = 6
   dp[2][2] = min(dp[1][2], dp[2][1]) + 1 = min(6,6) + 1 = 7

3. Final Answer:

   7 -> Option C

4. Quick Check:

   Minimum path sum = 7 [OK]

Hint: Fill dp row and column first, then inner cells with min of top and left [OK]

Common Mistakes:

• Not updating dp correctly for inner cells
• Adding max instead of min
• Incorrect initialization of first row or column