Unique Paths in Grid DP in DSA C - Build from Scratch

Unique Paths in Grid DP

📖 Scenario: Imagine you are helping a robot find how many different ways it can move from the top-left corner to the bottom-right corner of a grid. The robot can only move right or down.

🎯 Goal: Build a program that calculates the number of unique paths in a grid using dynamic programming.

📋 What You'll Learn

Create a 2D array to represent the grid

Use a variable to store the number of rows and columns

Use nested loops to fill the DP table with the number of unique paths

Print the total number of unique paths from top-left to bottom-right

💡 Why This Matters

🌍 Real World

Robots and games often need to find paths through grids or maps. This program helps count possible routes.

💼 Career

Understanding dynamic programming and grid traversal is useful for software engineers working on algorithms, robotics, and game development.

1

Create the grid size variables

Create two integer variables called rows and cols and set rows = 3 and cols = 3.

DSA C

# Your code here

Hint

Use int to declare the variables and assign the values 3 to both.

2

Create the DP table

Create a 2D integer array called dp with size rows by cols.

DSA C

int rows = 3;
int cols = 3;
// Your code here

Hint

Declare dp as a 2D array with fixed size 3 by 3.

3

Fill the DP table with unique paths

Use nested for loops with variables i and j to fill dp. Set dp[i][0] = 1 for all rows and dp[0][j] = 1 for all columns. For other cells, set dp[i][j] = dp[i-1][j] + dp[i][j-1].

DSA C

int rows = 3;
int cols = 3;
int dp[3][3];
// Your code here

Hint

Initialize the first row and first column with 1, then fill the rest by adding the top and left cells.

4

Print the total unique paths

Print the value of dp[rows-1][cols-1] using printf.

DSA C

int rows = 3;
int cols = 3;
int dp[3][3];

for (int i = 0; i < rows; i++) {
    dp[i][0] = 1;
}
for (int j = 0; j < cols; j++) {
    dp[0][j] = 1;
}

for (int i = 1; i < rows; i++) {
    for (int j = 1; j < cols; j++) {
        dp[i][j] = dp[i-1][j] + dp[i][j-1];
    }
}
// Your code here

Hint

Use printf("%d\n", dp[rows-1][cols-1]); to print the result.