package com.thealgorithms.dynamicprogramming; import java.util.Arrays; /** * Author: Siddhant Swarup Mallick * Github: https://github.com/siddhant2002 *
* Problem Description: * A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). * The robot can only move either down or right at any point in time. * The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). * How many possible unique paths are there? *
* Program Description: * This program calculates the number of unique paths possible for a robot to reach the bottom-right corner * of an m x n grid using dynamic programming. */ public final class UniquePaths { private UniquePaths() { } /** * Calculates the number of unique paths using a 1D dynamic programming array. * Time complexity O(n*m) * Space complexity O(min(n,m)) * * @param m The number of rows in the grid. * @param n The number of columns in the grid. * @return The number of unique paths. */ public static int uniquePaths(final int m, final int n) { if (m > n) { return uniquePaths(n, m); // Recursive call to handle n > m cases } int[] dp = new int[n]; // Create a 1D array to store unique paths for each column Arrays.fill(dp, 1); // Initialize all values to 1 (one way to reach each cell) for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { dp[j] = Math.addExact(dp[j], dp[j - 1]); // Update the number of unique paths for each cell } } return dp[n - 1]; // The result is stored in the last column of the array } /** * Calculates the number of unique paths using a 2D dynamic programming array. * Time complexity O(n*m) * Space complexity O(n*m) * * @param m The number of rows in the grid. * @param n The number of columns in the grid. * @return The number of unique paths. */ public static int uniquePaths2(final int m, final int n) { int[][] dp = new int[m][n]; // Create a 2D array to store unique paths for each cell for (int i = 0; i < m; i++) { dp[i][0] = 1; // Initialize the first column to 1 (one way to reach each cell) } for (int j = 0; j < n; j++) { dp[0][j] = 1; // Initialize the first row to 1 (one way to reach each cell) } for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) { dp[i][j] = Math.addExact(dp[i - 1][j], dp[i][j - 1]); // Update the number of unique paths for each cell } } return dp[m - 1][n - 1]; // The result is stored in the bottom-right cell of the array } }