Given N which denotes the initial position of the person on the number line. Also given L which is the probability of the person of going left. Find the probability of reaching all points on the number line after N moves from point N. Each move can be either to the left or to the right.

**Examples:**

Input:n = 2, l = 0.5

Output:0.2500 0.0000 0.5000 0.0000 0.2500

The person can reach n-1^{th}position and n+1^{th}position in 2 passes, hence the probability is 0. The person can reach 0th position by only moving 2 steps left from index 2, hence the probability of reaching 0th index is 05*0.5=0.25. Similarly for 2n index, the probability is 0.25.

Input:n = 3, l = 0.1

Output:0.0010 0.0000 0.0270 0.0000 0.2430 0.0000 0.7290

The person can reach n-1^{th}in three ways, i.e., (llr, lrl, rll) where l denotes left and r denotes right. Hence the probability of n-1^{th}index is 0.027. Similarly probabilties for all other points are also calculated.

**Approach: ** Construct an array **arr[n+1][2n+1]** where each row represents a pass and the columns represent the points on the line. The maximum a person can move from index N is to 0^{th} index at left or to 2n^{th} index at right. Initially the probabilities after one pass will be right for arr[1][n-1] and left for arr[1][n+1]. The n-1 moves which are left will be done, hence the two possible moves will either be n steps to the right or n steps to the left. So the recurrence relations for right and left moves for all will be:

arr[i][j] += (arr[i – 1][j – 1] * right)

arr[i][j] += (arr[i – 1][j + 1] * left)

The summation of probabilities for all possible moves for any index will be stored in arr[n][i].

Below is the implementation of the above approach:

// Java program to calculate the // probability of reaching all points // after N moves from point N import java.util.*; class GFG { // Function to calculate the probabilities static void printProbabilities(int n, double left) { double right = 1 - left; // Array where row represent the pass and the // coloumn represents the points on the line double[][] arr = new double[n + 1][2 * n + 1]; // Initially the person can reach left // or right with one move arr[1][n + 1] = right; arr[1][n - 1] = left; // Calculate probabilities for N-1 moves for (int i = 2; i <= n; i++) { // when the person moves from ith index in // right direction when i moves has been done for (int j = 1; j <= 2 * n; j++) { arr[i][j] += (arr[i - 1][j - 1] * right); } // when the person moves from ith index in // left direction when i moves has been done for (int j = 2 * n - 1; j >= 0; j--) { arr[i][j] += (arr[i - 1][j + 1] * left); } } // Calling function to print the array with probabilities printArray(arr, n); } // Function that prints the array static void printArray(double[][] arr, int n) { for (int i = 0; i < arr[0].length; i++) { System.out.printf("%5.4f ", arr[n][i]); } } // Driver Code public static void main(String[] args) { int n = 2; double left = 0.5; printProbabilities(n, left); } }

**Output:**

0.2500 0.0000 0.5000 0.0000 0.2500

**Time Complexity**: O(N^{2})

**Auxiliary Space**: O(N^{2})

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