Probability of reaching a point with 2 or 3 steps at a time

A person starts walking from position X = 0, find the probability to reach exactly on X = N if she can only take either 2 steps or 3 steps. Probability for step length 2 is given i.e. P, probability for step length 3 is 1 – P.
Examples :

Input : N = 5, P = 0.20
Output : 0.32
Explanation :-
There are two ways to reach 5.
2+3 with probability = 0.2 * 0.8 = 0.16
3+2 with probability = 0.8 * 0.2 = 0.16
So, total probability = 0.32.

It is a simple dynamic programming problem. It is simple extension of this problem :- count-ofdifferent-ways-express-n-sum-1-3-4
Below is the implementation of the above approach. 

C++

// CPP Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
#include <bits/stdc++.h>
using namespace std;

// Returns probability to reach N
float find_prob(int N, float P)
{
    double dp[N + 1];
    dp[0] = 1;
    dp[1] = 0;
    dp[2] = P;
    dp[3] = 1 - P;
    for (int i = 4; i <= N; ++i)
        dp[i] = (P)*dp[i - 2] + (1 - P) * dp[i - 3];

    return dp[N];
}

// Driver code
int main()
{
    int n = 5;
    float p = 0.2;
    cout << find_prob(n, p);
    return 0;
}

Java

// Java Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
import java.io.*;

class GFG {

    // Returns probability to reach N
    static float find_prob(int N, float P)
    {
        double dp[] = new double[N + 1];
        dp[0] = 1;
        dp[1] = 0;
        dp[2] = P;
        dp[3] = 1 - P;

        for (int i = 4; i <= N; ++i)
          dp[i] = (P) * dp[i - 2] +
                        (1 - P) * dp[i - 3];

        return ((float)(dp[N]));
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        float p = 0.2f;
        System.out.printf("%.2f",find_prob(n, p));
    }
}


/* This code is contributed by Nikita Tiwari.*/

Python3

# Python 3 Program to find
# probability to reach N with
# P probability to take 2
# steps (1-P) to take 3 steps

# Returns probability to reach N
def find_prob(N, P) :

    dp =[0] * (n + 1)
    dp[0] = 1
    dp[1] = 0
    dp[2] = P
    dp[3] = 1 - P

    for i in range(4, N + 1) :
        dp[i] = (P) * dp[i - 2] + (1 - P) * dp[i - 3]

    return dp[N]

# Driver code
n = 5
p = 0.2
print(round(find_prob(n, p), 2))

# This code is contributed by Nikita Tiwari.

C#

// C# Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
using System;

class GFG {

    // Returns probability to reach N
    static float find_prob(int N, float P)
    {
        double []dp = new double[N + 1];
        dp[0] = 1;
        dp[1] = 0;
        dp[2] = P;
        dp[3] = 1 - P;

        for (int i = 4; i <= N; ++i)
        dp[i] = (P) * dp[i - 2] +
                (1 - P) * dp[i - 3];

        return ((float)(dp[N]));
    }

    // Driver code
    public static void Main()
    {
        int n = 5;
        float p = 0.2f;
        Console.WriteLine(find_prob(n, p));
    }
}


/* This code is contributed by vt_m.*/

PHP

<?php
// PHP Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps

// Returns probability to reach N
function find_prob($N, $P)
{
    $dp;
    $dp[0] = 1;
    $dp[1] = 0;
    $dp[2] = $P;
    $dp[3] = 1 - $P;
    for ($i = 4; $i <= $N; ++$i)
        $dp[$i] = ($P) * $dp[$i - 2] +
                  (1 - $P) * $dp[$i - 3];

    return $dp[$N];
}

// Driver code
$n = 5;
$p = 0.2;
echo find_prob($n, $p);

// This code is contributed by mits.
?>

Javascript

<script>

// JavaScript Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps

   // Returns probability to reach N
    function find_prob(N, P)
    {
        let dp = [];
        dp[0] = 1;
        dp[1] = 0;
        dp[2] = P;
        dp[3] = 1 - P;

        for (let i = 4; i <= N; ++i)
          dp[i] = (P) * dp[i - 2] +
                        (1 - P) * dp[i - 3];

        return (dp[N]);
    }

// Driver Code
        let n = 5;
        let p = 0.2;
        document.write(find_prob(n, p));

      // This code is contributed by chinmoy1997pal.
</script>

Output :

0.32

Time Complexity: O(n)

Auxiliary Space: O(n)

Efficient approach : Space optimization O(1)

In previous approach the current value dp[i] is only depend on the previous 2 values of dp i.e. dp[i-2] and dp[i-3]. So to optimize the space complexity we can store previous 4 values of Dp in 4 variables  his way, the space complexity will be reduced from O(N) to O(1)

Implementation Steps:

• Initialize variables for dp[0], dp[1], dp[2], and dp[3] as 1, 0, P, and 1-P respectively.
• Iterate from i = 4 to N and use the formula dp[i] = (P)*dp[i – 2] + (1 – P) * dp[i – 3] to compute the current value of dp.
• After each iteration, update the values of dp0, dp1, dp2, and dp3 to dp1, dp2, dp3, and curr respectively.
• Return the final value of curr.

Implementation:

C++

// CPP Program to find probability to
// reach N with P probability to take
// 2 steps (1-P) to take 3 steps
#include <bits/stdc++.h>
using namespace std;

// Returns probability to reach N
float find_prob(int N, float P)
{
      // to store current value
    double curr;

      // store previous 4 values of DP
    double dp0 = 1, dp1=0, dp2=P, dp3= 1-P;

      // iterate over subproblems to get
      // current solution from previous computations
    for (int i = 4; i <= N; ++i){
        curr = (P)*dp2 + (1 - P) * dp1;

          // assigning values to iterate further
        dp0=dp1;
        dp1=dp2;
        dp2=dp3;
        dp3=curr;
    }

      // return final answer
    return curr;
}

// Driver code
int main()
{
    int n = 5;
    float p = 0.2;
    cout << find_prob(n, p);
    return 0;
}

Output

0.32

Time complexity: O(N)

Auxiliary Space: O(1)