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.
// 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 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.*/
# 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# 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.*/
<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>
<?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.
?>
Output
0.32
Time Complexity: O(n)
Auxiliary Space: O(n)
Efficient approach: Space optimization O(1)
In the previous approach, the current value dp[i] only depends 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 the 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:
// 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;
}
import java.util.*;
public class Main {
// Returns probability to reach N
static 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 (float) curr;
}
// Driver code
public static void main(String[] args) {
int n = 5;
float p = 0.2f;
System.out.println(find_prob(n, p));
}
}
# Function to find the probability to reach N with P probability to
# take 2 steps and (1-P) to take 3 steps
def find_prob(N, P):
# Initialize variables to store current and previous values
curr = 0.0
# Initialize previous 4 values of DP
dp0, dp1, dp2, dp3 = 1.0, 0.0, P, 1 - P
# Iterate over subproblems to calculate the
# current solution from previous computations
for i in range(4, N + 1):
curr = P * dp2 + (1 - P) * dp1
# Update values for the next iteration
dp0, dp1, dp2, dp3 = dp1, dp2, dp3, curr
# Round the final answer to 2 decimal places
return round(curr, 2)
# Driver code
if __name__ == "__main__":
n = 5
p = 0.2
print(find_prob(n, p))
using System;
class Program
{
// Function to find the probability to reach N with P probability to take 2 steps and (1-P) to take 3 steps
static double FindProbability(int N, double P)
{
double curr = 0.0;
double dp1 = 0.0, dp2 = P, dp3 = 1 - P;
// Iterate over subproblems to calculate the current solution from previous computations
for (int i = 4; i <= N; i++)
{
curr = P * dp2 + (1 - P) * dp1;
// Update values for the next iteration
dp1 = dp2;
dp2 = dp3;
dp3 = curr;
}
// Return the final answer
return curr;
}
static void Main()
{
int n = 5;
double p = 0.2;
Console.WriteLine(FindProbability(n, p));
}
}
// Returns probability to reach N
function find_prob(N, P) {
// to store current value
let curr;
// store previous 4 values of DP
let dp0 = 1, dp1 = 0, dp2 = P, dp3 = 1 - P;
// iterate over subproblems to get
// current solution from previous computations
for (let 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
let n = 5;
let p = 0.2;
console.log(find_prob(n, p).toFixed(2));
Output
0.32
Time complexity: O(N)
Auxiliary Space: O(1)
