Given a linear board of length N numbered from 1 to N, the task is to find the expected number of moves required to reach the Nth cell of the board, if we start at cell numbered 1 and at each step we roll a cubical dice to decide the next move. Also, we cannot go outside the bounds of the board. Note that the expected number of moves can be fractional.
Examples:
Input: N = 8
Output: 7
p1 = (1 / 6) | 1-step -> 6 moves expected to reach the end
p2 = (1 / 6) | 2-steps -> 6 moves expected to reach the end
p3 = (1 / 6) | 3-steps -> 6 moves expected to reach the end
p4 = (1 / 6) | 4-steps -> 6 moves expected to reach the end
p5 = (1 / 6) | 5-steps -> 6 moves expected to reach the end
p6 = (1 / 6) | 6-steps -> 6 moves expected to reach the end
If we are 7 steps away, then we can end up at 1, 2, 3, 4, 5, 6 steps
away with equal probability i.e. (1 / 6).
Look at the above simulation to understand better.
dp[N – 1] = dp[7]
= 1 + (dp[1] + dp[2] + dp[3] + dp[4] + dp[5] + dp[6]) / 6
= 1 + 6 = 7Input: N = 10
Output: 7.36111
Approach: This problem can be solved using dynamic programming. To solve the problem, decide the states of the DP first. One way will be to use the distance between the current cell and the Nth cell to define the states of DP. Let’s call this distance X. Thus dp[X] can be defined as the expected number of steps required to reach the end of the board of length X + 1 starting from the 1st cell.
Thus, the recurrence relation becomes:
dp[X] = 1 + (dp[X – 1] + dp[X – 2] + dp[X – 3] + dp[X – 4] + dp[X – 5] + dp[X – 6]) / 6
Now, for the base-cases:
dp[0] = 0
Let’s try to calculate dp[1].
dp[1] = 1 + 5 * (dp[1]) / 6 + dp[0] (Why? its because (5 / 6) is the probability it stays stuck at 1.)
dp[1] / 6 = 1 (since dp[0] = 0)
dp[1] = 6
Similarly, dp[1] = dp[2] = dp[3] = dp[4] = dp[5] = 6
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>#define maxSize 50using namespace std;// To store the states of dpdouble dp[maxSize];// To determine whether a state// has been solved beforeint v[maxSize];// Function to return the countdouble expectedSteps(int x){ // Base cases if (x == 0) return 0; if (x <= 5) return 6; // If a state has been solved before // it won't be evaluated again if (v[x]) return dp[x]; v[x] = 1; // Recurrence relation dp[x] = 1 + (expectedSteps(x - 1) + expectedSteps(x - 2) + expectedSteps(x - 3) + expectedSteps(x - 4) + expectedSteps(x - 5) + expectedSteps(x - 6)) / 6; return dp[x];}// Driver codeint main(){ int n = 10; cout << expectedSteps(n - 1); return 0;} |
Java
// Java implementation of the approach import java.io.*;class GFG { static int maxSize = 50; // To store the states of dp static double dp[] = new double[maxSize]; // To determine whether a state // has been solved before static int v[] = new int[maxSize]; // Function to return the count static double expectedSteps(int x) { // Base cases if (x == 0) return 0; if (x <= 5) return 6; // If a state has been solved before // it won't be evaluated again if (v[x] == 1) return dp[x]; v[x] = 1; // Recurrence relation dp[x] = 1 + (expectedSteps(x - 1) + expectedSteps(x - 2) + expectedSteps(x - 3) + expectedSteps(x - 4) + expectedSteps(x - 5) + expectedSteps(x - 6)) / 6; return dp[x]; } // Driver code public static void main (String[] args) { int n = 10; System.out.println(expectedSteps(n - 1)); }}// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approachmaxSize = 50# To store the states of dpdp = [0] * maxSize# To determine whether a state# has been solved beforev = [0] * maxSize# Function to return the countdef expectedSteps(x): # Base cases if (x == 0): return 0 if (x <= 5): return 6 # If a state has been solved before # it won't be evaluated again if (v[x]): return dp[x] v[x] = 1 # Recurrence relation dp[x] = 1 + (expectedSteps(x - 1) + expectedSteps(x - 2) + expectedSteps(x - 3) + expectedSteps(x - 4) + expectedSteps(x - 5) + expectedSteps(x - 6)) / 6 return dp[x]# Driver coden = 10print(round(expectedSteps(n - 1), 5))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System;class GFG { static int maxSize = 50; // To store the states of dp static double []dp = new double[maxSize]; // To determine whether a state // has been solved before static int []v = new int[maxSize]; // Function to return the count static double expectedSteps(int x) { // Base cases if (x == 0) return 0; if (x <= 5) return 6; // If a state has been solved before // it won't be evaluated again if (v[x] == 1) return dp[x]; v[x] = 1; // Recurrence relation dp[x] = 1 + (expectedSteps(x - 1) + expectedSteps(x - 2) + expectedSteps(x - 3) + expectedSteps(x - 4) + expectedSteps(x - 5) + expectedSteps(x - 6)) / 6; return dp[x]; } // Driver code public static void Main () { int n = 10; Console.WriteLine(expectedSteps(n - 1)); } } // This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approachvar maxSize = 50;// To store the states of dpvar dp = Array(maxSize);// To determine whether a state// has been solved beforevar v = Array(maxSize);// Function to return the countfunction expectedSteps(x){ // Base cases if (x == 0) return 0; if (x <= 5) return 6; // If a state has been solved before // it won't be evaluated again if (v[x]) return dp[x]; v[x] = 1; // Recurrence relation dp[x] = 1 + (expectedSteps(x - 1) + expectedSteps(x - 2) + expectedSteps(x - 3) + expectedSteps(x - 4) + expectedSteps(x - 5) + expectedSteps(x - 6)) / 6; return dp[x];}// Driver codevar n = 10;document.write( expectedSteps(n - 1).toFixed(5));// This code is contributed by noob2000.</script> |
7.36111
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient approach: Space optimization O(1)
To optimize space complexity we can use variables instead of arrays to store the states of the DP and determine whether a state has been solved before because in previous approach the current value is dependent upon the previous values stored in array.
Implementation steps:
- Handle base cases: If bis 0, return 0. If x is less than or equal to 5, return 6.
- Initialize the previous values prev1, prev2, prev3, prev4, prev5, and prev6 to 6.
- Initialize the current value curr.
- Iterate from 6 to x (exclusive) to compute the current value based on the previous values.
- Calculate the current value as 1 plus the sum of the previous values divided by 6.
- Update the previous values by shifting their assignments.
- Return the final computed current value as the result.
Implementation:
C++
#include <bits/stdc++.h>using namespace std;// Function to return the countdouble expectedSteps(int x){ // Base cases if (x == 0) return 0; if (x <= 5) return 6; // initializing previous values double prev1 = 6; // expectedSteps(x - 1) double prev2 = 6; // expectedSteps(x - 2) double prev3 = 6; // expectedSteps(x - 3) double prev4 = 6; // expectedSteps(x - 4) double prev5 = 6; // expectedSteps(x - 5) double prev6 = 6; // expectedSteps(x - 6) // current value double curr; // iterate over subproblem to get current // value from previous computations for (int i = 6; i < x; i++) { curr = 1 + (prev1 + prev2 + prev3 + prev4 + prev5 + prev6) / 6; // assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = prev4; prev4 = prev5; prev5 = prev6; prev6 = curr; } return curr;}// Driver codeint main(){ int n = 10; // function call cout << expectedSteps(n - 1); return 0;} |
Java
import java.util.Scanner;public class GFG { // Function to return the count static double expectedSteps(int x) { // Base cases if (x == 0) return 0; if (x <= 5) return 6; // Initializing previous values double prev1 = 6; // expectedSteps(x - 1) double prev2 = 6; // expectedSteps(x - 2) double prev3 = 6; // expectedSteps(x - 3) double prev4 = 6; // expectedSteps(x - 4) double prev5 = 6; // expectedSteps(x - 5) double prev6 = 6; // expectedSteps(x - 6) // Current value double curr = 0; // Iterate over subproblem to get the current // value from previous computations for (int i = 6; i < x; i++) { curr = 1 + (prev1 + prev2 + prev3 + prev4 + prev5 + prev6) / 6; // Assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = prev4; prev4 = prev5; prev5 = prev6; prev6 = curr; } return curr; } // Driver code public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = 10; // Function call System.out.println(expectedSteps(n - 1)); sc.close(); }} |
Python3
def expected_steps(x): # Base cases if x == 0: return 0 if x <= 5: return 6 # initializing previous values prev1 = 6 # expectedSteps(x - 1) prev2 = 6 # expectedSteps(x - 2) prev3 = 6 # expectedSteps(x - 3) prev4 = 6 # expectedSteps(x - 4) prev5 = 6 # expectedSteps(x - 5) prev6 = 6 # expectedSteps(x - 6) # current value curr = 0 # iterate over subproblem to get current # value from previous computations for i in range(6, x): curr = 1 + (prev1 + prev2 + prev3 + prev4 + prev5 + prev6) / 6 # assigning values to iterate further prev1 = prev2 prev2 = prev3 prev3 = prev4 prev4 = prev5 prev5 = prev6 prev6 = curr return currdef main(): n = 10 # function call print(expected_steps(n - 1))if __name__ == "__main__": main() |
C#
using System;namespace ExpectedStepsExample{ class GFG { // Function to return the count static double ExpectedSteps(int x) { // Base cases if (x == 0) return 0; if (x <= 5) return 6; // initializing previous values double prev1 = 6; // ExpectedSteps(x - 1) double prev2 = 6; // ExpectedSteps(x - 2) double prev3 = 6; // ExpectedSteps(x - 3) double prev4 = 6; // ExpectedSteps(x - 4) double prev5 = 6; // ExpectedSteps(x - 5) double prev6 = 6; // ExpectedSteps(x - 6) // current value double curr = 0; // iterate over subproblem to get the current // value from previous computations for (int i = 6; i < x; i++) { curr = 1 + (prev1 + prev2 + prev3 + prev4 + prev5 + prev6) / 6; // assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = prev4; prev4 = prev5; prev5 = prev6; prev6 = curr; } return curr; } // Driver code static void Main(string[] args) { int n = 10; // function call Console.WriteLine(ExpectedSteps(n - 1)); } }} |
Javascript
function expectedSteps(x) { // Base cases if (x === 0) return 0; if (x <= 5) return 6; // Initializing previous values let prev1 = 6; // expectedSteps(x - 1) let prev2 = 6; // expectedSteps(x - 2) let prev3 = 6; // expectedSteps(x - 3) let prev4 = 6; // expectedSteps(x - 4) let prev5 = 6; // expectedSteps(x - 5) let prev6 = 6; // expectedSteps(x - 6) // Current value let curr; // Iterate over subproblem to get current // value from previous computations for (let i = 6; i < x; i++) { curr = 1 + (prev1 + prev2 + prev3 + prev4 + prev5 + prev6) / 6; // Assigning values to iterate further prev1 = prev2; prev2 = prev3; prev3 = prev4; prev4 = prev5; prev5 = prev6; prev6 = curr; } return curr;}// Driver codeconst n = 10;// Function callconsole.log(expectedSteps(n - 1)); |
Output:
7.36111
Time Complexity: O(N)
Auxiliary Space: O(1)
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