Probability of collision between two trucks

• Difficulty Level : Medium
• Last Updated : 20 Oct, 2021

Given two strings S and T, where S represents the first lane in which vehicles move from left to right and T represents the second lane in which vehicles move from right to left. Vehicles can be either B (bike), C (car), or T (truck). The task is to find the probability of collision between two trucks.

Examples:

Input: S = “TCCBCTTB”, T = “BTCCBBTT”
Output: 0.194444
Explanation:
Total collision = 7
Total accident = 36
Therefore, the probability can be calculated by 7/36 = 0.19444.

Input: S = “BTT”, T = “BTCBT”
Output: 0.25000

Illustration:

S = "TCCBCTTB", T = "BTCCBBTT"

Possible cases   | Accidents | Collision
-----------------------------------------
TCCBCTTB         |           |
BTCCBBTT         |     8     |   1
|           |
TCCBCTTB        |           |
BTCCBBTT         |     7     |   3
|           |
TCCBCTTB       |           |
BTCCBBTT         |     6     |   1
|           |
TCCBCTTB      |           |
BTCCBBTT         |     5     |   0
|           |
TCCBCTTB     |           |
BTCCBBTT         |     4     |   0
|           |
TCCBCTTB    |           |
BTCCBBTT         |     3     |   0
|           |
TCCBCTTB   |           |
BTCCBBTT         |     2     |   1
|           |
TCCBCTTB  |           |
BTCCBBTT         |     1     |   1

Total number of accidents: 8+7+6+5+4+3+2+1=36
Total number of collision: 1+3+1+0+0+0+1+1=7
Probability: 7/36=0.19444

Approach: Follow the steps below to solve the problem:

• Find the total number of favorable outcomes as the total number of collisions(accidents between trucks) and the total number of possible outcomes(total number of collisions) as the total number of accidents.
• Initialize a variable answer equals to 0 to stores the count of collisions.
• Count the number of trucks in the string T and store it in a variable count.
• Iterate over the characters of the strings S and T simultaneously:
• If S[i] is equal to ‘T’, increment answer by count.
• If T[i] is equal to ‘T’, decrement the count by 1.
• Now, calculate the total number of possible outcomes (total number of accidents). It is the sum of all the length of overlapping if keep shifting string a towards right or string b towards left by one unit.
• Let the length of string be N and string b be M. Then, the total number of overlapping will be:
• Find the probability as the ratio of the count of the collisions and the count of the accidents.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include using namespace std; // Function to calculate total// number of accidentsdouble count_of_accident(string a,                         string b){    // String size    int n = a.size(), m = b.size();     if (n > m)        return (m * (m + 1)) / 2;    else        return (n * (n + 1))                   / 2               + (m - n) * n;} // Function to calculate count// of all possible collisiondouble count_of_collision(string a,                          string b){    int n = a.size(), m = b.size();     // Stores the count of collisions    int answer = 0;     // Total number of truck in lane b    int count_of_truck_in_lane_b = 0;    for (int i = 0; i < m; i++)        if (b[i] == 'T')            count_of_truck_in_lane_b++;     // Count total number of collisions    // while traversing the string a    for (int i = 0; i < n && i < m; i++) {        if (a[i] == 'T')            answer                += count_of_truck_in_lane_b;         if (b[i] == 'T')            count_of_truck_in_lane_b--;    }    return answer;} // Function to calculate the// probability of collisionsdouble findProbability(string a,                       string b){    // Evaluate total outcome that is    // all the possible accident    double total_outcome        = count_of_accident(a, b);     // Evaluate favourable outcome i.e.,    // count of collision of trucks    double favourable_outcome        = count_of_collision(a, b);     // Print desired probability    cout << favourable_outcome                / total_outcome;} // Driver Codeint main(){    string S = "TCCBCTTB", T = "BTCCBBTT";     // Function Call    findProbability(S, T);     return 0;}

Java

 // Java program for the above approachimport java.util.*; class GFG{ // Function to calculate total// number of accidentsstatic int count_of_accident(String a,                         String b){    // String size    int n = a.length(), m = b.length();     if (n > m)        return (m * (m + 1)) / 2;    else        return (n * (n + 1))                   / 2               + (m - n) * n;} // Function to calculate count// of all possible collisionstatic double count_of_collision(String a,                          String b){    int n = a.length(), m = b.length();     // Stores the count of collisions    double answer = 0;     // Total number of truck in lane b    int count_of_truck_in_lane_b = 0;    for (int i = 0; i < m; i++)        if (b.charAt(i) == 'T')            count_of_truck_in_lane_b++;     // Count total number of collisions    // while traversing the String a    for (int i = 0; i < n && i < m; i++)    {        if (a.charAt(i) == 'T')            answer                += count_of_truck_in_lane_b;         if (b.charAt(i) == 'T')            count_of_truck_in_lane_b--;    }    return answer;} // Function to calculate the// probability of collisionsstatic void findProbability(String a,                       String b){    // Evaluate total outcome that is    // all the possible accident    int total_outcome        = count_of_accident(a, b);     // Evaluate favourable outcome i.e.,    // count of collision of trucks    double favourable_outcome        = count_of_collision(a, b);     // Print desired probability    System.out.printf("%4f",favourable_outcome                / total_outcome);} // Driver Codepublic static void main(String[] args){    String S = "TCCBCTTB", T = "BTCCBBTT";     // Function Call    findProbability(S, T);}} // This code is contributed by 29AjayKumar

Python3

 # Python3 program for the above approach # Function to calculate total# number of accidentsdef count_of_accident(a, b):         n = len(a)    m = len(b)         if (n > m):        return (m * (m + 1)) / 2    else:        return ((n * (n + 1)) / 2 +                (m - n) * n) # Function to calculate count# of all possible collisiondef count_of_collision(a, b):         # Size of string    n = len(a)    m = len(b)     # Stores the count of collisions    answer = 0     # Total number of truck in lane b    count_of_truck_in_lane_b = 0         for i in range(0, m):        if (b[i] == 'T'):            count_of_truck_in_lane_b += 1     # Count total number of collisions    # while traversing the string a    i = 0    while (i < m and i < n):        if (a[i] == 'T'):            answer += count_of_truck_in_lane_b        if (b[i] == 'T'):            count_of_truck_in_lane_b -= 1                     i += 1             return answer # Function to calculate the# probability of collisionsdef findProbability(a, b):         # Evaluate total outcome that is    # all the possible accident    total_outcome = count_of_accident(a, b);     # Evaluate favourable outcome i.e.,    # count of collision of trucks    favourable_outcome = count_of_collision(a, b);     # Print desired probability    print(favourable_outcome / total_outcome)  # Driver Code if __name__ == "__main__" :         S = "TCCBCTTB"    T = "BTCCBBTT"     # Function Call    findProbability(S, T)     # This code is contributed by Virusbuddah_

C#

 // C# program for the above approachusing System; class GFG{ // Function to calculate total// number of accidentsstatic int count_of_accident(String a,                         String b){    // String size    int n = a.Length, m = b.Length;     if (n > m)        return (m * (m + 1)) / 2;    else        return (n * (n + 1))                   / 2               + (m - n) * n;} // Function to calculate count// of all possible collisionstatic double count_of_collision(String a,                          String b){    int n = a.Length, m = b.Length;     // Stores the count of collisions    double answer = 0;     // Total number of truck in lane b    int count_of_truck_in_lane_b = 0;    for (int i = 0; i < m; i++)        if (b[i] == 'T')            count_of_truck_in_lane_b++;     // Count total number of collisions    // while traversing the String a    for (int i = 0; i < n && i < m; i++)    {        if (a[i] == 'T')            answer                += count_of_truck_in_lane_b;         if (b[i] == 'T')            count_of_truck_in_lane_b--;    }    return answer;} // Function to calculate the// probability of collisionsstatic void findProbability(String a,                       String b){    // Evaluate total outcome that is    // all the possible accident    int total_outcome        = count_of_accident(a, b);     // Evaluate favourable outcome i.e.,    // count of collision of trucks    double favourable_outcome        = count_of_collision(a, b);     // Print desired probability    Console.Write("{0:F4}", favourable_outcome                / total_outcome);} // Driver Codepublic static void Main(String[] args){    String S = "TCCBCTTB", T = "BTCCBBTT";     // Function Call    findProbability(S, T);}}  // This code is contributed by sapnasingh4991

Javascript


Output:
0.194444

Time Complexity: O(N)
Auxiliary Space: O(1)

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