# 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 accidents``double` `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 collision``double` `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 collisions``double` `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 Code``int` `main()``{``    ``string S = ``"TCCBCTTB"``, T = ``"BTCCBBTT"``;` `    ``// Function Call``    ``findProbability(S, T);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG``{` `// Function to calculate total``// number of accidents``static` `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 collision``static` `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 collisions``static` `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 Code``public` `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 accidents``def` `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 collision``def` `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 collisions``def` `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 approach``using` `System;` `class` `GFG``{` `// Function to calculate total``// number of accidents``static` `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 collision``static` `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 collisions``static` `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 Code``public` `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)

My Personal Notes arrow_drop_up