Find total no of collisions taking place between the balls in which initial direction of each ball is given - GeeksforGeeks

24 Nov, 2021
Given N balls in a line. The initial direction of each ball is represented by the string consists of only ‘L’ and ‘R’ for the left and right direction respectively. It is given that both the balls reverse their direction after the collision and speed will remain the same before and after the collision. Calculate the total number of the collision that takes place.Examples: 
Input: str = "RRLL"
Output: 4

Explanation
After first collision: RLRL
After second collision: LRRL
After third collision: LRLR
After fourth collision: LLRR

Input: str = "RRLR"
Output:  2

Explanation
After first collision: RLRR
After second collision: LRRR
Approach:At each stage we have to find the no of substrings “RL”, change the substring “RL” to “LR”, and again count the no of substring “RL” in the resulting string and repeat in the following until no further substring “RL” is available. We are not going to count the no of substring “RL” on each stage as it will consume lots of time. Another way of doing this is, observe that following resulting strings we have seen on each stage –> “RRLL” -> “RLRL”-> “LRLR” ->”LLRR”. The initial string is “RRLL”. let’s have a 3rd ball whose direction is L. Now observe this L is shifting to the left side of the string. As we are exchanging “RL” to “LR”, so we are shifting that L towards the left side of the full string. Now similarly, if we continue this exchange, this L will face every R that is present on the left side of that string, so again forming “RL”. Hence, for this L, the total no of “RL” will be the total no of R which is present on the left side of that particular L. We will repeat this for each L. Hence in general, in order to count every possible substring “RL” at every stage, we will count the total no of represent of the left side of each L as these R and L will form “RL” in every stage which can be done in linear time complexity.Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation to Find total// no of collisions taking place between// the balls in which initial direction// of each ball is given #include <bits/stdc++.h>using namespace std; // Function to count no of collisionint count(string s){    int N, i, cnt = 0, ans = 0;     // length of the string    N = s.length();     for (i = 0; i < N; i++) {        if (s[i] == 'R')            cnt++;         if (s[i] == 'L')            ans += cnt;    }     return ans;} // Driver codeint main(){     string s = "RRLL";     cout << count(s) << endl;     return 0;}
// Java implementation to Find total// no of collisions taking place between// the balls in which initial direction// of each ball is givenimport java.io.*;class GFG {      // Function to count// no of collisionstatic int count(String s){  int N, i, cnt = 0, ans = 0;   // length of the string  N = s.length();   for (i = 0; i < N; i++)  {    if (s.charAt(i) == 'R')      cnt++;     if (s.charAt(i) == 'L')      ans += cnt;  }  return ans;}  // Driver codestatic public void main(String[] args){  String s = "RRLL";  System.out.println(count(s));}} // This code is contributed by Rutvik_56
# Python3 implementation to find total# no of collisions taking place between# the balls in which initial direction# of each ball is given # Function to count no of collisiondef count(s):     cnt, ans = 0, 0     # Length of the string    N = len(s)     for i in range(N):        if (s[i] == 'R'):            cnt += 1         if (s[i] == 'L'):            ans += cnt         return ans # Driver codes = "RRLL" print( count(s)) # This code is contributed by chitranayal
// C# implementation to Find total// no of collisions taking place between// the balls in which initial direction// of each ball is givenusing System; class GFG{ // Function to count no of collisionstatic int count(String s){    int N, i, cnt = 0, ans = 0;     // length of the string    N = s.Length;     for(i = 0; i < N; i++)    {        if (s[i] == 'R')            cnt++;         if (s[i] == 'L')            ans += cnt;    }    return ans;} // Driver codepublic static void Main(String[] args){    String s = "RRLL";     Console.Write(count(s));}} // This code is contributed by shivanisinghss2110
<script>  // Javascript implementation to Find total  // no of collisions taking place between  // the balls in which initial direction  // of each ball is given     // Function to count no of collision  function count(s)  {      let N, i, cnt = 0, ans = 0;       // length of the string      N = s.length;       for (i = 0; i < N; i++) {          if (s[i] == 'R')              cnt++;           if (s[i] == 'L')              ans += cnt;      }       return ans;  }     let s = "RRLL";    document.write(count(s)); // This code is contributed by divyesh072019.</script>
4
 
Time Complexity: O(N)
Auxiliary Space: O(1)
ukasp
shivanisinghss2110
rutvik_56
divyesh072019
as5853535
rishavmahato348
Arrays
Competitive Programming
Greedy
Mathematical
Pattern Searching
Strings
Arrays
Strings
Greedy
Mathematical
Pattern Searching
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