C++     Dynamic Programming     Medium     String     Two Pointers    

Problem Statement:

There are n dominoes in a line, and we place each domino vertically upright. In the beginning, we simultaneously push some of the dominoes either to the left or to the right.

After each second, each domino that is falling to the left pushes the adjacent domino on the left. Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right.

When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces.

For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.

You are given a string dominoes representing the initial state where:

  • dominoes[i] = 'L', if the ith domino has been pushed to the left,
  • dominoes[i] = 'R', if the ith domino has been pushed to the right, and
  • dominoes[i] = '.', if the ith domino has not been pushed.

Return a string representing the final state.

 

Example 1:

Input: dominoes = "RR.L"
Output: "RR.L"
Explanation: The first domino expends no additional force on the second domino.

Example 2:

Input: dominoes = ".L.R...LR..L.."
Output: "LL.RR.LLRRLL.."

 

Constraints:

  • n == dominoes.length
  • 1 <= n <= 105
  • dominoes[i] is either 'L', 'R', or '.'.

Solution:

Consider example

 
".L.R...LR..L.."

Now any domino which is pushed at t = 0 will remain as it is and we only need to worry about the standing ones (dots). Now a domino falls towards right if the nearest non-dot domino in the left side is pushed rightwards. (Situation A). This means the pattern R.... And a domino falls towards left if the nearest non-dot domino in the right side is pushed leftwards. (Situation B). This means the pattern ...L

How do we do this? Create a data structure nearest_left which denotes for any dot, the nearest non-dot character in its left side along with the distance from itself.

 
0: NULL
2: (L,1)
4: (R,1)
5: (R,2)
6: (R,3)
9: (R,1)
10: (R,2)
12: (L,1)
13: (L,2)

Similarly we have nearest_right.

 
0: (L,1)
2: (R,1)
4: (L,3)
5: (L,2)
6: (L,1)
9: (L,2)
10: (L,1)
12: NULL
13: NULL

Now situation A means nearest_left for that dot is a R and situation B means nearest_right for that dot is a L.

Now for any dot we can have:

  • Neither A nor B ==> Remains standing
  • A only ==> falls towards right
  • B only ==> falls toward left
  • Both A and B. In this case the direction of fall is decided by the nearer of the two. Remain standing if distances same.

For our example:

 
0: B only
2: Neither A nor B
4: Both A and B, A wins
5: Both A and B, Tie.
6: Both A and B, B wins
9: Both A and B, A wins
10: Both A and B, B wins
12: Neither A nor B
13: Neither A nor B

This gives us the answer:

 
"LL.RR.LLRRLL.."

Now let us implement this logic

 
class Solution {
public:
    string pushDominoes(string dominoes) {
        int n = dominoes.length();
        vector<pair<int,pair<char,int>>> nearest_left, nearest_right;
        int left_idx=-1;
        char left_ch='X';
        for (int i=0; i<n; i++)
        {
            if (dominoes[i]!='.')
            {
                left_idx = i;
                left_ch = dominoes[i];
            }
            if (dominoes[i]=='.')
                nearest_left.push_back({i,{left_ch,i-left_idx} });
        }
        int right_idx=n;
        char right_ch='X';
        for (int i=n-1; i>=0; i--)
        {
            if (dominoes[i]!='.')
            {
                right_idx = i;
                right_ch = dominoes[i];
            }
            if (dominoes[i]=='.')
                nearest_right.push_back({i,{right_ch,right_idx-i} });
        }
        reverse(nearest_right.begin(),nearest_right.end());
        for (int i=0; i<nearest_left.size(); i++)
        {
            auto nl = nearest_left[i], nr = nearest_right[i];
            int dom_idx = nl.first; // this wil be same as nr.first
            auto nls = nl.second, nrs = nr.second;
            if (nls.first=='R' && nrs.first=='L')
            {
                if (nls.second<nrs.second) dominoes[dom_idx] = 'R';
                if (nrs.second<nls.second) dominoes[dom_idx] = 'L';
                continue;
            }
            if (nls.first=='R')
            {
                dominoes[dom_idx] = 'R';
                continue;
            }    
            if (nrs.first=='L') 
            {
                dominoes[dom_idx] = 'L';
                continue;
            }
        }
        return dominoes;
    }
};