2017-08-31 Check your solution. You can now check your Ruby solution at The Book of Problems.

Here is a problem that I considered solved for some months: The Adjacent Coins Problem. This is a problem where you have to choose a coin that maximizes your gain or minimizes your loss, but you have to do it in linear time and constant memory. Then, around three months ago, vladon reviewed my code. I was impressed.

This weekend, I finally put some time together and came up with the following solution.

Problem

Consider N coins aligned in a row. Each coin is showing either heads or tails. The adjacency of these coins is the number of adjacent pairs of coins with the same side facing up.

Write a C++ function

int solution ( const vector < int > & A );

that, given a non-empty zero-indexed array A consisting of N integers representing the coins, returns the maximum possible adjacency that can be obtained by reversing exactly one coin (that is, one of the coins must be reversed). Consecutive elements of array A represent consecutive coins in the row. Array A contains only 0s and/or 1s:

0 represents a coin with heads facing up;

1 represents a coin with tails facing up.

For example, given array A consisting of eight numbers, such that:

A [ 0 ] = 1 A [ 1 ] = 1 A [ 2 ] = 0 A [ 3 ] = 1 A [ 4 ] = 0 A [ 5 ] = 0 A [ 6 ] = 1 A [ 7 ] = 1

the function should return 5. The initial adjacency is 3, as there are three pairs of adjacent coins with the same side facing up, namely (0, 1), (4, 5) and (6, 7). After reversing the coin represented by A[2] , the adjacency equals 5, as there are five pairs of adjacent coins with the same side facing up, namely: (0, 1), (1, 2), (2, 3), (4, 5) and (6, 7), and it is not possible to obtain a higher adjacency.

The same adjacency can be obtained by reversing the coin represented by A[3] .

Assume that:

N is an integer within the range [1..100,000];

each element of array A is an integer within the range [0..1].

Complexity:

expected worst-case time complexity is O(N);

expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Analysis

Given a sequence of coins, the adjacency of the sequence either changes or remains the same when we flip a coin because we either gain pairs, lose pairs, or keep the same amount of pairs. Consider the following sequence and the delta adjacency for each coin.

Index | Coin | Delta adjacency ------------------------------- 0 | 1 | -1 1 | 1 | 0 2 | 0 | 2 3 | 1 | 2 4 | 0 | 0 5 | 0 | 0 6 | 1 | 0 7 | 1 | -1

The adjacency for the sequence is 3. If we flip the first or last coin, we lose one pair and the adjacency becomes 2. If we flip coin 2 or 3, we gain two pairs and the adjacency becomes 5. Flipping one of the other coins destroys a pair and creates another pair, so the adjacency does not change if we flip any of them.

Because we must flip a coin, we choose the coin to flip in the following way. When we can gain pairs, we choose to flip a coin that causes the biggest gain. When we cannot gain pairs, we choose to flip a coin that causes the smallest loss. That way we get the maximum number of pairs possible. For the sequence, we flip coin 2, which gives the maximum possible adjacency 5. Coin 3 is an alternative.

Solution

When the given sequence consists of one coin we return 0. For this case, 0 is always the right answer because adjacency is 0 and remains 0 after flipping the coin.

When the sequence has more coins, we compute the adjacency of the sequence, compute the maximum delta adjacency, and return the sum of these two values.

We compute the adjacency of the sequence by counting adjacent pairs. We count adjacent pairs by iterating the sequence and inspecting the current coin and the previous. At each iteration, the current and previous coins either form a pair or not. Each time the current coin and the previous show the same side, we count one pair. Our implementation is the following.

int adjacency ( const vector < int > & A ) { int adjacency = 0 ; int previous = - 1 ; for ( int i = 0 ; i < A . size (); i ++ ) { if ( previous == A [ i ]) adjacency ++ ; previous = A [ i ]; } return adjacency ; }

We compute the maximum delta adjacency by computing the delta adjacency for each coin and keeping the maximum. For the first and last coins, we apply the following rule and keep the maximum.

The first or last coin | Adjacent coin |/ Delta adjacency || / || | vv v 00 -> -1 01 -> 1 10 -> 1 11 -> -1

When there are coins in the middle, we apply the following rule to each coin and update the maximum delta.

Coin to the left | A coin in the middle |/ Coin to the right ||/ Delta adjacency ||| / ||| | vvv v 000 -> -2 001 -> 0 010 -> 2 011 -> 0 100 -> 0 101 -> 2 110 -> 0 111 -> -2

Our implementation is the following.

int max ( int x , int y ) { if ( x < y ) return y ; return x ; } int maximum_delta_adjacency ( const vector < int > & A ) { int maximum ; int last_position = A . size () - 1 ; maximum = A [ 0 ] == A [ 1 ] ? - 1 : 1 ; maximum = max ( A [ last_position ] == A [ last_position - 1 ] ? - 1 : 1 , maximum ); int middle [ 2 ][ 2 ][ 2 ] = { { { - 2 , 0 }, { 2 , 0 } }, { { 0 , 2 }, { 0 , - 2 } } }; for ( int i = 1 ; i < last_position ; i ++ ) { maximum = max ( middle [ A [ i - 1 ]][ A [ i ]][ A [ i + 1 ]], maximum ); } return maximum ; }

The last touch is adding the adjacency and the maximum delta adjacency in function solution as follows.

int solution ( const vector < int > & A ) { if ( A . size () == 1 ) return 0 ; return adjacency ( A ) + maximum_delta_adjacency ( A ); }

We illustrate usage of solution with the following program.

#include <iostream> #include <vector> using namespace std ; int adjacency ( const vector < int > & A ) { ... } int max ( int x , int y ) { ... } int maximum_delta_adjacency ( const vector < int > & A ) { ... } int solution ( const vector < int > & A ) { ... } void print_vector ( const vector < int > & A ) { cout << A [ 0 ]; for ( int i = 1 ; i < A . size (); i ++ ) cout << ", " << A [ i ]; cout << "

" ; } void solve_vector ( const int vA [], const int N ) { vector < int > A ( & vA [ 0 ], & vA [ 0 ] + N ); print_vector ( A ); cout << "Maximum possible adjacency is " << solution ( A ) << ".

" ; } int main () { int vA [] = { 0 }; solve_vector ( vA , 1 ); int vB [] = { 0 , 0 }; solve_vector ( vB , 2 ); int vC [] = { 0 , 1 }; solve_vector ( vC , 2 ); int vD [] = { 0 , 1 , 0 , 1 , 0 , 0 , 1 , 0 }; solve_vector ( vD , 8 ); int vE [] = { 1 , 1 , 0 , 1 , 0 , 0 , 1 , 1 }; solve_vector ( vE , 8 ); return 0 ; }

Discussion

The first time I tried to solve the problem, I had the impression that a solution involved ranking the coins and selecting the highest rank. So I created the two ranking rules. During my second attempt, my biggest obstacle was figuring how to apply the rules to sequences of any length. Then, I figured that I could handle sequences of length 1, length 2, and the rest separately. Afterwards, I figured that I only had to handle sequences of length 1 and the rest.

What was your experience trying to solve the problem? Let me know in the comments.

A constant-memory solution

Thanks to sltkr for posting the following solution on reddit.

Your analysis makes sense. My take was that the “adjacency number” is just the number of coins minus the number of runs (where a “run” is a maximal sequence of equal values in the input. For example, “0100” consists of three runs: “0”, “1” and “00”, and the “adjacency number” is thus 4 - 3 = 1). Maximizing the adjacency number means minimizing the number of runs, and the only way to do that is by flipping coins at the edge of runs. To summarize: If we have a run of length 1 between other runs, then we can improve the answer by 2. Example: “00100” -> “00000”.

If we have a run of length 1 at the beginning or end of the sequence, we can only improve the answer by 1. Example: “1001” -> “0001”.

If we have several runs, all of length 2 or more, then the best we can do is flip a coin between runs without changing the answer. Example: “0011” -> “0001”.

If we have only one run (i.e. all coins shows the same value) then flipping a coin makes things worse! The best we can do is flip a coin at the edge, for an “improvement” of -1. Example: “00000” -> “00001”.

Exception to the above: if we only have 1 coin, then flipping it doesn’t make a difference. The answer is always 0. To be honest, I missed the case at first, and I see you made the same mistake. :-) In code, this becomes:

int solve ( int N ) { // assumes N > 0 int pairs = 0 ; // number of pairs in the input int improve = - 1 ; // maximum improvement possible int a , b , c ; // values of last three coins cin >> c ; for ( int i = 1 ; i < N ; ++ i ) { a = b ; b = c ; cin >> c ; if ( b == c ) { ++ pairs ; } else if ( i >= 2 && a != b ) { improve = max ( improve , 2 ); } else if ( i == 1 || i == N - 1 ) { improve = max ( improve , 1 ); } else { improve = max ( improve , 0 ); } } return N == 1 ? 0 : pairs + improve ; } int main () { int N = 0 ; cin >> N ; cout << solve ( N ) << endl ; }

Note that this function combines the calculation of the number of pairs in the input with the maximum improvement, and it only keeps track of the values of the last three coins in the input, thus making it a truly constant-memory solution.

Acknowledgements

Special thanks to Vladislav Yaroslavlev (vladon) for reviewing my original attempt at solving the problem.

Many thanks to María Rabelero and Jan Křetínský for their valuable feedback on an early version of this post and my solution.

Jan included a valuable insight in his review. When we look for the maximum gain or minimum loss, we can stop iterating the coins in the middle as soon as we find a sequence 010 or 101 .

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