This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

You have a necklace of N red, white, or blue beads (3<=N<=350) some of which are red, others blue, and others white, arranged at random. Here are two examples for n=29:

``````        1 2                             1 2
r b b r                         b r r b
r         b                     b         b
r         r                     b         r
r             r                 w             r
b             r                 w             w
b                 b             r                 r
b                 b             b                 b
b                 b             r                 b
r             r                 b             r
b             r                 r             r
b         r                     r         r
r         r                     r         b
r b r                           r r w
Figure A                        Figure B
``````

The beads considered first and second in the text that follows have been marked in the picture.

The configuration in Figure A may be represented as a string of b’s and r’s, where b represents a blue bead and r represents a red one, as follows: brbrrrbbbrrrrrbrrbbrbbbbrrrrb .

Suppose you are to break the necklace at some point, lay it out straight, and then collect beads of the same color from one end until you reach a bead of a different color, and do the same for the other end (which might not be of the same color as the beads collected before this).

Determine the point where the necklace should be broken so that the most number of beads can be collected.

### Example

For example, for the necklace in Figure A, 8 beads can be collected, with the breaking point either between bead 9 and bead 10 or else between bead 24 and bead 25.

In some necklaces, white beads had been included as shown in Figure B above. When collecting beads, a white bead that is encountered may be treated as either red or blue and then painted with the desired color. The string that represents this configuration can include any of the three symbols r, b and w.

Write a program to determine the largest number of beads that can be collected from a supplied necklace.

### INPUT FORMAT

 Line 1: N, the number of beads Line 2: a string of N characters, each of which is r, b, or w

``````29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
``````

### OUTPUT FORMAT

A single line containing the maximum of number of beads that can be collected from the supplied necklace.

``````11
``````

### OUTPUT EXPLANATION

Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.

``````                Two necklace copies joined here
v
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb|wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
******|*****
rrrrrb|bbbbb  <-- assignments
5xr .....#|#####  6xb
5+6 = 11 total
``````

Java

C++

Pascal

## ANALYSIS

Russ Cox

In this problem, the necklace size is small enough (350) that we might as well just try breaking the necklace at each point and see how many beads can be collected. This will take approximately O(n^2) time, but n is small enough that it won’t matter.

The code is slightly simple-minded in that we might count the same beads twice if they can be taken off either side of the break. This can only happen if all beads can be taken off the necklace, so we just check for that at the end.

``````
#include <stdio.h>
#include <string.h>
#include <assert.h>

#define MAXN 400

char necklace[MAXN];
int len;

/*
* Return n mod m.  The C % operator is not enough because
* its behavior is undefined on negative numbers.
*/
int
mod(int n, int m)
{
while(n < 0)
n += m;
return n%m;
}

/*
* Calculate number of beads gotten by breaking
* before character p and going in direction dir,
* which is 1 for forward and -1 for backward.
*/
int
nbreak(int p, int dir)
{
char color;
int i, n;

color = 'w';

/* Start at p if going forward, bead before if going backward */
if(dir > 0)
i = p;
else
i = mod(p-1, len);

/* We use "n<len" to cut off loops that go around the whole necklace */
for(n=0; n<len; n++, i=mod(i+dir, len)) {
/* record which color we're going to collect */
if(color == 'w' && necklace[i] != 'w')
color = necklace[i];

/*
* If we've chosen a color and see a bead
* not white and not that color, stop
*/
if(color != 'w' && necklace[i] != 'w' && necklace[i] != color)
break;
}
return n;
}

void
main(void)
{
FILE *fin, *fout;
int i, n, m;

assert(fin != NULL && fout != NULL);

fscanf(fin, "%d %s", &len, necklace);
assert(strlen(necklace) == len);

m = 0;
for(i=0; i<len; i++) {
n = nbreak(i, 1) + nbreak(i, -1);
if(n > m)
m = n;
}

/*
* If the whole necklace can be gotten with a good
* break, we'll sometimes count beads more than
* once.  this can only happen when the whole necklace
* can be taken, when beads that can be grabbed from
* the right of the break can also be grabbed from the left.
*/
if(m > len)
m = len;

fprintf(fout, "%d\n", m);
exit (0);
}

``````

Daniel Bundala

Dynamic Programming is good method for solving this problem in O(N). If we consider two copies of the string we easy transform cyclic configuration of the necklace to linear. Now we can compute for each breaking point how many beads of the same color can be collected on the left and on the right from the breaking point. I show how we can compute it only for the left side. For right side it is analogical. Let r[p] and b[p] be the number of red / blue beads that can be collected, when necklace is broken in point p. If we know this and color of next bead (c) we can compute r[p+1] and b[p+1].

`````` r[0] = b[0] = 0
If c = 'r' then r[p+1] = r[p] + 1 and b[p+1] = 0
because the length of the blue beads is 0.
if c = 'b' then b[p+1] = b[p] + 1 and r[p+1] = 0
if c = 'w' then both length of the red and length of blue beads
can be longer.
so r[p+1] = r[p]+1 and b[p+1] = b[p] + 1.
``````

The number of beads that can be collected in breaking point p is then max(left[r[p]], left[b[p]]) + max(right[r[p]], right[b[p]]). And the maximum from this value is answer for the problem.

``````#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

FILE *in,*out;

int main () {

int n;
char tmp[400], s[800];
fscanf(in, "%d %s", &n, tmp);

strcpy(s, tmp);
strcat(s, tmp);

int left[800][2], right[800][2];
left[0][0] = left[0][1] = 0;

for (int i=1; i<= 2 * n; i++){
if (s[i - 1] == 'r'){
left[i][0] = left[i - 1][0] + 1;
left[i][1] = 0;
} else if (s[i - 1] == 'b'){
left[i][1] = left[i - 1][1] + 1;
left[i][0] = 0;
} else {
left[i][0] = left[i - 1][0] + 1;
left[i][1] = left[i - 1][1] + 1;
}
}

right[2 * n][0] = right[2 * n][1] = 0;
for (int i=2 * n - 1; i >= 0; i--){
if (s[i] == 'r'){
right[i][0] = right[i + 1][0] + 1;
right[i][1] = 0;
} else if (s[i] == 'b'){
right[i][1] = right[i + 1][1] + 1;
right[i][0] = 0;
} else {
right[i][0] = right[i + 1][0] + 1;
right[i][1] = right[i + 1][1] + 1;
}
}

int m = 0;
for (int i=0; i<2 * n; i++)
m = max(m, max(left[i][0], left[i][1]) + max(right[i][0], right[i][1]));
m = min(m, n);
fprintf(out, "%d\n", m);
fclose(in); fclose(out);
return 0;
}
``````

Frank Takes

Holland’s Frank Takes has a potentially easier solution:

``````/* This solution simply changes the string s into ss, then for every starting
// symbol it checks if it can make a sequence simply by repeatedly checking
// if a sequence can be found that is longer than the current maximum one.
*/

#include <iostream>
#include <fstream>
using namespace std;

int main() {
fstream input, output;
input.open(inputFilename.c_str(), ios::in);
output.open(outputFilename.c_str(), ios::out);

int n, max=0, current, state, i, j;
string s;
char c;

input >> n >> s;
s = s+s;
for(i=0; i<n; i++) {
c = (char) s[i];
if(c == 'w')
state = 0;
else
state = 1;
j = i;
current = 0;
while(state <= 2) {
// dont go further in second string than starting position in first string
while(j<n+i && (s[j] == c || s[j] == 'w')) {
current++;
j++;
} // while
state++;
c = s[j];
} // while
if(current > max)
max = current;
} // for

output << max << endl;
return 0;
} // main
``````