Repetitions -[CSES]
- Time limit: 1.00 s
- Memory limit: 512 MB
You are given a DNA sequence: a string consisting of characters A, C, G, and T. Your task is to find the longest repetition in the sequence. This is a maximum-length substring containing only one type of character.
Input
The only input line contains a string of nn characters.
Output
Print one integer: the length of the longest repetition.
Constraints
- 1≤n≤1061≤n≤106
Example
Input:ATTCGGGA
Output:3
JAVA Solution :-
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.Arrays;
import java.util.HashMap;
class Main {
public static void main(String[] args) throws IOException {
Reader sc = new Reader();
String st = sc.readLine();
long max=1;
for(int i=0;i<st.length()-1;)
{
int count=1;
char ch = st.charAt(i);
for(int j=i+1;j<st.length();j++)
{
if(ch==st.charAt(j))
{
count++;
}
else{
break;
}
}
if(count>=max)
{
max=count;
}
i=i+count;
}
System.out.println(max);
// while(t →0)
// {
// solve(sc);
// }
}
public static void solve(Reader sc) throws IOException
{
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[10000000]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == ‘\n’) {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ‘ ‘) {
c = read();
}
boolean neg = (c == ‘-’);
if (neg)
c = read();
do {
ret = ret * 10 + c — ‘0’;
} while ((c = read()) >= ‘0’ && c <= ‘9’);
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ‘ ‘)
c = read();
boolean neg = (c == ‘-’);
if (neg)
c = read();
do {
ret = ret * 10 + c — ‘0’;
} while ((c = read()) >= ‘0’ && c <= ‘9’);
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ‘ ‘)
c = read();
boolean neg = (c == ‘-’);
if (neg)
c = read();
do {
ret = ret * 10 + c — ‘0’;
} while ((c = read()) >= ‘0’ && c <= ‘9’);
if (c == ‘.’) {
while ((c = read()) >= ‘0’ && c <= ‘9’) {
ret += (c — ‘0’) / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}
