Missing Number -[ CSES ]
- Time limit: 1.00 s
- Memory limit: 512 MB
You are given all numbers between 1,2,…,n1,2,…,n except one. Your task is to find the missing number.
Input
The first input line contains an integer nn.
The second line contains n−1n−1 numbers. Each number is distinct and between 11 and nn (inclusive).
Output
Print the missing number.
Constraints
- 2≤n≤2⋅1052≤n≤2⋅105
Example
Input:5
2 3 1 5
Output:4
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();
long n;
n =sc.nextLong();
long a , sum=0;
for(long i=0;i<n-1;i++)
{
a=sc.nextLong();
sum=sum+a;
}
System.out.println( (n * (n + 1) / 2)-sum);
// 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[64]; // 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();
}
}
}
