HackerEarth Two Arrays problem solution

In this HackerEarth Two Arrays problem solution Let's define the matching number for two arrays X and Y, each with size n, as the number of pairs of indices (i,j) such that Xi = Yj (1 <= i, j <= n).

You are given an integer K and two arrays A and B with sizes N and M respectively. Find the number of ways to pick two subarrays with equal size, one from array A and the other from array B, such that the matching number for these two subarrays is >= K.

Note: a subarray consists of one or more consecutive elements of an array.



HackerEarth Two Arrays problem solution


HackerEarth Two Arrays problem solution.

import java.io.*;
import java.util.*;
import java.math.*;
import java.util.concurrent.*;

public final class sol
{
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static FastScanner sc=new FastScanner(br);
static PrintWriter out=new PrintWriter(System.out);
static Random rnd=new Random();
static int[][] arr;
static int n,m,k;

static int get_sum(int i,int j,int len)
{
return arr[i][j]-arr[i][j-len]-arr[i-len][j]+arr[i-len][j-len];
}

public static void main(String args[]) throws Exception
{
n=sc.nextInt();m=sc.nextInt();k=sc.nextInt();int[] a=new int[n+1],b=new int[m+1];

for(int i=1;i<=n;i++)
{
a[i]=sc.nextInt();
}

for(int i=1;i<=m;i++)
{
b[i]=sc.nextInt();
}

arr=new int[n+1][m+1];

for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
arr[i][j]=(a[i]==b[j]?1:0);
}
}

long res=0;

for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
arr[i][j]+=arr[i][j-1];
}
}

for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
arr[i][j]+=arr[i-1][j];
}
}

for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
int low=1,high=Math.min(i,j);

while(low<high)
{
int mid=(low+high)>>1;

if(get_sum(i,j,mid)>=k)
{
high=mid;
}

else
{
low=mid+1;
}
}

if(get_sum(i,j,low)>=k)
{
res+=(Math.min(i,j)-low+1);
}
}
}

out.println(res);out.close();
}
}
class FastScanner
{
BufferedReader in;
StringTokenizer st;

public FastScanner(BufferedReader in) {
this.in = in;
}

public String nextToken() throws Exception {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}

public String next() throws Exception {
return nextToken().toString();
}

public int nextInt() throws Exception {
return Integer.parseInt(nextToken());
}

public long nextLong() throws Exception {
return Long.parseLong(nextToken());
}

public double nextDouble() throws Exception {
return Double.parseDouble(nextToken());
}
}


Post a Comment

0 Comments