TCS CodeVita 2016 Round1 Question: Min Product Array

The task is to find the minimum sum of Products of two arrays of the same size, given that k modifications are allowed on the first array. In each modification, one array element of the first array can either be increased or decreased by 2.

Note- the product sum is Summation (A[i]*B[i]) for all i from 1 to n where n is the size of both arrays

Input Format: 

1. First line of the input contains n and k delimited by whitespace
2. Second line contains the Array A (modifiable array) with its values delimited by spaces
3. Third line contains the Array B (non-modifiable array) with its values delimited by spaces

Output Format:

Output the minimum sum of products of the two arrays

Constraints:

1. 1 ≤ N ≤ 10^5
2. 0 ≤ |A[i]|, |B[i]| ≤ 10^5
3. 0 ≤ K ≤ 10^9

ANSWER:     SOLUTION IN JAVA.

 import java.util.*;
public class codevita {
    public static void main(String []args){
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt(),i,j,k,r,min=0,min1=0,sum,sum1;
    System.out.print("\t");
     k=sc.nextInt();
     System.out.println();
int []a=new int[n];
int []b=new int[n];
for( i=0;i<n;i++)
a[i]=sc.nextInt();
System.out.println();
for( i=0;i<n;i++)
b[i]=sc.nextInt();

for(i=0;i<n;i++){
    sum=0;
    a[i]-=2*k;
    for(j=0;j<n;j++){
           
sum+=a[j]*b[j];

    }
    if(i==0)
        min=sum;
    a[i]+=2*k;
    if(min>=sum){
        min=sum;
    }
}
for(i=0;i<n;i++){
    sum1=0;
    a[i]+=2*k;
    for(j=0;j<n;j++){
           
sum1+=a[j]*b[j];

    }
    if(i==0)
        min1=sum1;
    a[i]-=2*k;
    if(min1>=sum1){
        min1=sum1;
    }
}
if(min<min1)
    System.out.println(min);
else System.out.println(min1);

sc.close();
}
}

INPUT:
3 4
3 4 5
5 6 7

OUTPUT: 18