Minimum Product Array

         Minimum Product Array



TCS codevita 2016 round 1:
        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:
         First line of the input contains n and k delimited by white space Second line contains the Array A (modifiable array) with its values delimited by spaces 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 ≤ N ≤ 10^50 ≤ |A[i]|, |B[i]| ≤ 10^50 ≤ K ≤ 10^9

Sample       Input                   Output


1.                3 5                            -31

                   1 2 -3

                  -2 3 -5



2.               5 3                              25

                  2 3 4 5 4

                  3 4 2 3 2




Explanation for sample 1:

     Here total numbers are 3 and total modifications allowed are 5. So we modified A[2], which is -3 and increased it by 10 (as 5 modifications are allowed). Now final sum will be
(1 * -2) + (2 * 3) + (7 * -5)
-2 + 6 - 35
-31

-31 is our final answer.




Explanation for sample 2:

     Here total numbers are 5 and total modifications allowed are 3. So we modified A[1], which is 3 and decreased it by 6 (as 3 modifications are allowed).
Now final sum will be
(2 * 3) + (-3 * 4) + (4 * 2) + (5 * 3) + (4 * 2)
6 - 12 + 8 + 15 + 8
25

25 is our final answer. 




PROGRAM:

#include<stdio.h>
int main()
{
 int m[10],i,min,max,k,n,p=1,sum=0,u[10],mins;
 scanf("%d",&n);
 scanf("%d",&k);
 for(i=0;i<n;i++)
 scanf("%d",&u[i]);
 for(i=0;i<n;i++)
 scanf("%d",&m[i]);
 min=m[0];
 max=m[0];
 for(i=0;i<n;i++)
 {
  if(m[i]<min)
  min=m[i];
  if(m[i]>max)
  max=m[i];
 }
 if (min<0&&max<0)
 mins=min<max?min:max;
 else if(min>0&&max>0)
 mins=max>min?max:min;
 else
 {
 if((min-max)<-(2*min))
  mins=min;
  else 
  mins=max;
 }
 for(i=0;i<n;i++)
 {
  if(mins==m[i]&&mins>0)
   {

u[i]=u[i]-(2*k);
   goto x;
   }
   if(mins==m[i]&&mins<0)
    { 
    u[i]=u[i]+(2*k);
    goto x;
    }
  }
x:for(i=0;i<n;i++)
 {
 p=u[i]*m[i];
 sum=sum+p;
 }
 printf("%d",sum);
 return 0;
 
}

OUTPUT:

5 3                              

2 3 4 5 4

3 4 2 3 2

25

You can directly run it on a IDE: https://ide.geeksforgeeks.org/lP9ISaZ9yx
You can comment your feedback and doubts if any cheers!

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