CODEWITHSUNDEEP
arraysalgorithmdata-structures
Gsquare
Gsquare

Reputation: 700

Create least cost array

I have an array of size n+1 where I wish to store the least cost of purchasing n items(i^th index stores cost of i^th item).

There are m different sellers: each seller provides item L to item R (where L,R>=1 and L,R<=n) for a cost of C each.

To create the least cost array I did the following:

for (int i=1; i<=m; i++) {
    int L = L of i^th seller
    int R = R of i^th seller
    int C = selling price of i^th seller
    for (int j=L; j<=R; j++) {
        if (leastCost[j]==0 || leastCost[j]>C) {
            leastCost[j]=C
    }
}

Constructing this array is O(n*m) and accessing this array is O(1).

For very large values of n and m, is there a better way to construct the least cost array?

Perhaps a different approach of not storing it in an array and something else so that overall time complexity is reduced?

Upvotes: 5

Views: 273

Answers (3)

Sumeet
Sumeet

Reputation: 8292

Definitely, we can do better than O(n*m) and also the solution is very simple.

Following is the pseudocode to solve this problem:

Construct a MIN-HEAP of the sellers based on their costs c.

Construct another array x[1...n] with its each element set to 0 initially.

Do the following initializations:
 count=0

while(count < n)
{
 S = EXTRACT_MIN(Heap)
 if(count==0)
 {            
  for(j=S.L to S.R)
  {
   leastCost[j]=S.c
   ++count
   x[j]=S.R
  } 
 }
 else
 {
  for(j=S.L;j<=S.R;++j)
  {
   if(x[j]!=0)
   {
    i=x[j] and continue;
   }
   else 
   {
    leastCost[j]=S.c
    ++count
    x[j]=S.R
   }
  }
 }
}

Explanation

The best optimization that we can achieve in this problem is that we skip all those array indices that are already filled because they already have least cost.

x the helper array: The array x helps us to skip all the array indices which have already been filled because:

x[i] stores the index j such that i to j already have least cost filled in the array so this is the reason for using the conditional statement:

if(x[i]!=0)
{
 i=x[i] and continue;
}

So basically it is helping us to jump straight to the index which is not filled.

MIN-HEAP: It allows us to find the minimum cost c of all the costs present in time O(logm).

Time Complexity

Since we access the array leastCost at most n times therefore for accessing the array : O(n)

Building the heap takes O(m)

In Worst Case all the sellers will contribute to some indices and there will be exact m EXTRACT_MIN(Heap) operations and each takes O(logm) time so time for this is: O(m*logm).

Therefore total Time Complexity=O(n + mlogm + m) = O(n + mlogm).

By the way, If I was using C language, I would use struct for Seller and if Java then a class for Seller.Feel free for any queries.

Upvotes: 1

Pham Trung
Pham Trung

Reputation: 11294

This problem is a classic problem Range minimum query, which you can use Segment tree to obtain a solution which has time complexity O(m log n) to create the tree and O(log n) for each query.

More details:

We using a tree to represent the minimum price for each item from 1 to n.

For each seller, we update the tree:

tree.update(L, R, C);

Finally, to get the min value for item i, we query the tree:

tree.getMin(i, i);

As both update and getMin has time complexity O(log n), the time complexity for the whole program is O(max(m,n) log n). You don't need to modify anything from the original Segment tree implementation.

Upvotes: 0

Herokiller
Herokiller

Reputation: 2991

You can use some form of stack here

1st sort all sellers by L. and by C desc if L equal.

Then starting from 1st seller(with min Li)

If stack is empty put him to the stack. If there are no more sellers with

(Lj == Li)

then cost[Li] = C[i] and Li++ (in stack)

if there is entry with Lj == Li then

  • if Rj < Ri and Cj > Ci skip

  • if Rj < Ri and Cj < Ci add this seller to stack

  • if Rj > Ri and Cj > Ci then pop current seller(i), add this seller(j) and add seller(i)

Upvotes: 1

Related Questions