Find the min number in all contiguous subarrays of size L of a array of size n

Question

The minimum number in a subarray of size L. I have to find it for all the subarrays of the array. Is there any other way than scanning through all the subarrays individually

Answer 1

Scala version answer

def movingMin(windowSize: Int, array: Seq[Double]) = { 
    (1 to array.length - (windowSize - 1)).
        map{i => (array.take(i + (windowSize - 1)).takeRight(windowSize)).min}
}

Answer 2

I think your solution is OK , but to work properly it should be something like :

a[n]//the array
minimum[n-l+1]//fixed

minpos=position_minimum_in_subarray(a,0,l-1);
minimum[0]=a[minpos];
for(i=1;i<=n-l-1;i++)
{
    if(minpos=i-1)        
        minpos=position_minimum_in_subarray(a,i,i+l-1);                   
    else if(a[minpos]>a[i+l-1]) //fixed
        minpos=i+l-1; //fixed

    minimum[i] = a[minpos];
}

// Complexity Analysis :

//Time - O(n^2) in worse case(array is sorted) we will run
         "position_minimum_in_subarray" on each iteration

//Space - O(1) - "minimum array" is required for store the result

---

If you want to improve your time complexity, you can do it with additional space. For example you can store each sub array in some self-balancing BST (e.g. red-black tree) and fetch minimum on each iteration :

for (int i= 0; i<n; i++) {
    bst.add(a[i]);

    if (bst.length == l) {
        minimum[i-l] = bst.min;
        bst.remove(a[i - l]);            
    }
  }

 //It's still not O(n) but close.

 //Complexity Analysis :

 //Time - O(n*log(l)) = O(n*log(n)) - insert/remove in self-balancing tree
                                      is proportional to the height of tree (log)

 //Space - O(l) = O(n) 

Answer 3

You can use a double ended queue(Q) .Find a way such that the smallest element always appears at the front of the Q and the size of Q never exceeds L. Thus you are always inserting and deleting elements at most once making the solution O(n). I feel this is enough hint to get you going.