1. I explained the Naive approach which was taking 2 loops and considering every subarray once. It was an O(N^2 ) approach.
2. I optimised my approach to one loop making it O(N), by applying kadane alg...read more

1. Create a nested loop. The outer loop will go from i = 0 to i = n - k. This will cover the starting indices of all k-subarrays
2. The inner loop will go from j = i to j = i + k - 1. Thi...read more

1. Create a BST and insert the first k elements in it
2. Print the largest element in the tree
3. Now iterate through the remaining elements of the tree:
   1. Insert the current element in the BST
   2. Delete th...read more

1. Create a double ended queue. Note that the deque will store the indices of the elements, not the elements themselves
2. Insert the first k elements (the first subarray) in the deque. While i...read more

The maximum sum would be -1

The maximum sum would be -1

Send line elements is maximum sum is -1

/*

Time Complexity - O(N * K)

Space Complexity - O(1)

where N is the length of the array and K is the size of subarrays

*/

void printSubarrayMax(int *arr, int n, int k)

{

if (n == 0 || k == 0)

{

return;

...read more