Associate Engineer

First sort the array
second print the value at 2nd position

while (current != NULL) { 
// Store next 
next = current->next; 
// Reverse current node's pointer 
current->next = prev; 
// Move pointers one position ahead. 
prev = current; 
current = next; 
}

1. Read all the inputs.
->number of rows and columns say m and n.
->mxn matrix elements.
->base column number b. 
2. let i is the row index and j is the column index which starts from 0.
->hence first element of the base column b means (0,b).
3. start applying the sort logic on the column b=2.
4. for every swap required swap the elements in the other columns as well.
->when it's checked that (r1, b)needs to be swapped with (r2, b), don‟t just swap these 
two
->start from column 1, swap element at r1 with that of r2, then go on until the column n.
->repeat the swapping until all the m elements of the base column b are sorted.

1. Maintain sum of elements encountered so far in a variable (say sum).
2. If current sum is 0, we found a subarray starting from index 0 and ending at 
index current index
3. Check if current sum exists in the hash table or not.
4. If current sum already exists in the hash table then it indicates that this sum was 
the sum of some sub-array elements arr[0]…arr[i] and now the same sum is 
obtained for the current sub-array arr[0]…arr[j] which means that the sum of the 
sub-array arr[i+1]…arr[j] must be 0.
5. Insert current sum into the hash table

1. Create an array arr3[ ] of size n1 + n2. 
2. Simultaneously traverse arr1[ ] and arr2[ ]. 
o Pick smaller of current elements in arr1[ ] and arr2[ ], copy this 
smaller element to next position in arr3[ ] and move ahead in 
arr3[ ] and the array whose element is picked.
3. If there are remaining elements in arr1[ ] or arr2[ ], copy them also in 
arr3[ ].