SDE - 1

If last characters of two strings are same, nothing much to do. Ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1.
Else (If last characters are not same), we consider all operations on ‘str1’, consider all three operations on last character of first string, recursively compute minimum cost for all three operations and take minimum of three values. 
Insert: Recur for m and n-1
Remove: Recur for m-1 and n
Replace: Recur for m-1 and n-1

See, here each coin of a given denomination can come an infinite number of times. (Repetition allowed), this is what we call UNBOUNDED KNAPSACK. We have 2 choices for a coin of a particular denomination, either i) to include, or ii) to exclude. But here, the inclusion process is not for just once; we can include any denomination any number of times until N<0.

Basically, If we are at s[m-1], we can take as many instances of that coin ( unbounded inclusion ) i.e count(S, m, n – S[m-1] ) ; then we move to s[m-2]. After moving to s[m-2], we can’t move back and can’t make choices for s[m-1] i.e count(S, m-1, n ).

Finally, as we have to find the total number of ways, so we will add these 2 possible choices, i.e count(S, m, n – S[m-1] ) + count(S, m-1, n ) ; which will be our required answer.

The simple idea of Kadane's algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive sum compare it with max_so_far and update max_so_far if it is greater than max_so_far

Tip 1 : Read Galvin for OS thoroughly.
Tip 2: Do practice for SQL queries.
Tip 3: Discuss your problem in detail.