Suppose we have a list of numbers called nums and two values x and y, we have to find the maximum sum of two non-overlapping sublists in nums which have lengths x and y.
So, if the input is like nums = [3, 2, 10, -2, 7, 6] x = 3 y = 1, then the output will be 22, as the sublist with length 3 we select [3, 2, 10] and for the other we select [7].
To solve this, we will follow these steps −
- P := a list with single element 0
- for each x in A, do
- insert (last element of P + x) at the end of P
- Define a function solve() . This will take len1, len2
- Q := a list with element (P[i + len1] - P[i]) for each i in range 0 to size of P - len1
- prefix := a copy of Q
- for i in range 0 to size of prefix - 1, do
- prefix[i + 1] := maximum of prefix[i + 1] and prefix[i]
- ans := -infinity
- for i in range len1 to size of P - len2, do
- left := prefix[i - len1]
- right := P[i + len2] - P[i]
- ans := maximum of ans and (left + right)
- return ans
- From the main method do the following −
- return maximum of solve(len1, len2) , solve(len2, len1)
Let us see the following implementation to get better understanding −
Example
class Solution:
def solve(self, A, len1, len2):
P = [0]
for x in A:
P.append(P[-1] + x)
def solve(len1, len2):
Q = [P[i + len1] - P[i] for i in range(len(P) - len1)]
prefix = Q[:]
for i in range(len(prefix) - 1):
prefix[i + 1] = max(prefix[i + 1], prefix[i])
ans = float("-inf")
for i in range(len1, len(P) - len2):
left = prefix[i - len1]
right = P[i + len2] - P[i]
ans = max(ans, left + right)
return ans
return max(solve(len1, len2), solve(len2, len1))
ob = Solution()
nums = [3, 2, 10, -2, 7, 6]
x = 3
y = 1
print(ob.solve(nums, x, y))Input
[3, 2, 10, -2, 7, 6], 3, 1
Output
22