1031. Maximum Sum of Two Non-Overlapping Subarrays
Problem Setting¶
Given an array A
of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L
and M
. (For clarification, the L
-length subarray could occur before or after the M
-length subarray.)
Formally, return the largest V
for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1])
and either:
-
0 <= i < i + L - 1 < j < j + M - 1 < A.length
, or -
0 <= j < j + M - 1 < i < i + L - 1 < A.length
.
Example 1:¶
Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.
Example 2:¶
Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.
Example 3:¶
Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.
Solution¶
This problem can be solved by dynamic programming approach where
class Solution(object):
def maxSumTwoNoOverlap(self, A, L, M):
for i in range(1, len(A)): # xrange -> range by Hiro
A[i] += A[i - 1]
res, Lmax, Mmax = A[L + M - 1], A[L - 1], A[M - 1]
for i in range(L + M, len(A)): # xrange -> range by Hiro
Lmax = max(Lmax, A[i - M] - A[i - L - M])
Mmax = max(Mmax, A[i - L] - A[i - L - M])
res = max(res, Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L])
return res
Execution¶
A = [0,6,5,2,2,5,1,9,4]; L = 1; M = 2
s = Solution()
s.maxSumTwoNoOverlap(A, L, M)
A = [0,6,5,2,2,5,1,9,4]
A
for i in range(1, len(A)): # xrange -> range by Hiro
A[i] += A[i - 1]
A
L5:
-
res
: initialize the variable for result -
Lmax
:L
s elements of the cumulative list i.e., 0 in the example 1 -
Mmax
:M
's elements of the cumulative list i.e., 6 in the example 1
res, Lmax, Mmax = A[L + M - 1], A[L - 1], A[M - 1]
res, Lmax, Mmax
L6: the second loop starts from L+M
to the end of the list.
-
Lmax
is the case whenL
contiguous elements are taken first -
Mmax
is the case whenM
contiguous elements are taken first
Finally, it compares 2 cases (L
elements first taken or M
elements first taken) including the existing maximum result res
.
res = A[L + M - 1]
for i in range(L + M, len(A)):
print('='*40)
print('processing at {}th element'.format(i))
print('{} [{}] {}'.format(A[:i], A[i], A[i+1:]) )
Lmax = max(Lmax, A[i - M] - A[i - L - M])
Mmax = max(Mmax, A[i - L] - A[i - L - M])
res = max(res, Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L])
print('Lmax:', Lmax)
print('Mmax:', Mmax)
print('A[i] - A[i - M]:', A[i] - A[i - M])
print('A[i] - A[i - L]:', A[i] - A[i - L])
print('Current maximum result:', res)
Comments
Comments powered by Disqus