Problem 1 (Maximum Subarray). Given an array A[1, n] such that A[i] ∈ R, we want
the two indexes i, j ∈ {1, 2, . . . , n} ∧ i ≤ j that maximize the following function:
j
X
A[k]
k=i
Define B[1, n] as an array such that
B[i] =
n
X
A[i].
(1)
k=i
Observation 1. Given two indexes i, j such that i > j,
j−1
X
A[i] = B[i] − B[j].
(2)
k=i
So we can restate the problem as find the two indexes i, j ∈ {1, 2, . . . , n} ∧ i ≤ j that
maximize the function B[i] − B[j + 1].
The Algorithm
We start to scan the array B in reverse order, from the index n to 1. Suppose that
the sub-array B[i − 1, n] have been visited, and that in this sub-array the indexes k1 , k2
maximize B[K1 ] − B[K2 ] with K1 ≤ K2 ∈ {i − 1, i − 2, . . . , n} (in other words, we know
solution of the problem for this sub-array).
Suppose that B[m] with i ≤ m ≤ k2 is the element such that B[m] = minj=i,i−1,...,k2 {B[j]}.
Lemma 1. If B[i] > B[k1 ] − B[k2 ] + B[m] then the problem for the sub-array B[i, n]
has solution i, max{m − 1, 1}. Else the solution is still k1 , max{k2 − 1, 1}.
Proof.
B[i] > B[k1 ] − B[k2 ] − B[m]
⇒ B[i] − B[m] > B[k1 ] − B[k2 ],
(3)
So the cost of the new sub-array is improved than the old optimal sub-array, this means
that the cost of the new sub-array is also better than the cost of all sub-arrays that
contain indexes into the range i − 1 to n. We have to prove that is also better than subarrays that start with index i. This is surely true because the index m is the minimum
minj=i,i−2,...,n {B[j]} for construction; infact B[k2 ] = minj=k1 ,i−1,...,n {B[j]} (otherwise
k1 , k2 − 1 is not optimal for the old sub-array) and we know that B[m] ≤ B[k2 ] ∧ B[m] =
minj=i,i−1,...,k1 −1 {B[j]}.
In the case that B[i] ≤ B[k1 ] − B[k2 ] + B[m] we cannot improve the objective function
of the old sub-array.
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n
Lemma 2. The following algorithm solve the problem in O(n) steps and O( ) I/Os.
B
1
2
3
i n t A [ 1 , n ] // i n p u t
i n t B [ 1 , n ] // s u p p o r t a r r a y (A can be a l s o r e w r i t t e n
space saved )
i n t c , k 1 , k 2 , m, k , i ;
and
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5
6
7
8
9
10
11
12
13
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18
c = 0;
f o r ( i=n ; i >=1; i −−){
c = c + A[ i ] ;
B [ i ] = c ; } // t h e a r r a y B i s b u i l t
k 1 = n;
k 2 = n ; // i n i t i a l i z a t i o n t o max s u b s e q u e n c e A [ n , n ]
f o r ( i=n ; i >=1; i −−){
i f (B [ i ]<B [ k 2 ] ) {
m = i;
k = B [ k 1 ] − (B [ k 2 ] − B [m] ) ; }
i f (S [ i ] > k ){
k 1 = i;
k 2 = m;
k = B[ k 1 ]}}
The algorithm is correct. By induction. At the first step k1 = k2 = n are the maximum
subsequence. Then because m is at every step the minimum B[j], using Lemma 1 we
can prove that when i = 1 the solution is found.
Complexity. The array is scanned left to right exactly two times: one for build B and
n
the other for find the indexes. This imply also that the number of I/Os is , we can
B
reduce the accesses that are not consecutive (like when S[k1 ] is read at line 18) using
more variables.
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Roberto Ladu