The Maximum Contiguous Subsequence Sum Problem in Java

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63 The Maximum Contiguous Subsequence Sum Problem
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In this section, we consider the following problem:
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MAXIMUM CONTIGUOUS SUBSEQUENCE SUM PROBLEM GIVEN (POSSIBLY NEGATIVE) INTEGERS A 1, A , , A FIND ( A N D 1DEAJTIFY AI THE THE SEQUENCE CORRESPONDING TO) THE MAXIMUM VALUE OF I =i
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MAXIMUM CONTIGUOUS SUBSEQUENCE SUM IS ZERO IF ALL THE INTEGERS ARE NEGATIVE
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As an example, if the input is (-2, 1 1 , 4 , 13, -5, 2), the answer is 20, which represents the contiguous subsequence encompassing items 2 through 4 (shown in boldface type) As a second example, for the input { 1, -3,4, -2, -161, the answer is 7 for the subsequence encompassing the last four items
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I Each of N points can be paired with N - 1 points for a total of N(N - 1) pairs However this pairing double counts pairs A B and B, A, so we must divide by two
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Programming details are considered after the algorithm design Always consider emptiness
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There are lots of drastically different algorithms (in terms of efficiency) that can be used to solve the maximum contiguous subsequence sum problem
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In C++, arrays begin at 0, so a C++ program would represent the input as a sequence Ao, , A, _ This is a programming detail and not part of the algorithm design Before discussing the algorithms for this problem, we need to comment on the degenerate case, in which all input integers are negative The problem statement gives a maximum contiguous subsequence sum of 0 for this case You might wonder why we do this, rather than just returning the largest (ie, the smallest in magnitude) negative integer in the input The reason is that the empty subsequence, consisting of zero integers, is also a subsequence, and its sum is clearly 0 Because the empty subsequence is contiguous, there is always a contiguous subsequence whose sum is 0 This result is analogous to the empty set being a subset of any set Be aware that emptiness is always a possibility and that in many instances it is not a special case at all The maximum contiguous subsequence sum problem is interesting mainly because there are so many algorithms to solve it-and the performance of these algorithms varies drastically In this section we discuss three such algorithms The first is an obvious exhaustive search algorithm, but it is very inefficient The second is an improvement on the first, which is accomplished by a simple observation The third is a very efficient, but not obvious, algorithm We prove that its running time is linear In 8 we present a fourth algorithm, which has O(N log N ) running time That algorithm is not as efficient as the linear algorithm, but it is much more efficient than the other two It also is typical of the kinds of algorithms that result in O(N log N)running times The graphs shown in Figures 61 and 62 are representative of these four algorithms
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631 The Obvious O(N3)Algorithm
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A brute force algorithm is generally the least efficient but simplest method to code
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The simplest algorithm is a direct exhaustive search, or a brute force algorithm, as shown in Figure 64 Lines 10 and I I control a pair of loops that iterate over all possible subsequences For each possible subsequence, the value of its sum is computed at lines 13-15 If that sum is the best sum encountered, we update the value of maxsum,which is eventually returned at line 25 Two i n t s - s e q s t a r t and seqEnd (which are passed by reference)-are also updated whenever a new best sequence is encountered The direct exhaustive search algorithm has the merit of extreme simplicity; the less complex an algorithm is, the more likely it is to be programmed correctly However, exhaustive search algorithms are usually not as efficient as possible In the remainder of this section we show that the running time of the algorithm is cubic We count the number of times (as a function of the input size) the expressions in Figure 64 are evaluated We require only a
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