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    <p>Kadane's algorithm keeps track of the maximum value of a sequence ending at the current point, and the maximum value seen so far.</p> <p>Here is a Python implementation based on <a href="http://en.wikipedia.org/wiki/Maximum_subarray_problem" rel="nofollow">the wikipedia page</a>:</p> <pre><code>def kadane(A): max_ending_here = max_so_far = 0 for x in A: max_ending_here = max([x,0,max_ending_here+x]) max_so_far = max(max_so_far,max_ending_here) return max_so_far </code></pre> <p>We can modify the algorithm to keep track of the count of such sequences by adding two variables:</p> <ul> <li><strong>count_with_max_ending_here</strong> to count the number of sequences ending at the current point with values that sum to max_ending_here</li> <li><strong>count_with_max</strong> to count the number of sequences found so far with the maximum value</li> </ul> <p>Kadane's algorithm can be straightforwardly changed to keep track of these variables while staying at O(n) complexity:</p> <pre><code>def kadane_count(A): max_ending_here = max_so_far = 0 count_with_max_ending_here = 0 # Number of nontrivial sequences that sum to max_ending_here count_with_max = 0 for i,x in enumerate(A): new_max = max_ending_here+x if new_max&gt;=0: if count_with_max_ending_here==0: # Start a nontrivial sequence here count_with_max_ending_here=1 elif max_ending_here==0: # Can choose to carry on a nontrivial sequence, or start a new one here count_with_max_ending_here += 1 max_ending_here = new_max else: count_with_max_ending_here = 0 max_ending_here = 0 if max_ending_here &gt; max_so_far: count_with_max = count_with_max_ending_here max_so_far = max_ending_here elif max_ending_here == max_so_far: count_with_max += count_with_max_ending_here return count_with_max for A in [ [0,0,0,1],[0,0,0],[2,0,-2,2] ]: print kadane(A),kadane_count(A) </code></pre>
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