35 lines
1.3 KiB
Ruby
35 lines
1.3 KiB
Ruby
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#Given an array of integers(both positive and negative),find the subarray with maximum product.
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# Time Complexity: O(n),Auxiliary Space: O(1)
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#Note: It doesn’t work for arrays like {0, 0, -20, 0}, {0, 0, 0}.. etc
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def max_product(a)
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len=a.length
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max_ending_here=min_ending_here=max_so_far=1
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for i in 0...len
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# If this element is +ve, update max_ending_here.
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# Update minEndingHere only if minEndingHere is negative
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if a[i]>0
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max_ending_here=max_ending_here*a[i]
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min_ending_here = [min_ending_here*a[i],1].min
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# If this element is 0, then the maximum product
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# cannot end here, make both maxEndingHere and minEndingHere 0
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# Assumption: Output is alway greater than or equal to 1.
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elsif a[i]==0
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max_ending_here=min_ending_here=1
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# If this element is -ve,min_ending here is previous max_ending_here*current
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#max_ending_here is min of previous min_ending_here*current and 1
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else
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temp=max_ending_here
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max_ending_here=[min_ending_here*a[i],1].max
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min_ending_here=temp*a[i]
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end
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max_so_far=max_ending_here if max_so_far<max_ending_here
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end
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return max_so_far
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end
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max_product([-2, -3, 0, -2, -40]) # => 80
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