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# Given an array where all elements appear even number of times except one.
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# All repeating occurrences of elements appear in pairs and these pairs are not adjacent (there cannot be more than two consecutive occurrences of any element).
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# Find the element that appears odd number of times.
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# Note that input like {2, 2, 1, 2, 2, 1, 1} is valid as all repeating occurrences occur in pairs and these pairs are not adjacent.
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# Input like {2, 1, 2} is invalid as repeating elements don’t appear in pairs. Also, input like {1, 2, 2, 2, 2} is invalid as two pairs of 2 are adjacent.
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# Input like {2, 2, 2, 1} is also invalid as there are three consecutive occurrences of 2.
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#Time-complexity: O(logn),Auxiliary-space:O(1)
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#Algorithm:Binary-search
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def search_odd(a)
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lo=0
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hi=a.length-1
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while hi-lo>1
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mid=lo+(hi-lo)/2
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if mid%2==0
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if a[mid]==a[mid+1]
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lo=mid+2
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else
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hi=mid
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end
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else
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if a[mid]==a[mid-1]
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lo=mid+1
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else
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hi=mid
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end
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end
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end
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return a[lo]
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end
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search_odd([3,1, 1, 2, 2, 3, 3, 4, 4,600, 600, 4, 4]) # => [3]
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