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O(nlogn) majority element algorithm
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def majority_element(a): | |
n = len(a) | |
if n == 1: | |
return a[0] | |
if n == 0: | |
return None | |
k = n/2 | |
elem1 = majority_element(a[:k]) | |
elem2 = majority_element(a[k+1:]) | |
if elem1 == elem2: | |
return elem1 | |
count1 = a.count(elem1) | |
count2 = a.count(elem2) | |
if count1 > k: | |
return elem1 | |
elif count2 > k: | |
return elem2 | |
else: | |
return None | |
print "Should be 1:", majority_element([0,1,0,1,1]) | |
print "Should be 0:", majority_element([0,0]) | |
print "Should be None:", majority_element([0,1,1,1,0,0,2]) | |
print "Should be 0:", majority_element([0,1,1,1,0,0,0,0,2]) | |
print "Should be None:", majority_element([0,0,1,1]) | |
print "Should be None:", majority_element([0,0,1,1,2,2,1,1,2]) | |
print "Should be 2:", majority_element([0,2,2,2,2,2,0,1,1]) |
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