Majority element | Divide and Conquer | Leetcode 169

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  1. Deep Coding
  2. divide and conquer
  3. leetcode
  4. majority element
  5. leetcode solution
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Комментарии
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def divide_conquer(nums, l, r) :
if l == r:
return (nums[l], 1)
mid = (l + r) // 2
f = divide_conquer(nums, l, mid)
s = divide_conquer(nums, mid+1, r)
## not the same majority substract the extra
if f[0] != s[0] :
if f[1] > s[1] :
return (f[0], f[1]-s[1])
else :
return (s[0], s[1]-f[1])
else :
return (f[0], s[1]+f[1])
# print(divide_conquer(nums))
n = len(nums)
return divide_conquer(nums, 0, n-1)[0]
Here is the code in python based on your explanation and it works. Thanks

rafik
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Thankyou! Your recursion tree is very clear for me understand this algo!

harrybi
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why can't we make a hash map and update the count in the hash map with a key which is coming in each iteration of array !!
and iterate the hash to find the max occiring element?

pravinpoudel
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wtf was that where is code . why you wrote leetcode 169

gursimransingh