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October 23, 2022 05:53
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| Try to find max in list: | |
| l = [1,2,3,4,4] | |
| num = max(l, key=lambda x:l.count(x)) # num will be 4 | |
| the you can get the count of num. | |
| l.count(num) # this wil returns 2 | |
| in the other hand as @buddemat said(Here), it is better to: | |
| from collections import Counter | |
| l = [1,2,3,4,4] | |
| c = Counter(l) | |
| c.most_common()[0] | |
| will give you | |
| (4,2) # 4 is a maximum number and 2 is number of occurrence. | |
| First, find the maximum and use count | |
| l1 = [1, 2, 3, 4, 4] | |
| maximum = max(l1) | |
| count = l1.count(maximum) | |
| You can replace the maximum function by yourself. If you don't like to use the built-in function by following the ways | |
| def max(l1): | |
| maximum = l1[0] | |
| for el in l1: | |
| if maximum< el: | |
| maximum = el | |
| return maximum | |
| The above function will check each element of the list iteratively. | |
| def maximum(l1): | |
| if len(l1) == 1: | |
| return l1[0] | |
| if len(l1) == 2: | |
| if l1[0]> l1[1]: | |
| return l1[0] | |
| else: | |
| return l1[1] | |
| else: | |
| max1 = maximum(l1[:len(l1)//2]) | |
| max2 = maximum(l1[len(l1)//2:]) | |
| if max1 > max2: | |
| return max1 | |
| else: | |
| return max2 | |
| The above function is using the divide and conquer approach to find the maximum which time complexity is logn |
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