This question already has answers here:
Finding all possible combinations of numbers to reach a given sum
(32 answers)
Closed 2 years ago.
I have a number let say 27 and a list of NumbersList. How do I get some selected numbers from NumbersList whose some is 27. And also using the most greater numbers from NumbersList
It can be done by going threw all the possibilities but deep down I know there is a simple solution and I am overthinking it.
#available numbers
NumbersList = [1,5,9,4,6,8,12,12,1,3,6,8,7,8,2]
def GetNumbers(number):
global NumbersList
tmpList = []
#Some magical code
return tmpList
#now result have
result = GetNumbers(27)
# result is [12, 12, 3]
# 12+12+3 using most possible greater numbers from "NumbersList"
You can try
NumbersList = [1,5,9,4,6,8,12,12,1,3,6,8,7,8,2]
def GetNumbers(number):
result = []
for i in sorted(NumbersList, reverse=True):
if sum(result) + i <= number:
result.append(i)
return result
OR if the NumbersList is large then you can reduce iteration by
def GetNumbers(number):
result = []
for i in sorted(NumbersList, reverse=True):
sum_list = sum(result)
if sum_list + i == number:
result.append(i)
return result
elif sum_list + i < number:
result.append(i)
return result
print(GetNumbers(27))
Output
[12, 12, 3]
This is one way to do it
NumbersList = [1,5,9,4,6,8,12,12,1,3,6,8,7,8,2]
def GetNumbers(NumbersList, number):
tmpList = []
summ = 0
srtd = sorted(NumbersList)
for i in range(len(srtd)-1, 0, -1):
if summ + srtd[i] > number:
break
summ += srtd[i]
tmpList.append(srtd[i])
for i in range(len(NumbersList)):
if NumbersList[i] == number - summ:
tmpList.append(NumbersList[i])
if sum(tmpList) == number:
return tmpList
else:
return None
result = GetNumbers(NumbersList, 27)
print(result)
Certainly not the best, but does the work for this one.
First, you sort the list, find all the biggest numbers which does not exceeds the limit. Then you find the missing number.
If there is, return the array.
If there is not, we return None.
This code will not work for all cases, but a good one to start with.
Easyest way to do it, is to sort the list, loop it through backwards, then
Is there room for the next number?
If there is, you can add it to the list.
If there is not, skip.
At the end you should check the sum of yout tmpList. There is a possibility that its sum is != your intended number.
I'm learning Python and the simple ways to handle lists is presented as an advantage. Sometimes it is, but look at this:
>>> numbers = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> numbers.remove(max(numbers))
>>> max(numbers)
74
A very easy, quick way of obtaining the second largest number from a list. Except that the easy list processing helps write a program that runs through the list twice over, to find the largest and then the 2nd largest. It's also destructive - I need two copies of the data if I wanted to keep the original. We need:
>>> numbers = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> if numbers[0]>numbers[1]):
... m, m2 = numbers[0], numbers[1]
... else:
... m, m2 = numbers[1], numbers[0]
...
>>> for x in numbers[2:]:
... if x>m2:
... if x>m:
... m2, m = m, x
... else:
... m2 = x
...
>>> m2
74
Which runs through the list just once, but isn't terse and clear like the previous solution.
So: is there a way, in cases like this, to have both? The clarity of the first version, but the single run through of the second?
You could use the heapq module:
>>> el = [20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7]
>>> import heapq
>>> heapq.nlargest(2, el)
[90.8, 74]
And go from there...
Since #OscarLopez and I have different opinions on what the second largest means, I'll post the code according to my interpretation and in line with the first algorithm provided by the questioner.
def second_largest(numbers):
count = 0
m1 = m2 = float('-inf')
for x in numbers:
count += 1
if x > m2:
if x >= m1:
m1, m2 = x, m1
else:
m2 = x
return m2 if count >= 2 else None
(Note: Negative infinity is used here instead of None since None has different sorting behavior in Python 2 and 3 – see Python - Find second smallest number; a check for the number of elements in numbers makes sure that negative infinity won't be returned when the actual answer is undefined.)
If the maximum occurs multiple times, it may be the second largest as well. Another thing about this approach is that it works correctly if there are less than two elements; then there is no second largest.
Running the same tests:
second_largest([20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7])
=> 74
second_largest([1,1,1,1,1,2])
=> 1
second_largest([2,2,2,2,2,1])
=> 2
second_largest([10,7,10])
=> 10
second_largest([1,1,1,1,1,1])
=> 1
second_largest([1])
=> None
second_largest([])
=> None
Update
I restructured the conditionals to drastically improve performance; almost by a 100% in my testing on random numbers. The reason for this is that in the original version, the elif was always evaluated in the likely event that the next number is not the largest in the list. In other words, for practically every number in the list, two comparisons were made, whereas one comparison mostly suffices – if the number is not larger than the second largest, it's not larger than the largest either.
You could always use sorted
>>> sorted(numbers)[-2]
74
Try the solution below, it's O(n) and it will store and return the second greatest number in the second variable. UPDATE: I've adjusted the code to work with Python 3, because now arithmetic comparisons against None are invalid.
Notice that if all elements in numbers are equal, or if numbers is empty or if it contains a single element, the variable second will end up with a value of None - this is correct, as in those cases there isn't a "second greatest" element.
Beware: this finds the "second maximum" value, if there's more than one value that is "first maximum", they will all be treated as the same maximum - in my definition, in a list such as this: [10, 7, 10] the correct answer is 7.
def second_largest(numbers):
minimum = float('-inf')
first, second = minimum, minimum
for n in numbers:
if n > first:
first, second = n, first
elif first > n > second:
second = n
return second if second != minimum else None
Here are some tests:
second_largest([20, 67, 3, 2.6, 7, 74, 2.8, 90.8, 52.8, 4, 3, 2, 5, 7])
=> 74
second_largest([1, 1, 1, 1, 1, 2])
=> 1
second_largest([2, 2, 2, 2, 2, 1])
=> 1
second_largest([10, 7, 10])
=> 7
second_largest( [1, 3, 10, 16])
=> 10
second_largest([1, 1, 1, 1, 1, 1])
=> None
second_largest([1])
=> None
second_largest([])
=> None
Why to complicate the scenario? Its very simple and straight forward
Convert list to set - removes duplicates
Convert set to list again - which gives list in ascending order
Here is a code
mlist = [2, 3, 6, 6, 5]
mlist = list(set(mlist))
print mlist[-2]
You can find the 2nd largest by any of the following ways:
Option 1:
numbers = set(numbers)
numbers.remove(max(numbers))
max(numbers)
Option 2:
sorted(set(numbers))[-2]
The quickselect algorithm, O(n) cousin to quicksort, will do what you want. Quickselect has average performance O(n). Worst case performance is O(n^2) just like quicksort but that's rare, and modifications to quickselect reduce the worst case performance to O(n).
The idea of quickselect is to use the same pivot, lower, higher idea of quicksort, but to then ignore the lower part and to further order just the higher part.
This is one of the Simple Way
def find_second_largest(arr):
first, second = 0, 0
for number in arr:
if number > first:
second = first
first = number
elif number > second and number < first:
second = number
return second
If you do not mind using numpy (import numpy as np):
np.partition(numbers, -2)[-2]
gives you the 2nd largest element of the list with a guaranteed worst-case O(n) running time.
The partition(a, kth) methods returns an array where the kth element is the same it would be in a sorted array, all elements before are smaller, and all behind are larger.
there are some good answers here for type([]), in case someone needed the same thing on a type({}) here it is,
def secondLargest(D):
def second_largest(L):
if(len(L)<2):
raise Exception("Second_Of_One")
KFL=None #KeyForLargest
KFS=None #KeyForSecondLargest
n = 0
for k in L:
if(KFL == None or k>=L[KFL]):
KFS = KFL
KFL = n
elif(KFS == None or k>=L[KFS]):
KFS = n
n+=1
return (KFS)
KFL=None #KeyForLargest
KFS=None #KeyForSecondLargest
if(len(D)<2):
raise Exception("Second_Of_One")
if(type(D)!=type({})):
if(type(D)==type([])):
return(second_largest(D))
else:
raise Exception("TypeError")
else:
for k in D:
if(KFL == None or D[k]>=D[KFL]):
KFS = KFL
KFL = k
elif(KFS == None or D[k] >= D[KFS]):
KFS = k
return(KFS)
a = {'one':1 , 'two': 2 , 'thirty':30}
b = [30,1,2]
print(a[secondLargest(a)])
print(b[secondLargest(b)])
Just for fun I tried to make it user friendly xD
>>> l = [19, 1, 2, 3, 4, 20, 20]
>>> sorted(set(l))[-2]
19
O(n): Time Complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
To find the second largest number i used the below method to find the largest number first and then search the list if thats in there or not
x = [1,2,3]
A = list(map(int, x))
y = max(A)
k1 = list()
for values in range(len(A)):
if y !=A[values]:
k.append(A[values])
z = max(k1)
print z
Objective: To find the second largest number from input.
Input : 5
2 3 6 6 5
Output: 5
*n = int(raw_input())
arr = map(int, raw_input().split())
print sorted(list(set(arr)))[-2]*
def SecondLargest(x):
largest = max(x[0],x[1])
largest2 = min(x[0],x[1])
for item in x:
if item > largest:
largest2 = largest
largest = item
elif largest2 < item and item < largest:
largest2 = item
return largest2
SecondLargest([20,67,3,2.6,7,74,2.8,90.8,52.8,4,3,2,5,7])
list_nums = [1, 2, 6, 6, 5]
minimum = float('-inf')
max, min = minimum, minimum
for num in list_nums:
if num > max:
max, min = num, max
elif max > num > min:
min = num
print(min if min != minimum else None)
Output
5
Initialize with -inf. This code generalizes for all cases to find the second largest element.
max1= float("-inf")
max2=max1
for item in arr:
if max1<item:
max2,max1=max1,item
elif item>max2 and item!=max1:
max2=item
print(max2)
Using reduce from functools should be a linear-time functional-style alternative:
from functools import reduce
def update_largest_two(largest_two, x):
m1, m2 = largest_two
return (m1, m2) if m2 >= x else (m1, x) if m1 >= x else (x, m1)
def second_largest(numbers):
if len(numbers) < 2:
return None
largest_two = sorted(numbers[:2], reverse=True)
rest = numbers[2:]
m1, m2 = reduce(update_largest_two, rest, largest_two)
return m2
... or in a very concise style:
from functools import reduce
def second_largest(n):
update_largest_two = lambda a, x: a if a[1] >= x else (a[0], x) if a[0] >= x else (x, a[0])
return None if len(n) < 2 else (reduce(update_largest_two, n[2:], sorted(n[:2], reverse=True)))[1]
This can be done in [N + log(N) - 2] time, which is slightly better than the loose upper bound of 2N (which can be thought of O(N) too).
The trick is to use binary recursive calls and "tennis tournament" algorithm. The winner (the largest number) will emerge after all the 'matches' (takes N-1 time), but if we record the 'players' of all the matches, and among them, group all the players that the winner has beaten, the second largest number will be the largest number in this group, i.e. the 'losers' group.
The size of this 'losers' group is log(N), and again, we can revoke the binary recursive calls to find the largest among the losers, which will take [log(N) - 1] time. Actually, we can just linearly scan the losers group to get the answer too, the time budget is the same.
Below is a sample python code:
def largest(L):
global paris
if len(L) == 1:
return L[0]
else:
left = largest(L[:len(L)//2])
right = largest(L[len(L)//2:])
pairs.append((left, right))
return max(left, right)
def second_largest(L):
global pairs
biggest = largest(L)
second_L = [min(item) for item in pairs if biggest in item]
return biggest, largest(second_L)
if __name__ == "__main__":
pairs = []
# test array
L = [2,-2,10,5,4,3,1,2,90,-98,53,45,23,56,432]
if len(L) == 0:
first, second = None, None
elif len(L) == 1:
first, second = L[0], None
else:
first, second = second_largest(L)
print('The largest number is: ' + str(first))
print('The 2nd largest number is: ' + str(second))
You can also try this:
>>> list=[20, 20, 19, 4, 3, 2, 1,100,200,100]
>>> sorted(set(list), key=int, reverse=True)[1]
100
A simple way :
n=int(input())
arr = set(map(int, input().split()))
arr.remove(max(arr))
print (max(arr))
use defalut sort() method to get second largest number in the list.
sort is in built method you do not need to import module for this.
lis = [11,52,63,85,14]
lis.sort()
print(lis[len(lis)-2])
Just to make the accepted answer more general, the following is the extension to get the kth largest value:
def kth_largest(numbers, k):
largest_ladder = [float('-inf')] * k
count = 0
for x in numbers:
count += 1
ladder_pos = 1
for v in largest_ladder:
if x > v:
ladder_pos += 1
else:
break
if ladder_pos > 1:
largest_ladder = largest_ladder[1:ladder_pos] + [x] + largest_ladder[ladder_pos:]
return largest_ladder[0] if count >= k else None
def secondlarget(passinput):
passinputMax = max(passinput) #find the maximum element from the array
newpassinput = [i for i in passinput if i != passinputMax] #Find the second largest element in the array
#print (newpassinput)
if len(newpassinput) > 0:
return max(newpassinput) #return the second largest
return 0
if __name__ == '__main__':
n = int(input().strip()) # lets say 5
passinput = list(map(int, input().rstrip().split())) # 1 2 2 3 3
result = secondlarget(passinput) #2
print (result) #2
if __name__ == '__main__':
n = int(input())
arr = list(map(float, input().split()))
high = max(arr)
secondhigh = min(arr)
for x in arr:
if x < high and x > secondhigh:
secondhigh = x
print(secondhigh)
The above code is when we are setting the elements value in the list
as per user requirements. And below code is as per the question asked
#list
numbers = [20, 67, 3 ,2.6, 7, 74, 2.8, 90.8, 52.8, 4, 3, 2, 5, 7]
#find the highest element in the list
high = max(numbers)
#find the lowest element in the list
secondhigh = min(numbers)
for x in numbers:
'''
find the second highest element in the list,
it works even when there are duplicates highest element in the list.
It runs through the entire list finding the next lowest element
which is less then highest element but greater than lowest element in
the list set initially. And assign that value to secondhigh variable, so
now this variable will have next lowest element in the list. And by end
of loop it will have the second highest element in the list
'''
if (x<high and x>secondhigh):
secondhigh=x
print(secondhigh)
Max out the value by comparing each one to the max_item. In the first if, every time the value of max_item changes it gives its previous value to second_max. To tightly couple the two second if ensures the boundary
def secondmax(self, list):
max_item = list[0]
second_max = list[1]
for item in list:
if item > max_item:
second_max = max_item
max_item = item
if max_item < second_max:
max_item = second_max
return second_max
you have to compare in between new values, that's the trick, think always in the previous (the 2nd largest) should be between the max and the previous max before, that's the one!!!!
def secondLargest(lista):
max_number = 0
prev_number = 0
for i in range(0, len(lista)):
if lista[i] > max_number:
prev_number = max_number
max_number = lista[i]
elif lista[i] > prev_number and lista[i] < max_number:
prev_number = lista[i]
return prev_number
Most of previous answers are correct but here is another way !
Our strategy is to create a loop with two variables first_highest and second_highest. We loop through the numbers and if our current_value is greater than the first_highest then we set second_highest to be the same as first_highest and then the second_highest to be the current number. If our current number is greater than second_highest then we set second_highest to the same as current number
#!/usr/bin/env python3
import sys
def find_second_highest(numbers):
min_integer = -sys.maxsize -1
first_highest= second_highest = min_integer
for current_number in numbers:
if current_number == first_highest and min_integer != second_highest:
first_highest=current_number
elif current_number > first_highest:
second_highest = first_highest
first_highest = current_number
elif current_number > second_highest:
second_highest = current_number
return second_highest
print(find_second_highest([80,90,100]))
print(find_second_highest([80,80]))
print(find_second_highest([2,3,6,6,5]))
Best solution that my friend Dhanush Kumar came up with:
def second_max(loop):
glo_max = loop[0]
sec_max = float("-inf")
for i in loop:
if i > glo_max:
sec_max = glo_max
glo_max=i
elif sec_max < i < glo_max:
sec_max = i
return sec_max
#print(second_max([-1,-3,-4,-5,-7]))
assert second_max([-1,-3,-4,-5,-7])==-3
assert second_max([5,3,5,1,2]) == 3
assert second_max([1,2,3,4,5,7]) ==5
assert second_max([-3,1,2,5,-2,3,4]) == 4
assert second_max([-3,-2,5,-1,0]) == 0
assert second_max([0,0,0,1,0]) == 0
Below code will find the max and the second max numbers without the use of max function. I assume that the input will be numeric and the numbers are separated by single space.
myList = input().split()
myList = list(map(eval,myList))
m1 = myList[0]
m2 = myList[0]
for x in myList:
if x > m1:
m2 = m1
m1 = x
elif x > m2:
m2 = x
print ('Max Number: ',m1)
print ('2nd Max Number: ',m2)
Here I tried to come up with an answer.
2nd(Second) maximum element in a list using single loop and without using any inbuilt function.
def secondLargest(lst):
mx = 0
num = 0
sec = 0
for i in lst:
if i > mx:
sec = mx
mx = i
else:
if i > num and num >= sec:
sec = i
num = i
return sec
working on CheckIO exercises but stuck here. I need to design a function that'll find the sum of the elements with even indexes (0th, 2nd, 4th...) then multiply this summed number and the final element of the array together. The input is an array, the output is a number. Oh, and for an empty array, the result has to be zero.
def checkio(array):
sum = 0
if len(array) == 0:
return 0
else:
for i in array:
if array.index(i) % 2 == 0:
sum = sum + i
final = sum*(array[len(array)-1])
return final
for instance, with the array [-37,-36,-19,-99,29,20,3,-7,-64,84,36,62,26,-76,55,-24,84,49,-65,41], this function is returning -1476 when it should be giving out 1968.
As far as I can see the issue is that you are assuming all numbers in the array are unique. For example lets say I have the following array:
[0,33,33,22,22]
obviously in this array you need the 3nd and 5th elements (index 2 and 4).
with your current code however this will never happen and you will end up with a sum of 0. This is because the code:
array.index(i)
finds the first element that matches i, this would be the 2nd and 4th elements (index 1 and 3), which are odd indexes, and thus will not be added.
You can use List Comprehension. Like:
sum([i for i in L[::2]])*L[-1]
In your code array.index(i) is problem . So you can use for finding elements by using array[::2]
You can try with your code:
def checkio(array):
sum = 0
if len(array) == 0:
return 0
else:
for i in array[::2]:
sum = sum + i
final = sum*array[-1]
return final
Example:
L = [-37,-36,-19,-99,29,20,3,-7,-64,84,36,62,26,-76,55,-24,84,49,-65,41]
Output:
1968
Here is a working program I made.
def checkio(array):
listSum = 0
if array:
for i in range(0, len(array), 2):
listSum += array[i]
finalValue = listSum * array[-1]
return finalValue
else:
return 0
First, it checks to see if the array has any values. We can do that like this: if array:. If the array is empty, it will return 0 like you wanted.
Now, this is what checks every other element in your array: range(0, len(array), 2): This means that the value of i will start at 0, continue for the length of the array, and count by twos.
The sums are added here: listSum += array[i]. This takes the variable listSum and adds the value of the number at the index i in the array to it. the += operator is shorthand for listSum = listSum + array[i].
The last part of the function, takes the listSum variable, and multiplies it by array[-1] which gets the last value in the array and finnaly returns it.
When I ran your example array above, it returned 1968 as it should.
I think it is in order to to some beginner's explanation even if I'm repeating what other answers already said:
As for why your code does not work, let's change the original a little:
def checkio(array):
sum = 0
if len(array) == 0:
return 0
else:
for i in array:
print "pos of %s = %i" % (i, array.index(i))
if array.index(i) % 2 == 0:
sum = sum + i
final = sum*(array[len(array)-1])
return final
This yields
pos of -37 = 0
pos of -36 = 1
pos of -19 = 2
pos of -99 = 3
[...]
pos of 84 = 9
[...]
pos of 84 = 9
There's your problem, because index() yields the index of the first occurrence of an element and 84 appears twice. Your code only works when the elements in the array are unique, which they are not.
So, when not going for the all out python swagger using slicing:
def checkio(array):
# sum is a built-in, don't override it
result = 0
# "if len(array) != 0" is the same as "if array"
if array:
# enumerate is nice, but not really needed, see below
for i, x in enumerate(array):
# i is the index, x is the value
if i % 2 == 0:
# += is also nice
result += x
result *= array[-1]
return result
As for a more pythonic solution, you can do a lot with slicing.
array[::2]
is every second element of array and
array[-1]
is the last element. Hence
s = sum(array[::2]) * array[-1]
That does not handle an empty array, thus
# if there are no elements or only the last element, the sum is zero
if len(array) == 0:
return 0
else:
return sum(array[::2]) * array[-1]
or even
return sum(array[::2]) * array[-1] if array else 0
Which is python's equivalent of the ternary operator.
Just use enumerate to interate for each element with an index and keep track of the last element.
def checkio(array):
sum = 0
last_element = 0
for index, element in enumerate(array):
if index % 2 == 0:
sum += element
last_element = element
return last_element * sum