# Finding third largest element in the list without using len() and sort()

## Question:

Here is the code which I used to find the third largest element in the list without using any built in functions like max,sort,len.

``````list = [12, 45, 2, 41, 31, 10, 8, 6, 4]
#list = [35,10,45,9,8,5]
largest_1 = list
largest_2 = list
largest_3 = list
print (largest_1)
print (largest_2)
print (largest_3)
for each in list:
print ('Each element Before if Loop --->',each)
if each > largest_1:
print ('Each element inside if loop --->',each)
largest_1 = each
print('largest_1 element---->',largest_1)
elif largest_2 != largest_1 and largest_2 < each:
print ('Each element inside if loop --->',each)
largest_2 = each
print ('Largest_1 element is ---->',largest_1)
print ('Largest_2 element is ---->',largest_2)
elif largest_3 != largest_2  and largest_3 < each:
print ('Each element inside if loop --->',each)
largest_3 = each
print ('Largest_2 element is ---->',largest_2)
print ('Largest_3 element is ---->',largest_3)

print (largest_1)
print (largest_2)
print (largest_3)
``````

The same code is not working for the

``````list = [35,10,45,9,8,5]
``````

I am not getting what mistake I have done. How can I fix this?

I don’t know why you wouldn’t want to use `len()`, or `max()` – they’re literally built-in functions, not part of any library, and there’s no practical reason not to use them. That said, if you really want to do it otherwise, here’s another approach:

Take three variables, assign them `largest`, `second_largest`, and `third_largest`, and walk through the list.

``````largest = 0
second_largest = 0
third_largest = 0

for each in list:
if each >= largest:
# assign the new largest, and push the rest of them back down the chain
# we use >= instead of > to ensure that duplicate maximums still work.
#
largest, second_largest, third_largest = each, largest, second_largest
elif each >= second_largest:
second_largest, third_largest = each, second_largest
elif each > third_largest:
third_largest = each
print(third_largest)
``````

You could also store your top 3 max numbers in a dictionary, then print out the third largest one:

``````largest = {"first": 0, "second": 0, "third": 0}

lst = [12, 45, 2, 41, 31, 10, 8, 6, 4]

for number in lst:
if number > largest["first"]:
largest["third"] = largest["second"]
largest["second"] = largest["first"]
largest["first"] = number
elif number > largest["second"]:
largest["third"] = largest["second"]
largest["second"] = number
elif number > largest["third"]:
largest["third"] = number

print(largest)
# {'first': 45, 'second': 41, 'third': 31}

print(largest["third"])
# 31
``````

@venkat: Here another proposal to get the third largest number out of your list without using `len()` and `sort()`.

``````def find_largest(alist):
"""
Find the largest number in a list.

Return the largest number found and it index
"""
largest = alist
for item in alist[1:]:
if item > largest:
largest = item

idx =  alist.index(largest)
return (idx, largest)

#--
def get_third_largest(alist):
"""
Return the third largest number in a list.
"""
# Let make a copy of the input list so that any change to it may not affect the
# original data.
thisList = alist.copy()

index, largest = 0, 0
for item in range(3):
index, largest = find_largest(thisList)
if item != 2:
# delete the first two largest from the List
del thisList[index]
return largest

# Test of the algorithm
if __name__ == "__main__":
List = [12, 45, 2, 41, 31, 10, 8, 6, 4]
third = get_third_largest(List)
# print("Initial list: ", List)
# print("The third largest item in the list:")
print("tExpected: 31")
print("tResult:   %d" % third);

# --- Output---
# Expected: 31
# Result:   31
``````
``````list = [12, 45, 2, 41, 31, 10, 8, 6, 4]
c = 0
for i in list:
c+=1
for j in range(c):
for k in range(j):
if list[j] > list[k]:
list[j], list[k] = list[k],list[j]
else:
list[j],list[k] = list[j],list[k]
print(list," is the third largest element")
``````
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