Python program to find second largest number in a list
Last Updated :
17 May, 2023
Given a list of numbers, the task is to write a Python program to find the second largest number in the given list.
Examples:
Input: list1 = [10, 20, 4]
Output: 10
Input: list2 = [70, 11, 20, 4, 100]
Output: 70
Method 1: Sorting is an easier but less optimal method. Given below is an O(n) algorithm to do the same.
Python3
list1 = [10, 20, 4, 45, 99]
mx = max(list1[0], list1[1])
secondmax = min(list1[0], list1[1])
n = len(list1)
for i in range(2,n):
if list1[i] > mx:
secondmax = mx
mx = list1[i]
elif list1[i] > secondmax and \
mx != list1[i]:
secondmax = list1[i]
elif mx == secondmax and \
secondmax != list1[i]:
secondmax = list1[i]
print("Second highest number is : ",\
str(secondmax))
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Output
Second highest number is : 45
The time complexity of this code is O(n) as it makes a single linear scan of the list to find the second largest element. The operations performed in each iteration have a constant time complexity, so the overall complexity is O(n).
The space complexity is O(1) as the code only uses a constant amount of extra space.
Method 2: Sort the list in ascending order and print the second last element in the list.
Python3
list1 = [10, 20, 20, 4, 45, 45, 45, 99, 99]
list2 = list(set(list1))
list2.sort()
print("Second largest element is:", list2[-2])
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Output
Second largest element is: 45
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
Method 3: By removing the max element from the list
Python3
list1 = [10, 20, 4, 45, 99]
new_list = set(list1)
new_list.remove(max(new_list))
print(max(new_list))
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Time Complexity: O(n)
Auxiliary Space: O(1)
Method 4: Find the max list element on inputs provided by the user
Python3
list1 = [10, 20, 4, 45, 99]
print("Second largest element is:", sorted(list1)[-2])
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Output
Second largest element is: 45
Time Complexity: O(NlogN) where N is the number of elements in the Array.
Auxiliary Space: O(1)
Method 5: Traverse once to find the largest and then once again to find the second largest.
Python3
def findLargest(arr):
secondLargest = 0
largest = min(arr)
for i in range(len(arr)):
if arr[i] > largest:
secondLargest = largest
largest = arr[i]
else:
secondLargest = max(secondLargest, arr[i])
return secondLargest
print(findLargest([10, 20, 4, 45, 99]))
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Time Complexity: O(N)
Auxiliary Space: O(1)
Method 6: Using list comprehension
Python3
def secondmax(arr):
sublist = [x for x in arr if x < max(arr)]
return max(sublist)
if __name__ == '__main__':
arr = [10, 20, 4, 45, 99]
print(secondmax(arr))
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Method: Using lambda function
Python3
lst = [10, 20, 4, 45, 99]
maximum1 = max(lst)
maximum2 = max(lst, key=lambda x: min(lst)-1 if (x == maximum1) else x)
print(maximum2)
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Method: Using enumerate function
Python3
lst = [10, 20, 4, 45, 99]
m=max(lst)
x=[a for i,a in enumerate(lst) if a<m]
print(max(x))
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Method : Using heap
One approach is to use a heap data structure. A heap is a complete binary tree that satisfies the heap property: the value of each node is at least as great as the values of its children. This property allows us to efficiently find the largest or smallest element in the heap in O(1) time.
To find the second largest element in a list using a heap, we can first build a max heap using the elements in the list. Then, we can remove the root element (which is the largest element in the heap) and find the new root element, which will be the second largest element in the heap.
Here is an example of how this can be done in Python:
Python3
import heapq
def find_second_largest(numbers):
heap = [(-x, x) for x in numbers]
heapq.heapify(heap)
heapq.heappop(heap)
_, second_largest = heapq.heappop(heap)
return second_largest
numbers = [10, 20, 4, 45, 99]
print(find_second_largest(numbers))
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This approach has a time complexity of O(n log n) for building the heap and O(log n) for finding the second largest element, making it more efficient than the methods mentioned in the article which have a time complexity of O(n).
Method : Using numpy.argsort() function.
note: install numpy module using command “pip install numpy”
Here’s the implementation of finding the second largest number in a list using numpy.argsort() function.
Algorithm:
Create a numpy array from the given list.
Use numpy.argsort() function to find the indices that would sort the array.
Find the second last index from the sorted indices.
Return the element at that index from the original array.
Python3
import numpy as np
def find_second_largest(arr):
np_arr = np.array(arr)
sorted_indices = np.argsort(np_arr)
second_last_index = sorted_indices[-2]
return np_arr[second_last_index]
arr = [10, 20, 4, 45, 99]
print(find_second_largest(arr))
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Output:
45
The time complexity of the code is O(nlogn), where n is the number of elements in the list. This is because the numpy.argsort() function has a time complexity of O(nlogn) for sorting the input list, and selecting the second largest element from the sorted list takes constant time.
The auxiliary space of the code is O(n), where n is the number of elements in the list. This is because the numpy.argsort() function creates a sorted copy of the input list and stores it in memory.
Method : Using sorted()
Algorithm :
- The list of integers is initialized with values [2, 1, 8, 7, 3, 0].
- The original list is printed using the print() function and the list variable.
- The sorted() function is used to sort the list in descending order, and the sorted list is assigned to a new variable named list1.
- The sorted list is printed using the print() function and the list1 variable.
- The second largest number is printed using the print() function and the index of the second largest number in the sorted list (which is 1, since list indexes start from 0).
Python3
list = [2, 1, 8, 7, 3, 0]
print('The given list is:', list)
list1 = []
list1 = sorted(list, reverse=True)
print('The sorted list is:', list1)
print('The second largest number in the given list is:', list1[1])
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Output
The given list is: [2, 1, 8, 7, 3, 0]
The sorted list is: [8, 7, 3, 2, 1, 0]
The second largest number in the given list is: 7
Time Complexity : O(n log n)
Auxiliary Space : O(n)
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