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# Python program to find second largest number in a list

• Difficulty Level : Easy
• Last Updated : 01 Feb, 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

 `# Python program to find second largest` `# number in a list`   `# list of numbers - length of ` `# list should be at least 2` `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))`

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

 `# Python program to find largest number` `# in a list`   `# List of numbers` `list1 ``=` `[``10``, ``20``, ``20``, ``4``, ``45``, ``45``, ``45``, ``99``, ``99``]`   `# Removing duplicates from the list` `list2 ``=` `list``(``set``(list1))`   `# Sorting the  list` `list2.sort()`   `# Printing the second last element` `print``(``"Second largest element is:"``, list2[``-``2``])`

Output

`Second largest element is: 45`

Method 3: By removing the max element from the list

## Python3

 `# Python program to find second largest number` `# in a list`   `# List of numbers` `list1 ``=` `[``10``, ``20``, ``4``, ``45``, ``99``]`   `# new_list is a set of list1` `new_list ``=` `set``(list1)`   `# Removing the largest element from temp list` `new_list.remove(``max``(new_list))`   `# Elements in original list are not changed` `# print(list1)` `print``(``max``(new_list))`

Output

`45`

Method 4: Find the max list element on inputs provided by the user

## Python3

 `# Python program to find second largest` `# number in a list`   `# creating list of integer type` `list1 ``=` `[``10``, ``20``, ``4``, ``45``, ``99``]`   `'''` `# sort the list    ` `list1.sort()` `    `  `# print second maximum element` `print("Second largest element is:", list1[-2])`   `'''`   `# print second maximum element using sorted() method` `print``(``"Second largest element is:"``, ``sorted``(list1)[``-``2``])`

Output

`Second largest element is: 45`

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])`   `    ``# Returning second largest element` `    ``return` `secondLargest`     `# Calling above method over this array set` `print``(findLargest([``10``, ``20``, ``4``, ``45``, ``99``]))`

Output

`45`

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))`

Output

`45`

Method: Using lambda function

## Python3

 `# python code to print second largest element in list`   `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)`

Output

`45`

## Python3

 `lst ``=` `[``10``, ``20``, ``4``, ``45``, ``99``] ` `m``=``max``(lst)` `x``=``[a ``for` `i,a ``in` `enumerate``(lst) ``if` `a

Output

`45`

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):` `    ``# Build a max heap using the elements in the list` `    ``heap ``=` `[(``-``x, x) ``for` `x ``in` `numbers]` `    ``heapq.heapify(heap)` `    `  `    ``# Remove the root element (largest element)` `    ``heapq.heappop(heap)` `    `  `    ``# The new root element is the second largest element` `    ``_, second_largest ``=` `heapq.heappop(heap)` `    `  `    ``return` `second_largest`   `# Test the function` `numbers ``=` `[``10``, ``20``, ``4``, ``45``, ``99``]` `print``(find_second_largest(numbers))  ``# Output: 45`

Output

`45`

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).

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