# Find Second largest element in an array | Set 2

Given an array arr[] consisting of N integers, the task is to find the second largest element in the given array using N+log2(N) – 2 comparisons.

Examples:

Input : arr[] = {22, 33, 14, 55, 100, 12}
Output : 55

Input : arr[] = {35, 23, 12, 35, 19, 100}
Output : 35

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Sorting and Two-Traversal Approach: Refer to Find Second largest element in an array for the solutions using Sorting as well as traversing the array twice.

Efficient Approach:
Follow the steps below to solve the problem:

• Find the largest element from the given array and keep track of all the elements compared with the largest element.
• Split the current array into two equal length subarrays.
• For each subarray, recursively find the largest element, and return an array in which the first index contains the length of this array, second index element contains the largest element and the remaining array contains the elements with which the largest element has been compared.
• Now, from the two arrays returned by both the subarrays, compare the largest element of both the subarrays and return the array that contains the largest of the two.
• The final array returned by the findLargest() contains its size in the first index, the largest element of the array at the second index and the elements compared with the largest element in the remaining indices. Repeat the above steps using the findLargest() to find the second largest element of the array from the list of compared elements.

Analysis Of Algorithm:
It is clearly visible from the algorithm that the time complexity of the findLargest() algorithm is O(N) [N: size of the array]
Hence, (N-1) comparisons are performed.
Now, the size of the array returned by findLargest() is log2(N) + 2, out of which log2(N) elements are the ones with which the largest element is compared.
Hence, to find the second largest element, the largest among these log2(N) elements is calculated using log2(N) – 1 comparisons.

Therefore, the total number of comparisons = N – 1 + log2(N) – 1 = N + log2(N) – 2

Below is the implementation of the above approach:

## C++

 `// C++ program to implement  ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the largest  ` `// element in the array arr[]  ` `vector<``int``> findLargest(``int` `beg, ``int` `end, ` `                 ``vector<``int``> arr, ``int` `n) ` `{ ` `     `  `    ``// Base Condition  ` `    ``if` `(beg == end) ` `    ``{ ` `         `  `        ``// Initialize an empty list  ` `        ``vector<``int``> compared(n, 0); ` `        ``compared = 1; ` `        ``compared = arr[beg];  ` `        ``return` `compared;  ` `    ``} ` ` `  `    ``// Divide the array into two equal  ` `    ``// length subarrays and recursively  ` `    ``// find the largest among the two  ` `    ``vector<``int``> compared1 = findLargest( ` `                            ``beg, (beg + end) / 2, ` `                            ``arr, n); ` ` `  `    ``vector<``int``> compared2 = findLargest( ` `                            ``(beg + end) / 2 + 1,  ` `                             ``end, arr, n); ` ` `  `    ``if` `(compared1 > compared2) ` `    ``{ ` `        ``int` `k = compared1 + 1; ` ` `  `        ``// Store length of compared1[]  ` `        ``// in the first index  ` `        ``compared1 = k;  ` ` `  `        ``// Store the maximum element  ` `        ``compared1[k] = compared2;  ` ` `  `        ``// Return compared1 which  ` `        ``// contains the maximum element  ` `        ``return` `compared1;  ` `    ``} ` `    ``else` `    ``{ ` `        ``int` `k = compared2 + 1; ` ` `  `        ``// Store length of compared2[]  ` `        ``// in the first index  ` `        ``compared2 = k;  ` ` `  `        ``// Store the maximum element  ` `        ``compared2[k] = compared1;  ` ` `  `        ``// Return compared2[] which  ` `        ``// contains the maximum element  ` `        ``return` `compared2; ` `    ``} ` `} ` `         `  `// Function to print the second largest  ` `// element in the array arr[]  ` `void` `findSecondLargest(``int` `end, vector<``int``> arr) ` `{ ` ` `  `    ``// Find the largest element in arr[]  ` `    ``vector<``int``> compared1 = findLargest( ` `                            ``0, end - 1, arr, end);  ` ` `  `    ``// Find the second largest element  ` `    ``// in arr[]  ` `    ``vector<``int``> compared2 = findLargest( ` `                            ``2, compared1 + 2,  ` `                               ``compared1,  ` `                               ``compared1);  ` ` `  `    ``// Print the second largest element  ` `    ``cout << compared2;  ` `} ` ` `  `// Driver code     ` `int` `main() ` `{ ` `    ``int` `N = 10; ` ` `  `    ``vector<``int``> arr{ 20, 1990, 12, 1110, 1, ` `                     ``59, 12, 15, 120, 1110};  ` ` `  `    ``findSecondLargest(N, arr); ` ` `  `    ``return` `0; ` `} ` ` `  `// This code is contributed by divyeshrabadiya07 `

## Python3

 `# Python3 Program to implement ` `# the above approach ` ` `  `# Function to find the largest ` `#  element in the array arr[] ` `def` `findLargest(beg, end, arr, n): ` ` `  `    ``# Base Condition ` `    ``if``(beg ``=``=` `end): ` ` `  `        ``# Initialize an empty list ` `        ``compared ``=` `[``0``]``*``n ` `        ``compared[``0``] ``=` `1` `        ``compared[``1``] ``=` `arr[beg] ` `        ``return` `compared ` ` `  `    ``# Divide the array into two equal ` `    ``# length subarrays and recursively ` `    ``# find the largest among the two ` `    ``compared1 ``=` `findLargest(beg, (beg``+``end)``/``/``2``, ` `                            ``arr, n) ` ` `  `    ``compared2 ``=` `findLargest((beg``+``end)``/``/``2``+``1``, end, ` `                            ``arr, n) ` ` `  `    ``if``(compared1[``1``] > compared2[``1``]): ` `        ``k ``=` `compared1[``0``]``+``1` ` `  `    ``# Store length of compared1[] ` `    ``# in the first index ` `        ``compared1[``0``] ``=` `k ` ` `  `    ``# Store the maximum element ` `        ``compared1[k] ``=` `compared2[``1``] ` ` `  `        ``# Return compared1 which ` `        ``# contains the maximum element ` `        ``return` `compared1 ` ` `  `    ``else``: ` `        ``k ``=` `compared2[``0``]``+``1` ` `  `    ``# Store length of compared2[] ` `    ``# in the first index ` `        ``compared2[``0``] ``=` `k ` ` `  `    ``# Store the maximum element ` `        ``compared2[k] ``=` `compared1[``1``] ` ` `  `        ``# Return compared2[] which ` `        ``# contains the maximum element ` `        ``return` `compared2 ` ` `  ` `  `# Function to print the second largest ` `# element in the array arr[] ` `def` `findSecondLargest(end, arr): ` ` `  `    ``# Find the largest element in arr[] ` `    ``compared1 ``=` `findLargest(``0``, end``-``1``, arr, end) ` ` `  `    ``# Find the second largest element ` `    ``# in arr[] ` `    ``compared2 ``=` `findLargest(``2``, compared1[``0``]``+``2``, ` `                            ``compared1, ` `                            ``compared1[``0``]) ` ` `  `    ``# Print the second largest element ` `    ``print``(compared2[``1``]) ` ` `  ` `  `# Driver Code ` `N ``=` `10` ` `  `arr ``=` `[``20``, ``1990``, ``12``, ``1110``, ``1``, ``59``, ``12``, ``15``, ``120``, ``1110``] ` ` `  `findSecondLargest(N, arr) `

Output:

```1110
```

Time Complexity: O(N)
Auxiliary Space: O(logN) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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