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

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++

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// C++ program to implement 
// the above approach
#include <bits/stdc++.h>
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[0] = 1;
        compared[1] = 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[1] > compared2[1])
    {
        int 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
    {
        int 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[] 
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[0] + 2, 
                               compared1, 
                               compared1[0]); 
  
    // Print the second largest element 
    cout << compared2[1]; 
}
  
// 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

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Python3

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# 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)

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Output:

1110

Time Complexity: O(N)
Auxiliary Space: O(logN)

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Improved By : divyeshrabadiya07