Find the largest three distinct elements in an array

Given an array with all distinct elements, find the largest three elements. Expected time complexity is O(n) and extra space is O(1). 

Examples :

Input: arr[] = {10, 4, 3, 50, 23, 90}
Output: 90, 50, 23

Method 1:

Algorithm:

1) Initialize the largest three elements as minus infinite.
    first = second = third = -∞

2) Iterate through all elements of array.
   a) Let current array element be x.
   b) If (x > first)
      {
          // This order of assignment is important
          third = second
          second = first
          first = x   
       }
   c)  Else if (x > second and x != first)
      {
          third = second
          second = x 
      }
   d)  Else if (x > third and x != second)
      {
          third = x  
      }

3) Print first, second and third.

Below is the implementation of the above algorithm.

C++

// C++ program for find the largest
// three elements in an array
#include <bits/stdc++.h>
using namespace std;

// Function to print three largest elements
void print3largest(int arr[], int arr_size)
{
    int first, second, third;

    // There should be atleast three elements
    if (arr_size < 3)
    {
        cout << " Invalid Input ";
        return;
    }

    third = first = second = INT_MIN;
    for(int i = 0; i < arr_size; i++)
    {

        // If current element is
        // greater than first
        if (arr[i] > first)
        {
            third = second;
            second = first;
            first = arr[i];
        }

        // If arr[i] is in between first
        // and second then update second
        else if (arr[i] > second && arr[i] != first)
        {
            third = second;
            second = arr[i];
        }

        else if (arr[i] > third && arr[i] != second)
            third = arr[i];
    }

    cout << "Three largest elements are "
        << first << " " << second << " "
        << third << endl;
}

// Driver code
int main()
{
    int arr[] = { 12, 13, 1, 10, 34, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    print3largest(arr, n);
    return 0;
}

// This code is contributed by Anjali_Chauhan

C

// C program for find the largest
// three elements in an array
#include <limits.h> /* For INT_MIN */
#include <stdio.h>

/* Function to print three largest elements */
void print3largest(int arr[], int arr_size)
{
    int i, first, second, third;

    /* There should be atleast three elements */
    if (arr_size < 3) {
        printf(" Invalid Input ");
        return;
    }

    third = first = second = INT_MIN;
    for (i = 0; i < arr_size; i++) {
        /* If current element is greater than first*/
        if (arr[i] > first) {
            third = second;
            second = first;
            first = arr[i];
        }

        /* If arr[i] is in between first and second then update second  */
        else if (arr[i] > second) {
            third = second;
            second = arr[i];
        }

        else if (arr[i] > third)
            third = arr[i];
    }

    printf("Three largest elements are %d %d %d\n", first, second, third);
}

/* Driver program to test above function */
int main()
{
    int arr[] = { 12, 13, 1, 10, 34, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    print3largest(arr, n);
    return 0;
}
/*This code is edited by Ayush Singla(@ayusin51)*/

Java

// Java code to find largest three elements
// in an array

class PrintLargest {
    /* Function to print three largest elements */
    static void print3largest(int arr[], int arr_size)
    {
        int i, first, second, third;

        /* There should be atleast three elements */
        if (arr_size < 3) {
            System.out.print(" Invalid Input ");
            return;
        }

        third = first = second = Integer.MIN_VALUE;
        for (i = 0; i < arr_size; i++) {
            /* If current element is greater than
            first*/
            if (arr[i] > first) {
                third = second;
                second = first;
                first = arr[i];
            }

            /* If arr[i] is in between first and
            second then update second  */
            else if (arr[i] > second) {
                third = second;
                second = arr[i];
            }

            else if (arr[i] > third)
                third = arr[i];
        }

        System.out.println("Three largest elements are " + first + " " + second + " " + third);
    }

    /* Driver program to test above function*/
    public static void main(String[] args)
    {
        int arr[] = { 12, 13, 1, 10, 34, 1 };
        int n = arr.length;
        print3largest(arr, n);
    }
}
/*This code is contributed by Prakriti Gupta
and edited by Ayush Singla(@ayusin51)*/

Python3

# Python3 code to find largest three
# elements in an array
import sys

# Function to print three largest
# elements
def print3largest(arr, arr_size):

    # There should be atleast three
    # elements
    if (arr_size < 3):

        print(" Invalid Input ")
        return

    third = first = second = -sys.maxsize

    for i in range(0, arr_size):

        # If current element is greater
        # than first
        if (arr[i] > first):

            third = second
            second = first
            first = arr[i]


        # If arr[i] is in between first
        # and second then update second
        elif (arr[i] > second):

            third = second
            second = arr[i]

        elif (arr[i] > third):
            third = arr[i]

    print("Three largest elements are",
                  first, second, third)

# Driver program to test above function
arr = [12, 13, 1, 10, 34, 1]
n = len(arr)
print3largest(arr, n)

# This code is contributed by Smitha Dinesh Semwal
# and edited by Ayush Singla(@ayusin51).

C#

// C# code to find largest
// three elements in an array
using System;

class PrintLargest {

    // Function to print three
    // largest elements
    static void print3largest(int[] arr,
                              int arr_size)
    {
        int i, first, second, third;

        // There should be atleast three elements
        if (arr_size < 3) {
            Console.WriteLine("Invalid Input");
            return;
        }

        third = first = second = 000;
        for (i = 0; i < arr_size; i++) {
            // If current element is
            // greater than first
            if (arr[i] > first) {
                third = second;
                second = first;
                first = arr[i];
            }

            // If arr[i] is in between first
            // and second then update second
            else if (arr[i] > second && arr[i] != first) {
                third = second;
                second = arr[i];
            }

            else if (arr[i] > third && arr[i]!=second)
                third = arr[i];
        }

        Console.WriteLine("Three largest elements are " + first + " " + second + " " + third);
    }

    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 12, 13, 1, 10, 34, 1 };
        int n = arr.Length;
        print3largest(arr, n);
    }
}

// This code is contributed by KRV and edited by Ayush Singla(@ayusin51).

PHP

<?php
// PHP code to find largest
// three elements in an array

// Function to print
// three largest elements
function print3largest($arr, $arr_size)
{
    $i; $first; $second; $third;

    // There should be atleast
    // three elements
    if ($arr_size < 3)
    {
        echo " Invalid Input ";
        return;
    }

    $third = $first = $second = PHP_INT_MIN;
    for ($i = 0; $i < $arr_size ; $i ++)
    {
        // If current element is
        // greater than first
        if ($arr[$i] > $first)
        {
            $third = $second;
            $second = $first;
            $first = $arr[$i];
        }

        // If arr[i] is in between first
        // and second then update second
        else if ($arr[$i] > $second)
        {
            $third = $second;
            $second = $arr[$i];
        }

        else if ($arr[$i] > $third)
            $third = $arr[$i];
    }

    echo "Three largest elements are ",
       $first, " ", $second, " ", $third;
}


// Driver Code
$arr = array(12, 13, 1,
             10, 34, 1);
$n = count($arr);
print3largest($arr, $n);

// This code is contributed by anuj_67 and edited by Ayush Singla(@ayusin51).
?>

Javascript

<script>

// Javascript program for find the largest
// three elements in an array

// Function to print three largest elements
function print3largest(arr, arr_size)
{
    let first, second, third;

    // There should be atleast three elements
    if (arr_size < 3)
    {
        document.write(" Invalid Input ");
        return;
    }

    third = first = second = Number.MIN_VALUE;
    for(let i = 0; i < arr_size; i++)
    {

        // If current element is
        // greater than first
        if (arr[i] > first)
        {
            third = second;
            second = first;
            first = arr[i];
        }

        // If arr[i] is in between first
        // and second then update second
        else if (arr[i] > second)
        {
            third = second;
            second = arr[i];
        }

        else if (arr[i] > third)
            third = arr[i];
    }

    document.write("Three largest elements are "
        + first + " " + second + " "
        + third + "<br>");
}

// Driver code
    let arr = [ 12, 13, 1, 10, 34, 1 ];
    let n = arr.length;

    print3largest(arr, n);

// This is code is contributed by Mayank Tyagi

</script>

Output

Three largest elements are 34 13 12

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 2:

An efficient way to solve this problem is to use any O(nLogn) sorting algorithm & simply returning the last 3 largest elements.

C++

// C++ code to find largest three elements in an array
#include <bits/stdc++.h>
using namespace std;

void find3largest(int arr[], int n)
{
    sort(arr, arr + n); // It uses Tuned Quicksort with
    // avg. case Time complexity = O(nLogn)

    int check = 0, count = 1;
    for (int i = 1; i <= n; i++) {
        if (count < 4) {
            if (check != arr[n - i]) {
                // to handle duplicate values
                cout << arr[n - i] << " ";
                check = arr[n - i];
                count++;
            }
        }
        else
            break;
    }
}

// Driver code
int main()
{
    int arr[] = { 12, 45, 1, -1, 45, 54, 23, 5, 0, -10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    find3largest(arr, n);
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C code to find largest three elements in an array
#include <stdio.h>
#include <stdlib.h>

// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}

void find3largest(int arr[], int n)
{
    qsort(arr, n, sizeof(int),
          cmpfunc); // It uses Tuned Quicksort with
    // avg. case Time complexity = O(nLogn)

    int check = 0, count = 1;
    for (int i = 1; i <= n; i++) {
        if (count < 4) {
            if (check != arr[n - i]) {
                // to handle duplicate values
                printf("%d ", arr[n - i]);
                check = arr[n - i];
                count++;
            }
        }
        else
            break;
    }
}

// Driver code
int main()
{
    int arr[] = { 12, 45, 1, -1, 45, 54, 23, 5, 0, -10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    find3largest(arr, n);
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java code to find largest
// three elements in an array

import java.io.*;
import java.util.Arrays;

class GFG {
    void find3largest(int[] arr)
    {
        Arrays.sort(arr); // It uses Tuned Quicksort with
        // avg. case Time complexity = O(nLogn)
        int n = arr.length;
        int check = 0, count = 1;

        for (int i = 1; i <= n; i++) {

            if (count < 4) {
                if (check != arr[n - i]) {
                    // to handle duplicate values
                    System.out.print(arr[n - i] + " ");
                    check = arr[n - i];
                    count++;
                }
            }
            else
                break;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        GFG obj = new GFG();
        int[] arr = { 12, 45, 1, -1, 45, 54, 23, 5, 0, -10 };
        obj.find3largest(arr);
    }
}
// This code is contributed by Prashant Malik

Python3

# Python3 code to find largest
# three elements in an array
def find3largest(arr, n):
    arr = sorted(arr) # It uses Tuned Quicksort with
                      # avg. case Time complexity = O(nLogn)

    check = 0
    count = 1

    for i in range(1, n + 1):

        if(count < 4):
            if(check != arr[n - i]):

                # to handle duplicate values
                print(arr[n - i], end = " ")
                check = arr[n - i]
                count += 1
        else:
            break

# Driver code
arr = [12, 45, 1, -1, 45,
       54, 23, 5, 0, -10]
n = len(arr)
find3largest(arr, n)

# This code is contributed by mohit kumar

C#

// C# code to find largest
// three elements in an array
using System;

class GFG {
    void find3largest(int[] arr)
    {
        Array.Sort(arr); // It uses Tuned Quicksort with
        // avg. case Time complexity = O(nLogn)
        int n = arr.Length;
        int check = 0, count = 1;

        for (int i = 1; i <= n; i++) {
            if (count < 4) {
                if (check != arr[n - i]) {
                    // to handle duplicate values
                    Console.Write(arr[n - i] + " ");
                    check = arr[n - i];
                    count++;
                }
            }
            else
                break;
        }
    }

    // Driver code
    public static void Main()
    {
        GFG obj = new GFG();
        int[] arr = { 12, 45, 1, -1, 45,
                      54, 23, 5, 0, -10 };
        obj.find3largest(arr);
    }
}

// This code is contributed by Code_Mech

Javascript

<script>

// JavaScript code to find largest
// three elements in an array

     function find3largest(arr) {
        arr.sort((a,b)=>a-b); // It uses Tuned Quicksort with
        // avg. case Time complexity = O(nLogn)

        let check = 0, count = 1;

        for (let i = 1; i <= arr.length; i++) {

            if (count < 4) {
                if (check != arr[arr.length - i]) {
                    // to handle duplicate values
                    document.write(arr[arr.length - i] + " ");
                    check = arr[arr.length - i];
                    count++;
                }
            }
            else
                break;
        }
    }

    // Driver code

    let arr = [ 12, 45, 1, -1, 45, 54, 23, 5, 0, -10 ];
    find3largest(arr);

// This code is contributed by Surbhi Tyagi

</script>

Output

54 45 23 

Time Complexity: O(n log n)
Auxiliary Space: O(1)

Method 3:
We can use Partial Sort of C++ STL. partial_sort uses Heapselect, which provides better performance than Quickselect for small M. As a side effect, the end state of Heapselect leaves you with a heap, which means that you get the first half of the Heapsort algorithm “for free”. The complexity is “approximately” O(N log(M)), where M is distance(middle-first).

C++

#include <bits/stdc++.h>
using namespace std;
int main()
{
    vector<int> V = { 11, 65, 193, 36, 209, 664, 32 };
    partial_sort(V.begin(), V.begin() + 3, V.end(), greater<int>());

    cout << "first = " << V[0] << "\n";
    cout << "second = " << V[1] << "\n";
    cout << "third = " << V[2] << "\n";
    return 0;
}

Java

// java program to find
// three largest elements
// in array.
import java.io.*;
import java.util.Arrays;
class GFG{
  public static void main(String[] args)
  {
    int[] V = { 11, 65, 193, 36, 209, 664, 32 };

    // sorting the array
    Arrays.sort(V);

    // taking the length of array
    int x = V.length;

    System.out.println("first = " + V[x-1] );
    System.out.println("second = " + V[x-2]);
    System.out.println("third = " + V[x-3] );

  }
}

// This code is Contributed Machhaliya Muhammad

Python3

# Python program to implement
# the above approach

# Driver Code
V = [ 11, 65, 193, 36, 209, 664, 32 ];
V.sort()
V.reverse()

print(f"first =  {V[0]}");
print(f"second =  {V[1]}");
print(f"third =  {V[2]}");

# This code is contributed by Saurabh Jaiswal

C#

// C# program to implement
// the above approach
using System;

class GFG
{


// Driver Code
public static void Main()
{
    int[] V = { 11, 65, 193, 36, 209, 664, 32 };
    Array.Sort(V);
    Array.Reverse(V);


    Console.WriteLine("first = " + V[0] );
    Console.WriteLine("second = " + V[1]);
    Console.WriteLine("third = " + V[2] );

}
}

// This code is contributed by code_hunt.

Javascript

<script>
// Javascript program to implement
// the above approach


// Driver Code

    let V = [ 11, 65, 193, 36, 209, 664, 32 ];
    V.sort((a, b) => a - b)
    V.reverse()


    document.write("first = " + V[0] + "<br>");
    document.write("second = " + V[1] + "<br>");
    document.write("third = " + V[2] + "<br>");



// This code is contributed by Saurabh Jaiswal
</script>

Output

first = 664
second = 209
third = 193

Time Complexity: O(n log m) where m is distance(middle-first).
Auxiliary Space: O(1)