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 –
Below is 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) { third = second second = x } d) Else if (x > third) { third = x } 3) Print first, second and third.
Below is the implementation of 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) { third = second; second = arr[i]; } else if (arr[i] > third) 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) { third = second; second = arr[i]; } else if (arr[i] > third) 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). ?> |
Output :
Three largest elements are 34 13 12
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 contibuted by Rajput-Ji |
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 contibuted 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 contibuted 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 contibuted by Code_Mech |
Output :
54 45 23
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; } |
Output :
first = 664 second = 209 third = 193
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