Java Program for Third largest element in an array of distinct elements

03 Jan, 2022
Given an array of n integers, find the third largest element. All the elements in the array are distinct integers. Example :  
Input: arr[] = {1, 14, 2, 16, 10, 20}
Output: The third Largest element is 14

Explanation: Largest element is 20, second largest element is 16 
and third largest element is 14

Input: arr[] = {19, -10, 20, 14, 2, 16, 10}
Output: The third Largest element is 16

Explanation: Largest element is 20, second largest element is 19 
and third largest element is 16
 
Naive Approach: The task is to first find the largest element, followed by the second-largest element and then excluding them both find the third-largest element. The basic idea is to iterate the array twice and mark the maximum and second maximum element and then excluding them both find the third maximum element, i.e the maximum element excluding the maximum and second maximum. 
Algorithm: First, iterate through the array and find maximum.Store this as first maximum along with its index.Now traverse the whole array finding the second max, excluding the maximum element.Finally traverse the array the third time and find the third largest element i.e., excluding the maximum and second maximum. 
First, iterate through the array and find maximum.Store this as first maximum along with its index.Now traverse the whole array finding the second max, excluding the maximum element.Finally traverse the array the third time and find the third largest element i.e., excluding the maximum and second maximum. 
First, iterate through the array and find maximum.
Store this as first maximum along with its index.
Now traverse the whole array finding the second max, excluding the maximum element.
Finally traverse the array the third time and find the third largest element i.e., excluding the maximum and second maximum. 
Java
// Java program to find third // Largest element in an array// of distinct elements  class GFG{static void thirdLargest(int arr[],                         int arr_size){    /* There should be     atleast three elements */    if (arr_size < 3)    {        System.out.printf(" Invalid Input ");        return;    }      // Find first     // largest element    int first = arr[0];    for (int i = 1;             i < arr_size ; i++)        if (arr[i] > first)            first = arr[i];      // Find second     // largest element    int second = Integer.MIN_VALUE;    for (int i = 0;              i < arr_size ; i++)        if (arr[i] > second &&             arr[i] < first)            second = arr[i];      // Find third    // largest element    int third = Integer.MIN_VALUE;    for (int i = 0;              i < arr_size ; i++)        if (arr[i] > third &&             arr[i] < second)            third = arr[i];      System.out.printf("The third Largest " +                   "element is %d", third);}  // Driver codepublic static void main(String []args){    int arr[] = {12, 13, 1,                  10, 34, 16};    int n = arr.length;    thirdLargest(arr, n);}}  // This code is contributed// by Smitha
Output:  
The third Largest element is 13
Complexity Analysis: Time Complexity: O(n). As the array is iterated thrice and is done in a constant timeSpace complexity: O(1). No extra space is needed as the indices can be stored in constant space.
Time Complexity: O(n). As the array is iterated thrice and is done in a constant time
Space complexity: O(1). No extra space is needed as the indices can be stored in constant space.
Efficient Approach: The problem deals with finding the third largest element in the array in a single traversal. The problem can be cracked by taking help of a similar problem- finding the second maximum element. So the idea is to traverse the array from start to end and to keep track of the three largest elements up to that index (stored in variables). So after traversing the whole array, the variables would have stored the indices (or value) of the three largest elements of the array. 
Algorithm: Create three variables, first, second, third, to store indices of three largest elements of the array. (Initially all of them are initialized to a minimum value).Move along the input array from start to the end.For every index check if the element is larger than first or not. Update the value of first, if the element is larger, and assign the value of first to second and second to third. So the largest element gets updated and the elements previously stored as largest becomes second largest, and the second largest element becomes third largest.Else if the element is larger than the second, then update the value of second,and the second largest element becomes third largest.If the previous two conditions fail, but the element is larger than the third, then update the third.Print the value of third after traversing the array from start to end 
Create three variables, first, second, third, to store indices of three largest elements of the array. (Initially all of them are initialized to a minimum value).Move along the input array from start to the end.For every index check if the element is larger than first or not. Update the value of first, if the element is larger, and assign the value of first to second and second to third. So the largest element gets updated and the elements previously stored as largest becomes second largest, and the second largest element becomes third largest.Else if the element is larger than the second, then update the value of second,and the second largest element becomes third largest.If the previous two conditions fail, but the element is larger than the third, then update the third.Print the value of third after traversing the array from start to end 
Create three variables, first, second, third, to store indices of three largest elements of the array. (Initially all of them are initialized to a minimum value).
Move along the input array from start to the end.
For every index check if the element is larger than first or not. Update the value of first, if the element is larger, and assign the value of first to second and second to third. So the largest element gets updated and the elements previously stored as largest becomes second largest, and the second largest element becomes third largest.
Else if the element is larger than the second, then update the value of second,and the second largest element becomes third largest.
If the previous two conditions fail, but the element is larger than the third, then update the third.
Print the value of third after traversing the array from start to end 
Java
// Java program to find third Largest element in an arrayclass GFG {      static void thirdLargest(int arr[], int arr_size) {        /* There should be atleast three elements */        if (arr_size < 3) {            System.out.printf(" Invalid Input ");            return;        }          // Initialize first, second and third Largest element        int first = arr[0], second = Integer.MIN_VALUE,                            third = Integer.MIN_VALUE;          // Traverse array elements to find the third Largest        for (int i = 1; i < arr_size; i++) {            /* If current element is greater than first,        then update first, second and third */            if (arr[i] > first) {                third = second;                second = first;                first = arr[i];            } /* If arr[i] is in between first and second */            else if (arr[i] > second) {                third = second;                second = arr[i];            } /* If arr[i] is in between second and third */            else if (arr[i] > third) {                third = arr[i];            }        }          System.out.printf("The third Largest element is %d", third);    }      /* Driver program to test above function */    public static void main(String []args) {        int arr[] = {12, 13, 1, 10, 34, 16};        int n = arr.length;        thirdLargest(arr, n);    }}//This code is contributed by 29AjayKumar
Output:  
The third Largest element is 13
Complexity Analysis:Time Complexity: O(n). As the array is iterated once and is done in a constant timeSpace complexity: O(1). No extra space is needed as the indices can be stored in constant space.
Time Complexity: O(n). As the array is iterated once and is done in a constant time
Space complexity: O(1). No extra space is needed as the indices can be stored in constant space.
Please refer complete article on Third largest element in an array of distinct elements for more details!
Amazon
Arrays
Java
Java Programs
Searching
Amazon
Arrays
Searching
Java
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Introduction to Data Structures
Window Sliding Technique
Search, insert and delete in an unsorted array
Chocolate Distribution Problem
Find duplicates in O(n) time and O(1) extra space | Set 1
Split() String method in Java with examples
Arrays.sort() in Java with examples
Reverse a string in Java
Object Oriented Programming (OOPs) Concept in Java
For-each loop in Java