Find a sorted subsequence of size 3 three in linear time

Given an array of n integers, find the 3 elements such that a[i] < a[j] < a[k] and i < j < k in 0(n) time. If there are multiple such triplets, then print any one of them.

Examples:Input:  arr[] = {12, 11, 10, 5, 6, 2, 30}Output: 5, 6, 30Input:  arr[] = {1, 2, 3, 4}Output: 1, 2, 3 OR 1, 2, 4 OR 2, 3, 4Input:  arr[] = {4, 3, 2, 1}Output: No such triplet

Solution:
1) Create an auxiliary array smaller[0..n-1]. smaller[i] should store the index of a number which is smaller than arr[i] and is on left side of arr[i]. smaller[i] should contain -1 if there is no such element.
2) Create another auxiliary array greater[0..n-1]. greater[i] should store the index of a number which is greater than arr[i] and is on right side of arr[i]. greater[i] should contain -1 if there is no such element.
3) Finally traverse both smaller[] and greater[] and find the index i for which both smaller[i] and greater[i] are not -1.

// A function to fund a sorted subsequence of size 3void find3Numbers(int arr[], int n){   int max = n-1; //Index of maximum element from right side   int min = 0; //Index of minimum element from left side   int i;   // Create an array that will store index of a smaller   // element on left side. If there is no smaller element   // on left side, then smaller[i] will be -1.   int *smaller = new int[n];   smaller[0] = -1;  // first entry will always be -1   for (i = 1; i < n; i++)   {       if (arr[i] <= arr[min])       {          min = i;          smaller[i] = -1;       }       else          smaller[i] = min;   }   // Create another array that will store index of a   // greater element on right side. If there is no greater   // element on right side, then greater[i] will be -1.   int *greater = new int[n];   greater[n-1] = -1;  // last entry will always be -1   for (i = n-2; i >= 0; i--)   {       if (arr[i] >= arr[max])       {          max = i;          greater[i] = -1;       }       else          greater[i] = max;   }   // Now find a number which has both a greater number on   // right side and smaller number on left side   for (i = 0; i < n; i++)   {       if (smaller[i] != -1 && greater[i] != -1)       {          printf("%d %d %d", arr[smaller[i]],                 arr[i], arr[greater[i]]);          return;       }   }   // If we reach number, then there are no such 3 numbers   printf("No such triplet found");   // Free the dynamically alloced memory to avoid memory leak   delete [] smaller;   delete [] greater;   return;}