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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, 30 Input: arr[] = {1, 2, 3, 4} Output: 1, 2, 3 OR 1, 2, 4 OR 2, 3, 4 Input: arr[] = {4, 3, 2, 1} Output: No such triplet
Source: Amazon Interview Question
Hint: Use Auxiliary Space
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.
#include<stdio.h>
// A function to fund a sorted subsequence of size 3
void 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;
}
// Driver program to test above function
int main()
{
int arr[] = {12, 11, 10, 5, 6, 2, 30};
int n = sizeof(arr)/sizeof(arr[0]);
find3Numbers(arr, n);
return 0;
}
Output:
5 6 30
Time Complexity: O(n)
Auxliary Space: O(n)
Exercise:
1. Find a subsequence of size 3 such that arr[i] < arr[j] > arr[k].
2. Find a sorted subsequence of size 4 in linear time
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