Find Increasing Triplet Sub-sequence

Objective: Given an integer array A[1..n], find an instance of i,j,k where 0 < i < j < k <= n and A[i] < A[j] < A[k].

Example :

int arrA[] = { 10, 9, 4, 3, 2, 1, 7, 3, 1, 11, 2, 6, 9 };
Output :
Increasing triplets are

1, 7, 11
1, 2, 6
1, 3, 9

likewise you can find more triplets.

Approach:

Naive Solution : Brute-Force O(n^2) – Use two for loops and try every triplet till you find the increasing triplet.

Better Solution : O(n)

  • Cre­ate 2 Aux­i­lary Arrays say Lmin[] and Rmax[] of the same size as main array
  • Put Lmin[0]=0 and Rmax[Rmax.length-1] =Rmax.length-1
  • Tra­verse the main array and fill the Lmin array with the index posi­tion which has the min­i­mum value so far
  • Tra­verse the main array back­words and fill the Rmax array with the index posi­tion which has the max­imun value so far.
  • Now Traverse the main array and check for the element with the following condition and print it.

arrA[Lmin[i]] < arrA[i] && arrA[Rmax[i]] > arrA[i]

Code:

public class IncreasingTripletSubsequence {
// Find a sorted subsequence of size 3 in linear time
// Given an integer array A[1..n], find an instance of i,j,k where 0 < i < j
// < k <= n and A[i] < A[j] < A[k].
public void triplet(int[] arrA) {
int[] Lmin = new int[arrA.length];
int[] Rmax = new int[arrA.length];
int leftMinIndex = 0;
int leftMinValue = arrA[0];
int rightMaxValue = arrA[arrA.length 1];
int rightMaxIndex = arrA.length 1;
// traverse the main array and fill the Lmin array with the index
// position which has the minimum value so far
for (int i = 0; i < arrA.length; i++) {
if (leftMinValue > arrA[i]) {
leftMinIndex = i;
leftMinValue = arrA[i];
}
Lmin[i] = leftMinIndex;
}
// System.out.println(Arrays.toString(Lmin));
// traverse the main array backwords and fill the Rmax array with the
// index position which has the maximun value so far
for (int i = arrA.length 1; i >= 0; i) {
if (rightMaxValue < arrA[i]) {
rightMaxIndex = i;
rightMaxValue = arrA[i];
}
Rmax[i] = rightMaxIndex;
}
// Now Traverse the main array and check for the element with the
// following condition and print it.
// arrA[Lmin[i]] < arrA[i] && arrA[Rmax[i]] > arrA[i]
for (int i = 0; i < arrA.length; i++) {
if (arrA[Lmin[i]] < arrA[i] && arrA[Rmax[i]] > arrA[i]) {
System.out.println("Triplet " + arrA[Lmin[i]] + " " + arrA[i]
+ " " + arrA[Rmax[i]]);
return;
}
}
}
public static void main(String[] args) {
int arrA[] = { 10, 9, 4, 3, 2, 1, 7, 3, 1, 11, 2, 6, 9 };
IncreasingTripletSubsequence i = new IncreasingTripletSubsequence();
i.triplet(arrA);
}
}

Output:

Triplet 1 7 11