Find a sorted subsequence of size 3 in linear time

• Difficulty Level : Medium
• Last Updated : 27 Dec, 2021

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
Explanation: As 5 < 6 < 30, and they
appear in the same sequence in the array

Input: arr[] = {1, 2, 3, 4}
Output: 1, 2, 3 OR 1, 2, 4 OR 2, 3, 4
Explanation: As the array is sorted, for every i, j, k,
where i < j < k, arr[i] < arr[j] < arr[k]

Input: arr[] = {4, 3, 2, 1}
Output: No such triplet exists.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

METHOD 1:

Hint: Use Auxiliary Space.
Solution: So, the main motive is to find an element which has an element smaller than itself on the left side of the array and an element greater than itself on the right side of the array, if there is any such element then there exists a triplet that satisfies the criteria.

Approach: This can be solved in a very simple way. To find an element which has an element smaller than itself on its left side of the array, check if that element is the smallest element while traversing the array from the starting index i.e., (0), and to check if there is an element greater than itself on its right side of the array check whether that element is the largest element while traversing from the end of the array i.e., (n-1). If the element is not the smallest element from 0 to that index then it has an element smaller than itself on its left side, and similarly, if the element is not the largest element from that index to the last index then there is a larger element on its right side.

Algorithm

1. Create an auxiliary array smaller[0..n-1]. smaller[i] stores the index of a number which is smaller than arr[i] and is on the left side. The array contains -1 if there is no such element.
2. Create another auxiliary array greater[0..n-1]. greater[i] stores the index of a number which is greater than arr[i] and is on the right side of arr[i]. The array contains -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 equal to -1.

C++

// C++ program to find a sorted
// sub-sequence of size 3
#include <bits/stdc++.h>
using namespace std;

// A function to fund a sorted
// sub-sequence of size 3
void find3Numbers(int arr[], int n)
{
// Index of maximum element
// from right side
int max = n - 1;

// Index of minimum element
// from left side
int min = 0;
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];

// first entry will always be -1
smaller = -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];

// last entry will always be -1
greater[n - 1] = -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) {
cout << arr[smaller[i]]
<< " " << arr[i] << " "
<< arr[greater[i]];
return;
}
}

// If we reach number, then there are
// no such 3 numbers
cout << "No such triplet found";

// Free the dynamically allocated memory
// to avoid memory leak
delete[] smaller;
delete[] greater;

return;
}

// Driver code
int main()
{
int arr[] = { 12, 11, 10, 5, 6, 2, 30 };
int n = sizeof(arr) / sizeof(arr);
find3Numbers(arr, n);
return 0;
a greater number on
}

// This is code is contributed by rathbhupendra

C

// C program to find a sorted
// sub-sequence of size 3
#include <stdio.h>

// A function to fund a sorted
// sub-sequence of size 3
void find3Numbers(int arr[], int n)
{
// Index of maximum element
// from right side
int max = n - 1;

// Index of minimum element
// from left side
int min = 0;
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];

// first entry will always be -1
smaller = -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];

// last entry will always be -1
greater[n - 1] = -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 allocated 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);
find3Numbers(arr, n);
return 0;
}

Java

// Java program to find a sorted
// sub-sequence of size 3
import java.io.*;

class SortedSubsequence {
// A function to find a sorted
// sub-sequence of size 3
static void find3Numbers(int arr[])
{
int n = arr.length;

// Index of maximum element
// from right side
int max = n - 1;

// Index of minimum element
// from left side
int min = 0;
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];

// first entry will always be -1
smaller = -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];

// last entry will always be -1
greater[n - 1] = -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 greater number on right
// side and smaller number on left side
for (i = 0; i < n; i++) {
if (
smaller[i] != -1 && greater[i] != -1) {
System.out.print(
arr[smaller[i]] + " " + arr[i]
+ " " + arr[greater[i]]);
return;
}
}

// If we reach number, then there
// are no such 3 numbers
System.out.println("No such triplet found");
return;
}

public static void main(String[] args)
{
int arr[] = { 12, 11, 10, 5, 6, 2, 30 };
find3Numbers(arr);
}
}
/* This code is contributed by Devesh Agrawal*/

Python

# Python program to fund a sorted
# sub-sequence of size 3

def find3numbers(arr):
n = len(arr)

# Index of maximum element from right side
max = n-1

# Index of minimum element from left side
min = 0

# 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.
smaller = *10000
smaller = -1
for i in range(1, n):
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.
greater = *10000
greater[n-1] = -1

for i in range(n-2, -1, -1):
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 in range(0, n):
if smaller[i] != -1 and greater[i] != -1:
print arr[smaller[i]], arr[i], arr[greater[i]]
return

# If we reach here, then there are no such 3 numbers
print "No triplet found"
return

# Driver function to test above function
arr = [12, 11, 10, 5, 6, 2, 30]
find3numbers(arr)

# This code is contributed by Devesh Agrawal

C#

// C# program to find a sorted
// subsequence of size 3
using System;

class SortedSubsequence {

// A function to find a sorted
// subsequence of size 3
static void find3Numbers(int[] arr)
{
int n = arr.Length;

// Index of maximum element from right side
int max = n - 1;

// Index of minimum element from left side
int min = 0;
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];

// first entry will always be -1
smaller = -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];

// last entry will always be -1
greater[n - 1] = -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) {
Console.Write(
arr[smaller[i]] + " " + arr[i]
+ " " + arr[greater[i]]);
return;
}
}

// If we reach number, then there
// are no such 3 numbers
Console.Write("No such triplet found");
return;
}

// Driver code
public static void Main()
{
int[] arr = { 12, 11, 10, 5, 6, 2, 30 };
find3Numbers(arr);
}
}

/* This code is contributed by vt_m*/

PHP

<?php
// PHP program to find a sorted
// subsequence of size 3

// A function to fund a sorted
// subsequence of size 3
function find3Numbers(&\$arr, \$n)
{
// Index of maximum element from right side
\$max = \$n - 1;

// Index of minimum element from left side
\$min = 0;

// 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.
\$smaller = array_fill(0, \$n, NULL);
\$smaller = -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.
\$greater = array_fill(0, \$n, NULL);

// last entry will always be -1
\$greater[\$n - 1] = -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)
{
echo \$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");

return;
}

// Driver Code
\$arr = array(12, 11, 10, 5, 6, 2, 30);
\$n = sizeof(\$arr);
find3Numbers(\$arr, \$n);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript program to find a sorted
// sub-sequence of size 3

// A function to find a sorted
// sub-sequence of size 3
function find3Numbers(arr)
{
let n = arr.length;
// Index of maximum element
// from right side
let max = n - 1;
// Index of minimum element
// from left side
let min = 0;
let 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.
let smaller = new Array(n);
for(i=0;i<n;i++)
{
smaller[i]=0;
}
// first entry will always be -1
smaller = -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.
let greater = new Array(n);
for(i=0;i<n;i++)
{
greater[i]=0;
}
// last entry will always be -1
greater[n - 1] = -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 greater number on right
// side and smaller number on left side
for (i = 0; i < n; i++) {
if (
smaller[i] != -1 && greater[i] != -1) {
document.write(
arr[smaller[i]] + " " + arr[i]
+ " " + arr[greater[i]]);
return;
}
}

// If we reach number, then there
// are no such 3 numbers
document.write("No such triplet found <br>");
return;
}

let arr=[12, 11, 10, 5, 6, 2, 30 ]
find3Numbers(arr);

// This code is contributed by avanitrachhadiya2155

</script>

Complexity Analysis

• Time Complexity: O(n). As the array is traveled only once and there are no nested loops, the time complexity will be in the order of n.
• Auxiliary Space: O(n). Since two extra array is needed to store the index of the previous lesser element and next greater element so the space required will also be in the order of n

METHOD 2:

Solution: First find two elements arr[i] & arr[j] such that arr[i] < arr[j]. Then find a third element arr[k] greater than arr[j].
Approach: We can think of the problem in three simple terms.

1. First we only need to find two elements arr[i] < arr[j] and i < j. This can be done in linear time with just 1 loop over the range of the array. For instance, while keeping track of the min element, its easy to find any subsequent element that is greater than it. Thus we have our arr[i] & arr[j].
2. Secondly, consider this sequence – {3, 4, -1, 0, 2}. Initially min is 3, arr[i] is 3 and arr[j] is 4. While iterating over the array we can easily keep track of min and eventually update it to -1. And we can also update arr[i] & arr[j] to lower values i.e. -1 & 0 respectively.
3. Thirdly, as soon as we have arr[i] & arr[j] values, we can immediately start monitoring the subsequent elements in the same loop for an arr[k] > arr[j]. Thus we can find all three values arr[i] < arr[j] < arr[k] in a single pass over the array.

Algorithm: Iterate over the length of the array. Keep track of the min. As soon as the next iteration has an element greater than min, we have found our arr[j] and the min will be saved as arr[i]. Continue iterating until we find an element arr[k] which is greater than arr[j]. In case the next elements are of lower value, then we update min, arr[i] and arr[j] to these lower values, so as to give us the best chance to find arr[k].

C++

// C++ Program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the triplet
void find3Numbers(vector<int>& nums)
{

// If number of elements < 3
// then no triplets are possible
if (nums.size() < 3){
cout << "No such triplet found";
return;
}

// track best sequence length
// (not current sequence length)
int seq = 1;

// min number in array
int min_num = nums;

// least max number in best sequence
// i.e. track arr[j] (e.g. in
// array {1, 5, 3} our best sequence
// would be {1, 3} with arr[j] = 3)
int max_seq = INT_MAX;

// save arr[i]
int store_min = min_num;

// Iterate from 1 to nums.size()
for (int i = 1; i < nums.size(); i++)
{
if (nums[i] == min_num)
continue;

else if (nums[i] < min_num)
{
min_num = nums[i];
continue;
}

// this condition is only hit
// when current sequence size is 2
else if (nums[i] < max_seq) {

// update best sequence max number
// to a smaller value
// (i.e. we've found a
// smaller value for arr[j])
max_seq = nums[i];

// store best sequence start value
// i.e. arr[i]
store_min = min_num;
}

// Increase best sequence length &
// save next number in our triplet
else if (nums[i] > max_seq)
{
// We've found our arr[k]!
// Print the output
cout << "Triplet: " << store_min <<
", " << max_seq << ", " <<
nums[i] << endl;
return;
}
}

// No triplet found
cout << "No such triplet found";
}

// Driver Code
int main() {
vector<int> nums {1,2,-1,7,5};

// Function Call
find3Numbers(nums);
}

Java

// Java Program for above approach
class Main
{
// Function to find the triplet
public static void find3Numbers(int[] nums)
{

// If number of elements < 3
// then no triplets are possible
if (nums.length < 3){
System.out.print("No such triplet found");
return;
}

// track best sequence length
// (not current sequence length)
int seq = 1;

// min number in array
int min_num = nums;

// least max number in best sequence
// i.e. track arr[j] (e.g. in
// array {1, 5, 3} our best sequence
// would be {1, 3} with arr[j] = 3)
int max_seq = Integer.MIN_VALUE;

// save arr[i]
int store_min = min_num;

// Iterate from 1 to nums.size()
for (int i = 1; i < nums.length; i++)
{
if (nums[i] == min_num)
continue;

else if (nums[i] < min_num)
{
min_num = nums[i];
continue;
}

// this condition is only hit
// when current sequence size is 2
else if (nums[i] < max_seq) {

// update best sequence max number
// to a smaller value
// (i.e. we've found a
// smaller value for arr[j])
max_seq = nums[i];

// store best sequence start value
// i.e. arr[i]
store_min = min_num;
}

// Increase best sequence length &
// save next number in our triplet
else if (nums[i] > max_seq)
{
seq++;

// We've found our arr[k]!
// Print the output
if (seq == 3)
{
System.out.println("Triplet: " + store_min +
", " + max_seq + ", " + nums[i]);
return;
}
max_seq = nums[i];
}
}

// No triplet found
System.out.print("No such triplet found");
}

// Driver program
public static void main(String[] args)
{
int[] nums = {1,2,-1,7,5};

// Function Call
find3Numbers(nums);
}
}

// This code is contributed by divyesh072019

Python3

# Python3 Program for above approach
import sys

# Function to find the triplet
def find3Numbers(nums):

# If number of elements < 3
# then no triplets are possible
if (len(nums) < 3):
print("No such triplet found", end = '')
return

# Track best sequence length
# (not current sequence length)
seq = 1

# min number in array
min_num = nums

# Least max number in best sequence
# i.e. track arr[j] (e.g. in
# array {1, 5, 3} our best sequence
# would be {1, 3} with arr[j] = 3)
max_seq = -sys.maxsize - 1

# Save arr[i]
store_min = min_num

# Iterate from 1 to nums.size()
for i in range(1, len(nums)):
if (nums[i] == min_num):
continue
elif (nums[i] < min_num):
min_num = nums[i]
continue

# This condition is only hit
# when current sequence size is 2
elif (nums[i] < max_seq):

# Update best sequence max number
# to a smaller value
# (i.e. we've found a
# smaller value for arr[j])
max_seq = nums[i]

# Store best sequence start value
# i.e. arr[i]
store_min = min_num

# Increase best sequence length &
# save next number in our triplet
elif (nums[i] > max_seq):
if seq == 1:
store_min = min_num
seq += 1

# We've found our arr[k]!
# Print the output
if (seq == 3):
print("Triplet: " + str(store_min) +
", " + str(max_seq) + ", " +
str(nums[i]))

return

max_seq = nums[i]

# No triplet found
print("No such triplet found", end = '')

# Driver Code
if __name__=='__main__':

nums = [ 1, 2, -1, 7, 5 ]

# Function Call
find3Numbers(nums)

# This code is contributed by rutvik_56

C#

// C# Program for above approach
using System;
class GFG {

// Function to find the triplet
static void find3Numbers(int[] nums)
{

// If number of elements < 3
// then no triplets are possible
if (nums.Length < 3){
Console.Write("No such triplet found");
return;
}

// track best sequence length
// (not current sequence length)
int seq = 1;

// min number in array
int min_num = nums;

// least max number in best sequence
// i.e. track arr[j] (e.g. in
// array {1, 5, 3} our best sequence
// would be {1, 3} with arr[j] = 3)
int max_seq = Int32.MinValue;

// save arr[i]
int store_min = min_num;

// Iterate from 1 to nums.size()
for (int i = 1; i < nums.Length; i++)
{
if (nums[i] == min_num)
continue;

else if (nums[i] < min_num)
{
min_num = nums[i];
continue;
}

// this condition is only hit
// when current sequence size is 2
else if (nums[i] < max_seq) {

// update best sequence max number
// to a smaller value
// (i.e. we've found a
// smaller value for arr[j])
max_seq = nums[i];

// store best sequence start value
// i.e. arr[i]
store_min = min_num;
}

// Increase best sequence length &
// save next number in our triplet
else if (nums[i] > max_seq)
{
seq++;

// We've found our arr[k]!
// Print the output
if (seq == 3)
{
Console.WriteLine("Triplet: " + store_min +
", " + max_seq + ", " + nums[i]);
return;
}
max_seq = nums[i];
}
}

// No triplet found
Console.Write("No such triplet found");
}

static void Main() {
int[] nums = {1,2,-1,7,5};

// Function Call
find3Numbers(nums);
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// Javascript program for above approach

// Function to find the triplet
function find3Numbers(nums)
{

// If number of elements < 3
// then no triplets are possible
if (nums.length < 3)
{
document.write("No such triplet found");
return;
}

// Track best sequence length
// (not current sequence length)
let seq = 1;

// min number in array
let min_num = nums;

// Least max number in best sequence
// i.e. track arr[j] (e.g. in
// array {1, 5, 3} our best sequence
// would be {1, 3} with arr[j] = 3)
let max_seq = Number.MIN_VALUE;

// Save arr[i]
let store_min = min_num;

// Iterate from 1 to nums.size()
for(let i = 1; i < nums.length; i++)
{
if (nums[i] == min_num)
continue;

else if (nums[i] < min_num)
{
min_num = nums[i];
continue;
}

// This condition is only hit
// when current sequence size is 2
else if (nums[i] < max_seq)
{

// Update best sequence max number
// to a smaller value
// (i.e. we've found a
// smaller value for arr[j])
max_seq = nums[i];

// Store best sequence start value
// i.e. arr[i]
store_min = min_num;
}

// Increase best sequence length &
// save next number in our triplet
else if (nums[i] > max_seq)
{
seq++;

// We've found our arr[k]!
// Print the output
if (seq == 3)
{
document.write("Triplet: " + store_min +
", " + max_seq + ", " +
nums[i]);
return;
}
max_seq = nums[i];
}
}

// No triplet found
document.write("No such triplet found");
}

// Driver code
let nums = [1, 2, -1, 7, 5];

// Function Call
find3Numbers(nums);

// This code is contributed by suresh07

</script>
Output
Triplet: 1, 2, 7

Complexity Analysis:

Time Complexity: O(n). As the array is traveled only once and there are no nested loops, the time complexity will be in the order of n.
Auxiliary Space: O(1).

Exercise:

1. Find a sub-sequence of size 3 such that arr[i] arr[k].
2. Find a sorted sub-sequence of size 4 in linear time