# Sorted subsequence of size 3 in linear time using constant space

Given an array, the task is to find three elements of this array such that they are in sorted form i.e. for any three elements a[i], a[j] and a[k], they follow this relationship : a[i] < a[j] < a[k].

This problem is already solved below in linear time using linear space, you can read about that here:

Find a sorted subsequence of size 3 in linear time

In this post we will solve the problem without using any extra space.
Examples:

```Input : arr[] = {12, 11, 10, 5, 2, 6, 30}
Output : 5 6 30
or 2 6 30
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

As we are looking for sequence of length 3, at each index we can maintain smallest value we’ve got so far and second smallest value after smallest value’s index, now if we reach to an index whose value is larger than second smallest value, then we found our solution because we already maintained a sorted pair and we just found an element which is larger than both, so we found a 3 length sorted subsequence of array.

Please see below code for better understanding :

## C++

 `// C/C++ program to find a sorted subsequence of ` `// size 3 using constant space ` `#include ` `using` `namespace` `std; ` ` `  `// A function to fund a sorted subsequence of size 3 ` `void` `find3Numbers(``int` `arr[], ``int` `n) ` `{ ` `    ``// Initializing small and large(second smaller) ` `    ``// by INT_MAX ` `    ``int` `small = INT_MAX, large = INT_MAX; ` `    ``int` `i; ` `    ``for` `(i = 0; i < n; i++) ` `    ``{ ` `        ``// Update small for smallest value of array ` `        ``if` `(arr[i] <= small) ` `            ``small = arr[i]; ` ` `  `        ``// Update large for second smallest value of ` `        ``// array after occurrence of small ` `        ``else` `if` `(arr[i] <= large) ` `            ``large = arr[i]; ` ` `  `        ``// If we reach here, we found 3 numbers in ` `        ``// increasing order : small, large and arr[i] ` `        ``else` `            ``break``; ` `    ``} ` ` `  `    ``if` `(i == n) ` `    ``{ ` `        ``printf``(``"No such triplet found"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// last and second last will be same, but first ` `    ``// element can be updated retrieving first element ` `    ``// by looping upto i ` `    ``for` `(``int` `j = 0; j <= i; j++) ` `    ``{ ` `        ``if` `(arr[j] < large) ` `        ``{ ` `            ``small = arr[j]; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``printf``(``"%d %d %d"``, small, large, arr[i]); ` `    ``return``; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = {5, 7, 4, 8}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``find3Numbers(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find a sorted subsequence of ` `// size 3 using constant space ` ` `  `class` `GFG ` `{ ` `    ``// A function to fund a sorted subsequence of size 3 ` `    ``static` `void` `find3Numbers(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``// Initializing small and large(second smaller) ` `        ``// by INT_MAX ` `        ``int` `small = +``2147483647``, large = +``2147483647``; ` `        ``int` `i; ` `        ``for` `(i = ``0``; i < n; i++) ` `        ``{ ` `            ``// Update small for smallest value of array ` `            ``if` `(arr[i] <= small) ` `                ``small = arr[i]; ` `     `  `            ``// Update large for second smallest value of ` `            ``// array after occurrence of small ` `            ``else` `if` `(arr[i] <= large) ` `                ``large = arr[i]; ` `     `  `            ``// If we reach here, we found 3 numbers in ` `            ``// increasing order : small, large and arr[i] ` `            ``else` `                ``break``; ` `        ``} ` `     `  `        ``if` `(i == n) ` `        ``{ ` `            ``System.out.print(``"No such triplet found"``); ` `            ``return``; ` `        ``} ` `     `  `        ``// last and second last will be same, but first ` `        ``// element can be updated retrieving first element ` `        ``// by looping upto i ` `        ``for` `(``int` `j = ``0``; j <= i; j++) ` `        ``{ ` `            ``if` `(arr[j] < large) ` `            ``{ ` `                ``small = arr[j]; ` `                ``break``; ` `            ``} ` `        ``} ` `     `  `        ``System.out.print(small+``" "``+large+``" "``+arr[i]); ` `        ``return``; ` `    ``} ` `     `  `    ``// Driver program ` `    ``public` `static` `void` `main(String arg[]) ` `    ``{ ` `        ``int` `arr[] = {``5``, ``7``, ``4``, ``8``}; ` `        ``int` `n = arr.length; ` `        ``find3Numbers(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python3 program to find a sorted subsequence  ` `# of size 3 using constant space ` ` `  `# Function to fund a sorted subsequence of size 3 ` `def` `find3Numbers(arr, n): ` ` `  `    ``# Initializing small and large(second smaller) ` `    ``# by INT_MAX ` `    ``small ``=` `+``2147483647` `    ``large ``=` `+``2147483647` `     `  `    ``for` `i ``in` `range``(n): ` `     `  `        ``# Update small for smallest value of array ` `        ``if` `(arr[i] <``=` `small): ` `            ``small ``=` `arr[i] ` ` `  `        ``# Update large for second smallest value of ` `        ``# array after occurrence of small ` `        ``elif` `(arr[i] <``=` `large): ` `            ``large ``=` `arr[i] ` ` `  `        ``# If we reach here, we found 3 numbers in ` `        ``# increasing order : small, large and arr[i] ` `        ``else``: ` `            ``break` ` `  `    ``if` `(i ``=``=` `n): ` `     `  `        ``print``(``"No such triplet found"``) ` `        ``return` `     `  `    ``# last and second last will be same, but ` `    ``# first element can be updated retrieving  ` `    ``# first element by looping upto i ` `    ``for` `j ``in` `range``(i ``+` `1``): ` `     `  `        ``if` `(arr[j] < large): ` `         `  `            ``small ``=` `arr[j] ` `            ``break` ` `  `    ``print``(small,``" "``,large,``" "``,arr[i]) ` `    ``return` ` `  `# Driver program ` `arr``=` `[``5``, ``7``, ``4``, ``8``] ` `n ``=` `len``(arr) ` `find3Numbers(arr, n) ` ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# program to find a sorted subsequence of ` `// size 3 using constant space ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// A function to fund a sorted subsequence ` `    ``// of size 3 ` `    ``static` `void` `find3Numbers(``int` `[]arr, ``int` `n) ` `    ``{ ` `         `  `        ``// Initializing small and large(second smaller) ` `        ``// by INT_MAX ` `        ``int` `small = +2147483647, large = +2147483647; ` `        ``int` `i; ` `        ``for` `(i = 0; i < n; i++) ` `        ``{ ` `             `  `            ``// Update small for smallest value of array ` `            ``if` `(arr[i] <= small) ` `                ``small = arr[i]; ` `     `  `            ``// Update large for second smallest value of ` `            ``// array after occurrence of small ` `            ``else` `if` `(arr[i] <= large) ` `                ``large = arr[i]; ` `     `  `            ``// If we reach here, we found 3 numbers in ` `            ``// increasing order : small, large and arr[i] ` `            ``else` `                ``break``; ` `        ``} ` `     `  `        ``if` `(i == n) ` `        ``{ ` `            ``Console.Write(``"No such triplet found"``); ` `            ``return``; ` `        ``} ` `     `  `        ``// last and second last will be same, but first ` `        ``// element can be updated retrieving first element ` `        ``// by looping upto i ` `        ``for` `(``int` `j = 0; j <= i; j++) ` `        ``{ ` `            ``if` `(arr[j] < large) ` `            ``{ ` `                ``small = arr[j]; ` `                ``break``; ` `            ``} ` `        ``} ` `     `  `        ``Console.Write(small + ``" "` `+ large + ``" "` `+ arr[i]); ` `        ``return``; ` `    ``} ` `     `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = {5, 7, 4, 8}; ` `        ``int` `n = arr.Length; ` `        ``find3Numbers(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal `

## PHP

 ` `

Output:

```5 7 8
```

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Improved By : nitin mittal, vt_m

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