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] where i < j < k. This problem must be solved using constant space or no extra space.

This problem is already solved in linear time using linear space: Find a sorted sub-sequence of size 3 in Linear Time

Examples:

Input: arr[] = {12, 11, 10, 5, 2, 6, 30} 
Output: 5 6 30 
        or 2 6 30
Explanation: Answer is 5, 6, 30 because 5 < 6 < 30 
and they occur in this sequence in the array.

Input: arr[] = {5, 7, 4, 8}
Output: 5 7 8
Explanation: Answer is 5 ,7 ,8 because 5 < 7 < 8 
and they occur in the same sequence in the array 

Solution: The objective is to find three elements a, b and c such that a < b < c, and the elements must occur in the same sequence in the array.

Approach: The problem deals with finding three elements a, b, c where a < b < c and they must appear in the same order as in the array. So the intuition at any step must be followed as such. One of the variable (small) should store the smallest element of the array, and the second variable large will be assigned a value when there is already a smaller value present in the (small) variable. This will lead to the formation of a pair of two variables which will form the first two elements of the required sequence. Similarly if another value can be found in the array that is assigned when the first two variables are already assigned and has a smaller value than the present element the search for the third value would be complete. This completes the triplet a, b and c such that a < b < c, in similar sequence to the array.



Algorithm

  1. Create three variables, small– Stores the smallest element,large– stores the second element of the sequence, i– loop counter
  2. Move along the input array from start till end.
  3. If the present element is smaller than or equal to variable small, update the variable.
  4. Else if the present element is smaller than or equal to variable large, update the variable. So here we get a pair (small, large) at this instant, where small < large and they occur in the required sequence.
  5. Else if the previous two cases fail to match, break the loop as we have got a pair where the present element is greater than both variables small and large. Store the index in variable i.
  6. If the break statement has not been encountered then it is guaranteed that no such triplet exists.
  7. Else there is a triplet satisfies the criteria but the variable small might have been updated to a new smaller value.
  8. So Traverse the array from start to index i.
  9. Re-assign the variable small to any element less than large, it is guaranteed that there exists one.
  10. Print the values small, large and ith array element

Implementation

C++

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// C/C++ program to find a sorted sub-sequence of
// size 3 using constant space
#include <bits/stdc++.h>
using namespace std;
  
// A function to fund a sorted sub-sequence 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[0]);
    find3Numbers(arr, n);
    return 0;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// C# program to find a sorted sub-sequence of
// size 3 using constant space
using System;
  
class GFG {
      
    // A function to fund a sorted sub-sequence
    // 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);
    }
}
<br>
  
// This code is contributed by nitin mittal

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PHP

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<?php
// PHP program to find a sorted 
// subsequence of size 3 using 
// constant space
  
// A function to fund a sorted
// subsequence of size 3
function find3Numbers($arr, $n)
{
      
    // Initializing small and 
    // large(second smaller)
    // by INT_MAX
    $small = PHP_INT_MAX; 
    $large = PHP_INT_MAX;
    $i;
    for($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 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)
    {
        echo "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 = 0; $j <= $i; $j++)
    {
        if ($arr[$j] < $large)
        {
            $small = $arr[$j];
            break;
        }
    }
  
    echo $small," ", $large," ", $arr[$i];
    return;
}
  
    // Driver Code
    $arr = array(5, 7, 4, 8);
    $n = count($arr);
    find3Numbers($arr, $n);
  
// This code is contributed by anuj_67.
?>

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Output:

5 7 8

Complexity Analysis:

  • Time Complexity: O(n).
    As the array is traversed only twice, the time complexity is O(2*n) which is equal to O(n).
  • Space Complexity: O(1).
    Since only three elements are stored, the space complexity is constant or O(1).
  • This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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    Improved By : nitin mittal, vt_m, andrew1234

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