Check if a non-contiguous subsequence same as the given subarray exists or not

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Given an array arr[] consisting of N integers and two integer values L and R, indicating the starting and ending indices of a subarray, the task is to check if there exists a non-contiguous subsequence which is same as the given subarray or not. If found to be true, print “Yes”. Otherwise, print “No”.

A non-contiguous subsequence contains at least two consecutive characters from non-consecutive indices.

Examples:

Input: arr[] = {1, 7, 12, 1, 7, 5, 10, 11, 42}, L = 3, R = 6
Output: Yes
Explanation: The non-contiguous subsequence {arr[0], arr[1], arr[5], arr[6]} is same as the subarray {arr[3], .., arr[6]}.

Input: arr[] = {0, 1, 2, -2, 5, 10}, L = 1, R = 3

Naive Approach: The simplest approach is to generate all possible subsequences of the given array and check if any subsequence generated is equal to the given subarray or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Time Complexity: O(N*2^N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the key observation that there will always be a non-contiguous subsequence if there is at least one occurrence of the first element of the given subarray in the range [0, L – 1] and at least one occurrence of the last element of a subarray in the range [R + 1, N].

Proof of Logic:

If the 1^st element of the subarray {arr[L], … arr[R]} also occurs at any index K (K < L), then one such non-contiguous subsequence can be {arr[K], arr[L + 1], …., arr[R]}.
If the last element of the subarray {arr[L], … arr[R]} also occurs at any index K (K > R), then one such non-contiguous subsequence can be strong>{arr[L], arr[L + 1], …., arr[R – 1], arr[K]}.
If none of the above two conditions are satisfied, then no such non-contiguous subsequence exists.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Utility Function to check whether
// a subsequence same as the given
// subarray exists or not
bool checkSubsequenceUtil(
    int arr[], int L, int R, int N)
{
    // Check if first element of the
    // subarray is also present before
    for (int i = 0; i < L; i++)
        if (arr[i] == arr[L])
            return true;
 
    // Check if last element of the
    // subarray is also present later
    for (int i = R + 1; i < N; i++)
        if (arr[i] == arr[R])
            return true;
 
    // If above two conditions are
    // not satisfied, then no such
    // subsequence exists
    return false;
}
 
// Function to check and print if a
// subsequence which is same as the
// given subarray is present or not
void checkSubsequence(int arr[], int L,
                      int R, int N)
{
    if (checkSubsequenceUtil(arr, L,
                             R, N)) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 7, 12, 1, 7,
                  5, 10, 11, 42 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int L = 3, R = 6;
 
    // Function Call
    checkSubsequence(arr, L, R, N);
}

Java

// Java program for the above approach 
class GFG{
      
// Utility Function to check whether
// a subsequence same as the given
// subarray exists or not
static boolean checkSubsequenceUtil(int arr[], int L,
                                    int R, int N)
{
     
    // Check if first element of the
    // subarray is also present before
    for(int i = 0; i < L; i++)
        if (arr[i] == arr[L])
            return true;
  
    // Check if last element of the
    // subarray is also present later
    for(int i = R + 1; i < N; i++)
        if (arr[i] == arr[R])
            return true;
             
    // If above two conditions are
    // not satisfied, then no such
    // subsequence exists
    return false;
}
  
// Function to check and print if a
// subsequence which is same as the
// given subarray is present or not
static void checkSubsequence(int arr[], int L,
                             int R, int N)
{
    if (checkSubsequenceUtil(arr, L,
                             R, N)) 
    {
        System.out.print("YES\n");
    }
    else
    {
        System.out.print("NO\n");
    }
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 7, 12, 1, 7,
                  5, 10, 11, 42 };
    int N = arr.length;
    int L = 3, R = 6;
  
    // Function Call
    checkSubsequence(arr, L, R, N);
}
}
  
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach 
  
# Utility Function to check whether
# a subsequence same as the given
# subarray exists or not
def checkSubsequenceUtil(arr, L, R, N):
         
    # Check if first element of the
    # subarray is also present before
    for i in range(L):
        if (arr[i] == arr[L]):
            return True
  
    # Check if last element of the
    # subarray is also present later
    for i in range(R + 1, N, 1):
        if (arr[i] == arr[R]):
            return True
  
    # If above two conditions are
    # not satisfied, then no such
    # subsequence exists
    return False
 
# Function to check and print if a
# subsequence which is same as the
# given subarray is present or not
def checkSubsequence(arr, L, R, N):
     
    if (checkSubsequenceUtil(arr, L,R, N)):
        print("YES")
    else:
        print("NO")
         
# Driver Code
arr = [ 1, 7, 12, 1, 7,
        5, 10, 11, 42 ]
N = len(arr) 
L = 3
R = 6
  
# Function Call
checkSubsequence(arr, L, R, N)
 
# This code is contributed by susmitakundugoaldanga

C#

// C# program for the above approach 
using System;
 
class GFG{
       
// Utility Function to check whether
// a subsequence same as the given
// subarray exists or not
static bool checkSubsequenceUtil(int[] arr, int L,
                                 int R, int N)
{
     
    // Check if first element of the
    // subarray is also present before
    for(int i = 0; i < L; i++)
        if (arr[i] == arr[L])
            return true;
   
    // Check if last element of the
    // subarray is also present later
    for(int i = R + 1; i < N; i++)
        if (arr[i] == arr[R])
            return true;
              
    // If above two conditions are
    // not satisfied, then no such
    // subsequence exists
    return false;
}
   
// Function to check and print if a
// subsequence which is same as the
// given subarray is present or not
static void checkSubsequence(int[] arr, int L,
                             int R, int N)
{
    if (checkSubsequenceUtil(arr, L,
                             R, N)) 
    {
        Console.Write("YES\n");
    }
    else
    {
        Console.Write("NO\n");
    }
}
   
// Driver code
public static void Main()
{
    int[] arr = { 1, 7, 12, 1, 7,
                  5, 10, 11, 42 };
                   
    int N = arr.Length;
    int L = 3, R = 6;
   
    // Function Call
    checkSubsequence(arr, L, R, N); 
}
}
 
// This code is contributed by code_hunt

Javascript

<script>
// javascript program to implement
// the above approach
 
// Utility Function to check whether
// a subsequence same as the given
// subarray exists or not
function checkSubsequenceUtil(arr, L,
                                    R, N)
{
      
    // Check if first element of the
    // subarray is also present before
    for(let i = 0; i < L; i++)
        if (arr[i] == arr[L])
            return true;
   
    // Check if last element of the
    // subarray is also present later
    for(let i = R + 1; i < N; i++)
        if (arr[i] == arr[R])
            return true;
              
    // If above two conditions are
    // not satisfied, then no such
    // subsequence exists
    return false;
}
   
// Function to check and print if a
// subsequence which is same as the
// given subarray is present or not
function checkSubsequence(arr, L,
                             R, N)
{
    if (checkSubsequenceUtil(arr, L,
                             R, N))
    {
        document.write("YES\n");
    }
    else
    {
        document.write("NO\n");
    }
}
 
// Driver code
 
    let arr = [ 1, 7, 12, 1, 7,
                  5, 10, 11, 42 ];
    let N = arr.length;
    let L = 3, R = 6;
   
    // Function Call
    checkSubsequence(arr, L, R, N);
     
    // This code is contributed by souravghosh0416.
</script>

Output:

YES

Time Complexity: O(N)
Auxiliary Space: O(1)

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