Subarray of length K whose concatenation forms a palindrome
Last Updated :
01 Oct, 2021
Given an array arr[], consisting of N integers in the range [0, 9], the task is to find a subarray of length K from which we can generate a number which is a Palindrome Number. If no such subarray exists, print -1.
Note: The elements in the array are in the range of 0 to 10.
Examples:
Input: arr[] = {1, 5, 3, 2, 3, 5, 4}, K = 5
Output: 5, 3, 2, 3, 5
Explanation:
Number generated by concatenating all elements of the subarray, i.e. 53235, is a palindrome.
Input: arr[] = {2, 3, 5, 1, 3}, K = 4
Output: -1
Naive Approach: The simplest approach to solve the problem is to generate all subarrays of length K and for each subarray, concatenate all the elements from the subarray and check if the number formed is a Palindrome Number or not.
Time Complexity: O(N3)
Auxiliary Space: O(K)
Efficient Approach: The problem can be solved using the Window-Sliding technique. Follow the steps below to solve the problem:
- Make a palindrome function to check if the given subarray (Window-Sliding) is palindrome or not.
- Iterate over the array, and for each subarray call the palindrome function.
- If found to be true, return the starting index of that subarray, and print the array from starting index to the next k index.
- If no such subarray found which is a palindrome, print -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number
// is Palindrome or not
// here i is the starting index
// and j is the last index of the subarray
bool palindrome(vector<int> a, int i, int j)
{
while(i<j)
{
// If the integer at i is not equal to j
// then the subarray is not palindrome
if(a[i] != a[j])
return false;
// Otherwise
i++;
j--;
}
// all a[i] is equal to a[j]
// then the subarray is palindrome
return true;
}
// Function to find a subarray whose
// concatenation forms a palindrome
// and return its starting index
int findSubArray(vector<int> arr, int k)
{
int n= sizeof(arr)/sizeof(arr[0]);
// Iterating over subarray of length k
// and checking if that subarray is palindrome
for(int i=0; i<=n-k; i++){
if(palindrome(arr, i, i+k-1))
return i;
}
// If no subarray is palindrome
return -1;
}
// Driver Code
int main()
{
vector<int> arr = { 2, 3, 5, 1, 3 };
int k = 4;
int ans = findSubArray(arr, k);
if (ans == -1)
cout << -1 << "\n";
else {
for (int i = ans; i < ans + k;
i++)
cout << arr[i] << " ";
cout << "\n";
}
return 0;
}
// This code is contributed by Prafulla Shekhar
// Java program for the above approach
import java.io.*;
class GFG
{
// Function to check if a number
// is Palindrome or not
// here i is the starting index
// and j is the last index of the subarray
public static boolean palindrome(int[] a, int i, int j)
{
while(i<j)
{
// If the integer at i is not equal to j
// then the subarray is not palindrome
if(a[i] != a[j])
return false;
// Otherwise
i++;
j--;
}
// all a[i] is equal to a[j]
// then the subarray is palindrome
return true;
}
// Function to find a subarray whose
// concatenation forms a palindrome
// and return its starting index
static int findSubArray(int []arr, int k)
{
int n= arr.length;
// Iterating over subarray of length k
// and checking if that subarray is palindrome
for(int i=0; i<=n-k; i++){
if(palindrome(arr, i, i+k-1))
return i;
}
// If no subarray is palindrome
return -1;
}
// Driver code
public static void main (String[] args)
{
int []arr = { 2, 3, 5, 1, 3 };
int k = 4;
int ans = findSubArray(arr, k);
if (ans == -1)
System.out.print(-1 + "\n");
else
{
for(int i = ans; i < ans + k; i++)
System.out.print(arr[i] + " ");
System.out.print("\n");
}
}
}
// This code is contributed by Prafulla Shekhar
# Python3 program for the above approach
# Function to check if a number
# is Palindrome or not here i is
# the starting index and j is the
# last index of the subarray
def palindrome(a, i, j):
while(i < j):
# If the integer at i is not equal to j
# then the subarray is not palindrome
if (a[i] != a[j]):
return False
# Otherwise
i += 1
j -= 1
# all a[i] is equal to a[j]
# then the subarray is palindrome
return True
# Function to find a subarray whose
# concatenation forms a palindrome
# and return its starting index
def findSubArray(arr, k):
n = len(arr)
# Iterating over subarray of length k
# and checking if that subarray is palindrome
for i in range(n - k + 1):
if (palindrome(arr, i, i + k - 1)):
return i
return -1
# Driver code
arr = [ 2, 3, 5, 1, 3 ]
k = 4
ans = findSubArray(arr, k)
if (ans == -1):
print(-1)
else:
for i in range(ans,ans + k):
print(arr[i], end = " ")
# This code is contributed by avanitrachhadiya2155
// C# program for the above approach
using System;
class GFG{
// Function to check if a number
// is Palindrome or not here i is
// the starting index and j is the
// last index of the subarray
public static bool palindrome(int[] a, int i,
int j)
{
while (i < j)
{
// If the integer at i is not equal to j
// then the subarray is not palindrome
if (a[i] != a[j])
return false;
// Otherwise
i++;
j--;
}
// All a[i] is equal to a[j]
// then the subarray is palindrome
return true;
}
// Function to find a subarray whose
// concatenation forms a palindrome
// and return its starting index
static int findSubArray(int[] arr, int k)
{
int n = arr.Length;
// Iterating over subarray of length k
// and checking if that subarray is palindrome
for(int i = 0; i <= n - k; i++)
{
if (palindrome(arr, i, i + k - 1))
return i;
}
// If no subarray is palindrome
return -1;
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 2, 3, 5, 1, 3 };
int k = 4;
int ans = findSubArray(arr, k);
if (ans == -1)
Console.Write(-1 + "\n");
else
{
for(int i = ans; i < ans + k; i++)
Console.Write(arr[i] + " ");
Console.Write("\n");
}
}
}
// This code is contributed by aashish1995
<script>
// Javascript program for
// the above approach
// Function to check if a number
// is Palindrome or not
// here i is the starting index
// and j is the last index of the subarray
function palindrome(a, i, j)
{
while(i<j)
{
// If the integer at i is not equal to j
// then the subarray is not palindrome
if(a[i] != a[j])
return false;
// Otherwise
i++;
j--;
}
// all a[i] is equal to a[j]
// then the subarray is palindrome
return true;
}
// Function to find a subarray whose
// concatenation forms a palindrome
// and return its starting index
function findSubArray(arr, k)
{
let n= arr.length;
// Iterating over subarray of length k
// and checking if that subarray is palindrome
for(let i=0; i<=n-k; i++){
if(palindrome(arr, i, i+k-1))
return i;
}
// If no subarray is palindrome
return -1;
}
// Driver Code
let arr = [ 2, 3, 5, 1, 3 ];
let k = 4;
let ans = findSubArray(arr, k);
if (ans == -1)
document.write(-1 + "\n");
else
{
for(let i = ans; i < ans + k; i++)
document.write(arr[i] + " ");
document.write("<br/>");
}
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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