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 subarraybool 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 indexint 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 Codeint 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
// Java program for the above approachimport 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
# 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 subarraydef 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 = 4ans = 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#
// C# program for the above approachusing 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 subarraypublic 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 indexstatic 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 codepublic 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 |
Javascript
<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> |
Output
-1
Time Complexity: O(N)
Auxiliary Space: O(1)
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