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Count numbers from a range whose cube is a Palindrome

  • Difficulty Level : Expert
  • Last Updated : 04 Mar, 2021

Given an array Q[][] consisting of N queries of the form {L, R}, the task for each query is to find the total count of numbers from the range [L, R], whose cube is a palindrome.

Examples:

Input: Q[][] = {{2, 10}, {10, 20}}
Output: 
2
1
Explanation: 
Query 1: The numbers from the range [2, 10], whose cube is a palindrome are 2, 7
Query 2: The only number from the range [10, 20], whose cube is a palindrome is 11  

Input: Q[][] = {{1, 50}, {13, 15}}
Output: 
4
0
Explanation:
Query 1: The numbers from the range [1, 50], whose cube is a palindrome are {1, 2, 7, 11}.
Query 2: No such number exists in the range [13, 15].

 

Approach: The simplest idea to solve the problem is to use Inclusion-Exclusion Principle and Prefix Sum Array technique to solve this problem. Follow the steps below to solve the given problem:



  • Initialize an array arr[], to store at every ith index, whether cube of i is a palindrome or not.
  • Traverse the array arr[] and for every ith index,  check if the cube of i is a palindrome or not.
    • If found to be true, then set arr[i] = 1.
    • Otherwise, set arr[i] = 0.
  • Convert the array arr[] to a prefix sum array.
  • Traverse the array Q[][], and count the number from the range [L, R] whose cube is a palindrome by calculating arr[R] – arr[L-1].

Below is the implementation of the above approach.

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
int arr[10005];
 
// Function to check if n is
// a pallindrome number or not
int isPalindrome(int n)
{
    // Temporarily store n
    int temp = n;
 
    // Stores reverse of n
    int res = 0;
 
    // Iterate until temp reduces to 0
    while (temp != 0) {
        // Extract the last digit
        int rem = temp % 10;
 
        // Add to the start
        res = res * 10 + rem;
 
        // Remove the last digit
        temp /= 10;
    }
 
    // If the number and its
    // reverse are equal
    if (res == n) {
        return 1;
    }
 
    // Otherwise
    else
        return 0;
}
 
// Function to precompute and store
// the count of numbers whose cube
// is a palindrome number
void precompute()
{
    // Iterate upto 10^4
    for (int i = 1; i <= 10000; i++) {
 
        // Check if i*i*i is a
        // pallindrome or not
        if (isPalindrome(i * i * i))
            arr[i] = 1;
        else
            arr[i] = 0;
    }
 
    // Convert arr[] to prefix sum array
    for (int i = 1; i <= 10000; i++) {
        arr[i] = arr[i] + arr[i - 1];
    }
}
 
// Driver Code
int main()
{
    // Given queries
    vector<pair<int, int> > Q = { { 2, 7 }, { 10, 25 } };
 
    precompute();
 
    for (auto it : Q) {
 
        // Using inclusion-exclusion
        // principle, count required numbers
        cout << arr[it.second] - arr[it.first - 1] << "\n";
    }
 
    return 0;
}

Java




// Java program of the above approach
import java.io.*;
import java.util.*;
public class Pair {
  private final int key;
  private final int value;
 
  public Pair(int aKey, int aValue)
  {
    key = aKey;
    value = aValue;
  }
 
  public int key() { return key; }
  public int value() { return value; }
}
 
class GFG {
  static int[] arr = new int[10005];
 
  // Function to check if n is
  // a pallindrome number or not
  static int isPalindrome(int n)
  {
     
    // Temporarily store n
    int temp = n;
 
    // Stores reverse of n
    int res = 0;
 
    // Iterate until temp reduces to 0
    while (temp != 0)
    {
       
      // Extract the last digit
      int rem = temp % 10;
 
      // Add to the start
      res = res * 10 + rem;
 
      // Remove the last digit
      temp /= 10;
    }
 
    // If the number and its
    // reverse are equal
    if (res == n) {
      return 1;
    }
 
    // Otherwise
    else
      return 0;
  }
 
  // Function to precompute and store
  // the count of numbers whose cube
  // is a palindrome number
  static void precompute()
  {
     
    // Iterate upto 10^4
    for (int i = 1; i <= 10000; i++) {
 
      // Check if i*i*i is a
      // pallindrome or not
      if (isPalindrome(i * i * i)!= 0)
        arr[i] = 1;
      else
        arr[i] = 0;
    }
 
    // Convert arr[] to prefix sum array
    for (int i = 1; i <= 10000; i++) {
      arr[i] = arr[i] + arr[i - 1];
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
     
    // Given queries
    ArrayList<Pair> Q = new ArrayList<Pair>();
    Pair pair = new Pair(2, 7);
    Q.add(pair);
    Pair pair2 = new Pair(10, 25);
    Q.add(pair2);
 
    precompute();
 
    for (int i = 0; i < Q.size(); i++)  {
 
      // Using inclusion-exclusion
      // principle, count required numbers
      System.out.println(arr[Q.get(i).value()] - arr[Q.get(i).key()-1]);
    }
  }
}
 
// This code is contributed by Dharanendra L V

 
 

Output: 
2
1

 

Time Complexity: O(N)
Auxiliary Space: O(maxm), where maxm denotes the maximum value of R in a query 

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