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Count numbers from a given range that can be expressed as sum of digits raised to the power of count of digits
  • Difficulty Level : Hard
  • Last Updated : 25 Mar, 2021

Given an array arr[] consisting of queries of the form {L, R}, the task for each query is to count the numbers in the range [L, R] that can be expressed as the sum of its digits raised to the power of count of digits.

Examples:

Input: arr[][] = {{8, 11}}
Output: 2
Explanation:
From the given range [1, 9], the numbers that can be expressed as the sum of its digit raised to the power of count of digits are: 

  1. 8: Sum of digits = 8, Count of digit = 1. Therefore, 81 is equal to the given number.
  2. 9: Sum of digits = 9, Count of digit = 1. Therefore, 91 is equal to the given number.

Therefore, the count of such numbers from the given range is 2.

Input: arr[][] = {{10, 100}, {1, 400}}
Output: 0 11 



Naive Approach: The simplest approach is to iterate over the range arr[i][0] to arr[i][1] for each query and print the count of such numbers. 

Time Complexity: O(Q*(R – L)*log10R), where R and L denotes the limits of the longest range.
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by precomputing and storing all the numbers whether they can be expressed as the sum of its digit raised to the power of count of digits or not. Finally, print the count for each query efficiently. 
Follow the steps below to solve the problem:

  • Initialize an auxiliary array, say ans[], to store at ans[i], whether i can be expressed as the sum of its digit raised to the power of count of digits.
  • Iterate over the range [1, 106] and update the array ans[] accordingly.
  • Convert the array ans[] to a prefix sum array.
  • Traverse the given array of queries arr[] and for each query {arr[i][0], arr[i][1]}, print the value of (ans[arr[i][1]] – ans[arr[i][1] – 1]) as the resultant count of numbers that can be expressed as the sum of its digit raised to the power of count of digits.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
#define R 100005
int arr[R];
 
// Function to check if a number N can be
// expressed as sum of its digits raised
// to the power of the count of digits
bool canExpress(int N)
{
    int temp = N;
 
    // Stores the number of digits
    int n = 0;
 
    while (N != 0) {
        N /= 10;
        n++;
    }
 
    // Stores the resultant number
    N = temp;
 
    int sum = 0;
 
    while (N != 0) {
        sum += pow(N % 10, n);
        N /= 10;
    }
 
    // Return true if both the
    // numbers are same
    return (sum == temp);
}
 
// Function to precompute and store
// for all numbers whether they can
// be expressed
void precompute()
{
    // Mark all the index which
    // are plus perfect number
    for (int i = 1; i < R; i++) {
 
        // If true, then update the
        // value at this index
        if (canExpress(i)) {
            arr[i] = 1;
        }
    }
 
    // Compute prefix sum of the array
    for (int i = 1; i < R; i++) {
        arr[i] += arr[i - 1];
    }
}
 
// Function to count array elements that
// can be expressed as the sum of digits
// raised to the power of count of digits
void countNumbers(int queries[][2], int N)
{
    // Precompute the results
    precompute();
 
    // Traverse the queries
    for (int i = 0; i < N; i++) {
 
        int L1 = queries[i][0];
        int R1 = queries[i][1];
 
        // Print the resultant count
        cout << (arr[R1] - arr[L1 - 1])
             << ' ';
    }
}
 
// Driver Code
int main()
{
    int queries[][2] = {
        { 1, 400 },
        { 1, 9 }
    };
    int N = sizeof(queries)
            / sizeof(queries[0]);
    countNumbers(queries, N);
 
    return 0;
}

C#




// C# program for the above approach
using System;
 
class GFG{
     
static int R = 100005;
static int[] arr = new int[R];
 
// Function to check if a number N can be
// expressed as sum of its digits raised
// to the power of the count of digits
public static bool canExpress(int N)
{
    int temp = N;
 
    // Stores the number of digits
    int n = 0;
 
    while (N != 0)
    {
        N /= 10;
        n++;
    }
 
    // Stores the resultant number
    N = temp;
 
    int sum = 0;
 
    while (N != 0)
    {
        sum += ((int)Math.Pow(N % 10, n));
        N /= 10;
    }
 
    // Return true if both the
    // numbers are same
      if (sum == temp)
      return true;
   
    return false;
}
 
// Function to precompute and store
// for all numbers whether they can
// be expressed
public static void precompute()
{
 
    // Mark all the index which
    // are plus perfect number
    for(int i = 1; i < R; i++)
    {
         
        // If true, then update the
        // value at this index
        if (canExpress(i))
        {
            arr[i] = 1;
        }
    }
 
    // Compute prefix sum of the array
    for(int i = 1; i < R; i++)
    {
        arr[i] += arr[i - 1];
    }
}
 
// Function to count array elements that
// can be expressed as the sum of digits
// raised to the power of count of digits
public static void countNumbers(int[,] queries, int N)
{
     
    // Precompute the results
    precompute();
 
    // Traverse the queries
    for(int i = 0; i < N; i++)
    {
        int L1 = queries[i, 0];
        int R1 = queries[i, 1];
 
        // Print the resultant count
        Console.Write((arr[R1] - arr[L1 - 1]) + " ");
    }
}
 
// Driver Code
static public void Main()
{
    int[,] queries = { { 1, 400 }, { 1, 9 } };
    int N = queries.GetLength(0);
 
    // Function call
    countNumbers(queries, N);
}
}
 
// This code is contributed by Dharanendra L V.

Python3




# Python 3 program for the above approach
R = 100005
arr = [0 for i in range(R)]
 
# Function to check if a number N can be
# expressed as sum of its digits raised
# to the power of the count of digits
def canExpress(N):
    temp = N
 
    # Stores the number of digits
    n = 0
    while (N != 0):
        N //= 10
        n += 1
 
    # Stores the resultant number
    N = temp
    sum = 0
    while (N != 0):
        sum += pow(N % 10, n)
        N //= 10
 
    # Return true if both the
    # numbers are same
    return (sum == temp)
 
# Function to precompute and store
# for all numbers whether they can
# be expressed
def precompute():
   
    # Mark all the index which
    # are plus perfect number
    for i in range(1, R, 1):
       
        # If true, then update the
        # value at this index
        if(canExpress(i)):
            arr[i] = 1
 
    # Compute prefix sum of the array
    for i in range(1,R,1):
        arr[i] += arr[i - 1]
 
# Function to count array elements that
# can be expressed as the sum of digits
# raised to the power of count of digits
def countNumbers(queries, N):
   
    # Precompute the results
    precompute()
 
    # Traverse the queries
    for i in range(N):
        L1 = queries[i][0]
        R1 = queries[i][1]
 
        # Print the resultant count
        print((arr[R1] - arr[L1 - 1]),end = " ")
 
# Driver Code
if __name__ == '__main__':
    queries = [[1, 400],[1, 9]]
    N = len(queries)
    countNumbers(queries, N)
 
    # This code is contributed by SURENDRA_GANGWAR.

Javascript




<script>
// javascript program for the above approach
    var R = 100005;
    var arr = Array(R).fill(0);
 
    // Function to check if a number N can be
    // expressed as sum of its digits raised
    // to the power of the count of digits
    function canExpress(N) {
        var temp = N;
 
        // Stores the number of digits
        var n = 0;
 
        while (N != 0) {
            N = parseInt(N/10);
            n++;
        }
 
        // Stores the resultant number
        N = temp;
 
        var sum = 0;
 
        while (N != 0) {
            sum += Math.pow(N % 10, n);
            N = parseInt(N/10);
        }
 
        // Return true if both the
        // numbers are same
        return (sum == temp);
    }
 
    // Function to precompute and store
    // for all numbers whether they can
    // be expressed
    function precompute() {
 
        // Mark all the index which
        // are plus perfect number
        for (var i = 1; i < R; i++) {
 
            // If true, then update the
            // value at this index
            if (canExpress(i)) {
                arr[i] = 1;
            }
        }
 
        // Compute prefix sum of the array
        for (var i = 1; i < R; i++) {
            arr[i] += arr[i - 1];
        }
    }
 
    // Function to count array elements that
    // can be expressed as the sum of digits
    // raised to the power of count of digits
    function countNumbers(queries , N) {
        // Precompute the results
        precompute();
 
        // Traverse the queries
        for (var i = 0; i < N; i++) {
 
            var L1 = queries[i][0];
            var R1 = queries[i][1];
 
            // Prvar the resultant count
            document.write((arr[R1] - arr[L1 - 1]) + " ");
        }
    }
 
    // Driver Code
    var queries = [ [ 1, 400 ], [ 1, 9 ] ];
    var N = queries.length;
 
    // function call
    countNumbers(queries, N);
 
// This code is contributed by Princi Singh
</script>
Output: 
12 9

 

Time Complexity: O(Q + 106
Auxiliary Space: O(106
 

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