Given a range of number [L, R], the task is to find all numbers X in the given range such that X = sum of digits raised to setbit count of X i.e., if there are N setbits in binary representation of X and X = x1x2x3… then X = (x1)N + (x2)N + (x3)N + . . .
Examples:
Input: L = 0, R = 10000
Output: 1, 2, 4, 8, 4150, 9474
Explanation: For 2 (binary = 10): setbit count = 1. and 2 = 2^1.
So this is a required number. Same for the other numbers also.Input: L = 10000, R = 1000000
Output: -1
Explanation: There are no such numbers in the given range.
Approach: The given problem can be solved by checking for all numbers in the range [L, R] and if they satisfy the condition or not. It can be done with the help of Brian Kernighan’s Algorithm.
Follow the steps mentioned below to solve the problem.
- Run a loop from L to R and in every iteration check the number is index number or not.
- First calculate number of set bits in the decimal number from its binary representation.
- Then, Initialize Original = N, Res = 0, Index = Count of Set bits.
- Run a loop while N > 0
- Find last digit from the number say (L),
- Add LIndex to Res.
- Remove last digit from the number.
- If Original = Res, this will be one of the required numbers.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;
#define ll long long// Function to return Number of// set bits in any decimal numberint countSetBits(ll N)
{ int Count = 0;
while (N) {
N = N & (N - 1);
Count++;
}
return Count;
}// Function to check whether the// number is index number or notbool check(int Index, ll N)
{ ll Original = N, Res = 0;
if(N == 0)
return false;
while (N != 0) {
int L = N % 10;
Res += pow(L, Index);
N = N / 10;
}
return Original == Res;
}// Function to find the numbersvector<int> findNum(int l, int r)
{ // Vector to store the numbers
vector<int> ans;
for (ll i = l; i <= r; i++) {
int BitCount = countSetBits(i);
if (check(BitCount, i))
ans.push_back(i);
}
return ans;
}// Driver Codeint main()
{ int L = 0, R = 10000;
// Function call
vector<int> res = findNum(L, R);
if(res.size()==0)
cout << -1 << endl;
for (int x : res)
cout << x << " ";
return 0;
} |
Java
// Java program for the above approachimport java.util.*;
class GFG {
// Function to return Number of
// set bits in any decimal number
static int countSetBits(long N)
{
int Count = 0;
while (N != 0) {
N = N & (N - 1);
Count++;
}
return Count;
}
// Function to check whether the
// number is index number or not
static boolean check(int Index, long N)
{
long Original = N, Res = 0;
if(N == 0)
return false;
while (N != 0) {
long L = N % 10;
Res += Math.pow(L, Index);
N = N / 10;
}
return Original == Res;
}
// Function to find the numbers
static Vector<Integer> findNum(int l, int r)
{
// Vector to store the numbers
Vector<Integer> ans = new Vector<Integer>();
for (int i = l; i <= r; i++) {
int BitCount = countSetBits(i);
if (check(BitCount, i))
ans.add(i);
}
return ans;
}
// Driver Code
public static void main (String[] args) {
int L = 0, R = 10000;
// Function call
Vector<Integer> res = findNum(L, R);
if(res.size()==0)
System.out.println(-1);
res.forEach((x) -> System.out.print(x + " "));
}
}// This code is contributed by hrithikgarg03188. |
Python3
# Python code to implement the above approach# Function to return Number of# set bits in any decimal numberdef countSetBits(N):
Count = 0
while (N):
N = N & (N - 1)
Count += 1
return Count
# Function to check whether the# number is index number or notdef check(Index, N):
Original,Res = N,0
if(N == 0):
return False
while (N != 0):
L = N % 10
Res += pow(L, Index)
N = N // 10
return Original == Res
# Function to find the numbersdef findNum(l, r):
# Vector to store the numbers
ans = []
for i in range(l,r + 1):
BitCount = countSetBits(i)
if (check(BitCount, i)):
ans.append(i)
return ans
# Driver CodeL,R = 0,10000
# Function callres = findNum(L, R)
if(len(res)==0):
print(-1)
for x in res:
print(x , end = " ")
# This code is contributed by shinjanpatra |
C#
// C# program for the above approachusing System;
using System.Collections.Generic;
public class GFG
{ // Function to return Number of
// set bits in any decimal number
public static int countSetBits(long N)
{
int Count = 0;
while (N != 0) {
N = N & (N - 1);
Count++;
}
return Count;
}
// Function to check whether the
// number is index number or not
public static bool check(int Index, long N)
{
long Original = N, Res = 0;
if(N == 0)
return false;
while (N != 0) {
long L = N % 10;
Res += (long)Math.Pow(L, Index);
N = N / 10;
}
return Original == Res;
}
// Function to find the numbers
public static List<int> findNum(int l, int r)
{
// Vector to store the numbers
List<int> ans = new List<int>();
for (int i = l; i <= r; i++) {
int BitCount = countSetBits(i);
if (check(BitCount, i))
ans.Add(i);
}
return ans;
}
// Driver Code
public static void Main (String[] args) {
int L = 0, R = 10000;
// Function call
List<int> res = findNum(L, R);
if(res.Count == 0)
Console.WriteLine(-1);
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
}// This code is contributed by phasing17 |
Javascript
<script> // JavaScript program for the above approach
// Function to return Number of
// set bits in any decimal number
const countSetBits = (N) => {
let Count = 0;
while (N) {
N = N & (N - 1);
Count++;
}
return Count;
}
// Function to check whether the
// number is index number or not
const check = (Index, N) => {
let Original = N, Res = 0;
while (N != 0) {
let L = N % 10;
Res += Math.pow(L, Index);
N = parseInt(N / 10);
}
return Original == Res;
}
// Function to find the numbers
const findNum = (l, r) => {
// Vector to store the numbers
let ans = [];
for (let i = l; i <= r; i++) {
let BitCount = countSetBits(i);
if (check(BitCount, i))
ans.push(i);
}
return ans;
}
// Driver Code
let L = 0, R = 10000;
// Function call
let res = findNum(L, R);
for (let x in res)
document.write(`${res[x]} `);
// This code is contributed by rakeshsahni</script> |
Output
1 2 4 8 4150 9474
Time Complexity: O(R * d) where d is the maximum number of bits in a number
Auxiliary Space: O(1)