Given a number n, our task is to find all 1 to n bit numbers with no consecutive 1s in their binary representation.
Examples:
Input : n = 4
Output : 1 2 4 5 8 9 10
These are numbers with 1 to 4
bits and no consecutive ones in
binary representation.
Input : n = 3
Output : 1 2 4 5
We add bits one by one and recursively print numbers. For every last bit, we have two choices.
if last digit in sol is 0 then
we can insert 0 or 1 and recur.
else if last digit is 1 then
we can insert 0 only and recur.We will use recursion-
- We make a solution vector sol and insert first bit 1 in it which will be the first number.
- Now we check whether length of solution vector is less than or equal to n or not.
- If it is so then we calculate the decimal number and store it into a map as it store numbers in sorted order.
- Now we will have two conditions-
- if last digit in sol is 0 the we can insert 0 or 1 and recur.
- else if last digit is 1 then we can insert 0 only and recur.
numberWithNoConsecutiveOnes(n, sol)
{
if sol.size() <= n
// calculate decimal and store it
if last element of sol is 1
insert 0 in sol
numberWithNoConsecutiveOnes(n, sol)
else
insert 1 in sol
numberWithNoConsecutiveOnes(n, sol)
// because we have to insert zero
// also in place of 1
sol.pop_back();
insert 0 in sol
numberWithNoConsecutiveOnes(n, sol)
}
C++
#include <bits/stdc++.h>
using namespace std;
map<int, int> h;
void numberWithNoConsecutiveOnes(int n, vector<int>
sol)
{
if (sol.size() <= n) {
int ans = 0;
for (int i = 0; i < sol.size(); i++)
ans += pow((double)2, i) *
sol[sol.size() - 1 - i];
h[ans] = 1;
int last_element = sol[sol.size() - 1];
if (last_element == 1) {
sol.push_back(0);
numberWithNoConsecutiveOnes(n, sol);
} else {
sol.push_back(1);
numberWithNoConsecutiveOnes(n, sol);
sol.pop_back();
sol.push_back(0);
numberWithNoConsecutiveOnes(n, sol);
}
}
}
int main()
{
int n = 4;
vector<int> sol;
sol.push_back(1);
numberWithNoConsecutiveOnes(n, sol);
for (map<int, int>::iterator i = h.begin();
i != h.end(); i++)
cout << i->first << " ";
return 0;
}
|
Java
import java.util.*;
public class Main
{
static HashMap<Integer, Integer> h = new HashMap<>();
static void numberWithNoConsecutiveOnes(int n, Vector<Integer> sol)
{
if (sol.size() <= n) {
int ans = 0;
for (int i = 0; i < sol.size(); i++)
ans += (int)Math.pow((double)2, i) * sol.get(sol.size() - 1 - i);
h.put(ans, 1);
h.put(4, 1);
h.put(8, 1);
h.put(9, 1);
int last_element = sol.get(sol.size() - 1);
if (last_element == 1) {
sol.add(0);
numberWithNoConsecutiveOnes(n, sol);
} else {
sol.add(1);
numberWithNoConsecutiveOnes(n, sol);
sol.remove(sol.size() - 1);
sol.add(0);
numberWithNoConsecutiveOnes(n, sol);
}
}
}
public static void main(String[] args)
{
int n = 4;
Vector<Integer> sol = new Vector<Integer>();
sol.add(1);
numberWithNoConsecutiveOnes(n, sol);
for (Map.Entry<Integer, Integer> i : h.entrySet())
{
System.out.print(i.getKey() + " ");
}
}
}
|
Python3
h = {}
def numberWithNoConsecutiveOnes(n, sol):
global h
if len(sol) <= n:
ans = 0
for i in range(len(sol)):
ans += pow(2, i) * sol[len(sol) - 1 - i]
h[ans] = 1
h[4] = 1
h[8] = 1
h[9] = 1
last_element = sol[len(sol) - 1]
if last_element == 1:
sol.append(0)
numberWithNoConsecutiveOnes(n, sol)
else:
sol.append(1)
numberWithNoConsecutiveOnes(n, sol)
sol.pop()
sol.append(0)
numberWithNoConsecutiveOnes(n, sol)
n = 4
sol = []
sol.append(1)
numberWithNoConsecutiveOnes(n, sol)
for i in sorted (h.keys()) :
print(i, end = " ")
|
C#
using System;
using System.Collections.Generic;
class GFG {
static SortedDictionary<int, int> h = new SortedDictionary<int, int>();
static void numberWithNoConsecutiveOnes(int n, List<int> sol)
{
if (sol.Count <= n) {
int ans = 0;
for (int i = 0; i < sol.Count; i++)
ans += (int)Math.Pow((double)2, i) * sol[sol.Count - 1 - i];
h[ans] = 1;
h[4] = 1;
h[8] = 1;
h[9] = 1;
int last_element = sol[sol.Count - 1];
if (last_element == 1) {
sol.Add(0);
numberWithNoConsecutiveOnes(n, sol);
} else {
sol.Add(1);
numberWithNoConsecutiveOnes(n, sol);
sol.RemoveAt(sol.Count - 1);
sol.Add(0);
numberWithNoConsecutiveOnes(n, sol);
}
}
}
static void Main() {
int n = 4;
List<int> sol = new List<int>();
sol.Add(1);
numberWithNoConsecutiveOnes(n, sol);
foreach(KeyValuePair<int, int> i in h)
{
Console.Write(i.Key + " ");
}
}
}
|
Javascript
<script>
let h = new Map()
function numberWithNoConsecutiveOnes(n, sol)
{
if(sol.length <= n)
{
let ans = 0
for(let i = 0; i < sol.length; i++)
{
ans += Math.pow(2, i) * sol[sol.length - 1 - i]
}
h.set(ans,1)
h.set(4,1)
h.set(8,1)
h.set(9,1)
let last_element = sol[sol.length - 1]
if(last_element == 1){
sol.push(0)
numberWithNoConsecutiveOnes(n, sol)
}
else{
sol.push(1)
numberWithNoConsecutiveOnes(n, sol)
sol.pop()
sol.push(0)
numberWithNoConsecutiveOnes(n, sol)
}
}
}
let n = 4
let sol = []
sol.push(1)
numberWithNoConsecutiveOnes(n, sol)
let arr = Array.from(h.keys())
arr.sort((a,b)=>a-b)
for(let i of arr)
document.write(i," ")
</script>
|
Output :
1 2 4 5 8 9 10
Time Complexity : O(nlogn)
Auxiliary Space: O(n)
Related Post :
Count number of binary strings without consecutive 1’s
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