Dado 
y 
. La tarea es encontrar el número de strings binarias de longitud n de 2 n tal que satisfagan f(string de bits) = k.
Dónde,
f(x) = Number of times a set bit is adjacent to
another set bit in a binary string x.
For Example:
f(011101101) = 3
f(010100000) = 0
f(111111111) = 8
Ejemplos :
Input : n = 5, k = 2 Output : 6 Explanation There are 6 ways to form bit strings of length 5 such that f(bit string s) = 2, These possible strings are:- 00111, 01110, 10111, 11011, 11100, 11101
Método 1 (Fuerza bruta): El enfoque más simple es resolver el problema recursivamente, pasando el índice actual, el valor de f (string de bits) formado hasta el índice actual y el último bit que colocamos en la string binaria formada hasta (índice actual – 1). Si alcanzamos el índice donde el índice actual = n y el valor de f (string de bits) = k , contamos de esta manera, de lo contrario no lo hacemos.
A continuación se muestra la implementación del enfoque de fuerza bruta:
C++
// C++ program to find the number of Bit Strings
// of length N with K adjacent set bits
#include <bits/stdc++.h>
using namespace std;
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
int waysToKAdjacentSetBits(int n, int k, int currentIndex,
int adjacentSetBits, int lastBit)
{
/* Base Case when we form bit string of length n */
if (currentIndex == n) {
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
int noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1) {
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k,currentIndex + 1,
adjacentSetBits, 0);
}
else if (!lastBit) {
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 0);
}
return noOfWays;
}
// Driver Code
int main()
{
int n = 5, k = 2;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
int totalWays = waysToKAdjacentSetBits(n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(n, k, 1, 0, 0);
cout << "Number of ways = " << totalWays << "\n";
return 0;
}
Java
// Java program to find the number of Bit Strings
// of length N with K adjacent set bits
import java.util.*;
class solution
{
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
static int waysToKAdjacentSetBits(int n, int k, int currentIndex,
int adjacentSetBits, int lastBit)
{
// Base Case when we form bit string of length n
if (currentIndex == n) {
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
int noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1) {
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k,currentIndex + 1,
adjacentSetBits, 0);
}
else if (lastBit == 0) {
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 0);
}
return noOfWays;
}
// Driver Code
public static void main(String args[])
{
int n = 5, k = 2;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
int totalWays = waysToKAdjacentSetBits(n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(n, k, 1, 0, 0);
System.out.println("Number of ways = "+totalWays);
}
}
//This code is contributed by
// Surendra _Gangwar
Python 3
# Python 3 program to find the number of Bit
# Strings of length N with K adjacent set bits
# Function to find the number of Bit Strings
# of length N with K adjacent set bits
def waysToKAdjacentSetBits(n, k, currentIndex,
adjacentSetBits, lastBit):
# Base Case when we form bit string of length n
if (currentIndex == n):
# if f(bit string) = k, count this way
if (adjacentSetBits == k):
return 1;
return 0
noOfWays = 0
# Check if the last bit was set, if it was set
# then call for next index by incrementing the
# adjacent bit count else just call the next
# index with same value of adjacent bit count
# and either set the bit at current index or
# let it remain unset
if (lastBit == 1):
# set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
# unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k,currentIndex + 1,
adjacentSetBits, 0);
elif (lastBit != 1):
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 0);
return noOfWays;
# Driver Code
n = 5; k = 2;
# total ways = (ways by placing 1st bit as 1 +
# ways by placing 1st bit as 0)
totalWays = (waysToKAdjacentSetBits(n, k, 1, 0, 1) +
waysToKAdjacentSetBits(n, k, 1, 0, 0));
print("Number of ways =", totalWays);
# This code is contributed by Akanksha Rai
C#
// C# program to find the number of Bit Strings
// of length N with K adjacent set bits
using System;
class GFG
{
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
static int waysToKAdjacentSetBits(int n, int k, int currentIndex,
int adjacentSetBits, int lastBit)
{
/* Base Case when we form bit
string of length n */
if (currentIndex == n)
{
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
int noOfWays = 0;
/* Check if the last bit was set, if it was
set then call for next index by incrementing
the adjacent bit count else just call the next
index with same value of adjacent bit count and
either set the bit at current index or let it
remain unset */
if (lastBit == 1)
{
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k,currentIndex + 1,
adjacentSetBits, 0);
}
else if (lastBit != 1)
{
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 0);
}
return noOfWays;
}
// Driver Code
public static void Main()
{
int n = 5, k = 2;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
int totalWays = waysToKAdjacentSetBits(n, k, 1, 0, 1) +
waysToKAdjacentSetBits(n, k, 1, 0, 0);
Console.WriteLine("Number of ways = " + totalWays);
}
}
// This code is contributed
// by Akanksha Rai
PHP
<?php
// PHP program to find the number of
// Bit Strings of length N with K
// adjacent set bits
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
function waysToKAdjacentSetBits($n, $k, $currentIndex,
$adjacentSetBits, $lastBit)
{
/* Base Case when we form bit
string of length n */
if ($currentIndex == $n)
{
// if f(bit string) = k, count this way
if ($adjacentSetBits == $k)
return 1;
return 0;
}
$noOfWays = 0;
/* Check if the last bit was set, if it
was set then call for next index by
incrementing the adjacent bit count else
just call the next index with same value
of adjacent bit count and either set the
bit at current index or let it remain
unset */
if ($lastBit == 1)
{
// set the bit at currentIndex
$noOfWays += waysToKAdjacentSetBits($n, $k, $currentIndex + 1,
$adjacentSetBits + 1, 1);
// unset the bit at currentIndex
$noOfWays += waysToKAdjacentSetBits($n, $k,$currentIndex + 1,
$adjacentSetBits, 0);
}
else if (!$lastBit)
{
$noOfWays += waysToKAdjacentSetBits($n, $k, $currentIndex + 1,
$adjacentSetBits, 1);
$noOfWays += waysToKAdjacentSetBits($n, $k, $currentIndex + 1,
$adjacentSetBits, 0);
}
return $noOfWays;
}
// Driver Code
$n = 5;
$k = 2;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
$totalWays = waysToKAdjacentSetBits($n, $k, 1, 0, 1) +
waysToKAdjacentSetBits($n, $k, 1, 0, 0);
echo "Number of ways = ", $totalWays, "\n";
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript program to find the number of Bit Strings
// of length N with K adjacent set bits
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
function waysToKAdjacentSetBits(n, k, currentIndex,
adjacentSetBits, lastBit)
{
/* Base Case when we form bit string of length n */
if (currentIndex == n) {
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
let noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1) {
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(n, k,currentIndex + 1,
adjacentSetBits, 0);
}
else if (!lastBit) {
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(n, k, currentIndex + 1,
adjacentSetBits, 0);
}
return noOfWays;
}
// Driver Code
let n = 5, k = 2;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
let totalWays = waysToKAdjacentSetBits(n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(n, k, 1, 0, 0);
document.write("Number of ways = " + totalWays);
</script>
Producción:
Number of ways = 6
Método 2 (eficiente): en el método 1, hay subproblemas superpuestos para eliminar para los cuales podemos aplicar Programación Dinámica (Memoización).
Para optimizar el método 1, podemos aplicar la memorización a la solución recursiva anterior de modo que,
DP[i][j][k] = Number of ways to form bit string of length i with
f(bit string till i) = j where the last bit is k,
which can be 0 or 1 depending on whether the
last bit was set or not
A continuación se muestra la implementación del enfoque eficiente:
C++
// C++ program to find the number of Bit Strings
// of length N with K adjacent set bits
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
int waysToKAdjacentSetBits(int dp[][MAX][2], int n, int k,
int currentIndex, int adjacentSetBits, int lastBit)
{
/* Base Case when we form bit
string of length n */
if (currentIndex == n) {
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
if (dp[currentIndex][adjacentSetBits][lastBit] != -1) {
return dp[currentIndex][adjacentSetBits][lastBit];
}
int noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1) {
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits, 0);
}
else if (!lastBit) {
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits, 0);
}
dp[currentIndex][adjacentSetBits][lastBit] = noOfWays;
return noOfWays;
}
// Driver Code
int main()
{
int n = 5, k = 2;
/* dp[i][j][k] represents bit strings of length i
with f(bit string) = j and last bit as k */
int dp[MAX][MAX][2];
memset(dp, -1, sizeof(dp));
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
int totalWays = waysToKAdjacentSetBits(dp, n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(dp, n, k, 1, 0, 0);
cout << "Number of ways = " << totalWays << "\n";
return 0;
}
Java
// Java program to find the number of Bit Strings
// of length N with K adjacent set bits
class solution
{
static final int MAX=1000;
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
static int waysToKAdjacentSetBits(int dp[][][], int n, int k,
int currentIndex, int adjacentSetBits, int lastBit)
{
/* Base Case when we form bit
string of length n */
if (currentIndex == n) {
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
if (dp[currentIndex][adjacentSetBits][lastBit] != -1) {
return dp[currentIndex][adjacentSetBits][lastBit];
}
int noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1) {
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits, 0);
}
else if (lastBit==0) {
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1,
adjacentSetBits, 0);
}
dp[currentIndex][adjacentSetBits][lastBit] = noOfWays;
return noOfWays;
}
// Driver Code
public static void main(String args[])
{
int n = 5, k = 2;
/* dp[i][j][k] represents bit strings of length i
with f(bit string) = j and last bit as k */
int dp[][][]= new int[MAX][MAX][2];
//initialize the dp
for(int i=0;i<MAX;i++)
for(int j=0;j<MAX;j++)
for(int k1=0;k1<2;k1++)
dp[i][j][k1]=-1;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
int totalWays = waysToKAdjacentSetBits(dp, n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(dp, n, k, 1, 0, 0);
System.out.print( "Number of ways = " + totalWays + "\n");
}
}
Python3
# Python3 program to find the number of Bit Strings # of length N with K adjacent set bits MAX = 1000 # Function to find the number of Bit Strings # of length N with K adjacent set bits def waysToKAdjacentSetBits(dp, n, k, currentIndex, adjacentSetBits, lastBit): """ Base Case when we form bit string of length n """ if currentIndex == n: # if f(bit string) = k, count this way if adjacentSetBits == k: return 1 return 0 if dp[currentIndex][adjacentSetBits][lastBit] != -1: return dp[currentIndex][adjacentSetBits][lastBit] noOfWays = 0 """ Check if the last bit was set, if it was set then call for next index by incrementing the adjacent bit count else just call the next index with same value of adjacent bit count and either set the bit at current index or let it remain unset """ if lastBit == 1: # set the bit at currentIndex noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1, adjacentSetBits + 1, 1) # unset the bit at currentIndex noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1, adjacentSetBits, 0) elif not lastBit: noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1, adjacentSetBits, 1) noOfWays += waysToKAdjacentSetBits(dp, n, k, currentIndex + 1, adjacentSetBits, 0) dp[currentIndex][adjacentSetBits][lastBit] = noOfWays return noOfWays n, k = 5, 2 """ dp[i][j][k] represents bit strings of length i with f(bit string) = j and last bit as k """ dp = [[[-1 for i in range(2)] for i in range(MAX)] for j in range(MAX)] """ total ways = (ways by placing 1st bit as 1 + ways by placing 1st bit as 0) """ totalWays = waysToKAdjacentSetBits(dp, n, k, 1, 0, 1) + waysToKAdjacentSetBits(dp, n, k, 1, 0, 0) print( "Number of ways =", totalWays) # This code is contributed by decode2207.
C#
using System;
// C# program to find the number
// of Bit Strings of length N
// with K adjacent set bits
class GFG
{
static readonly int MAX=1000;
// Function to find the number
// of Bit Strings of length N
// with K adjacent set bits
static int waysToKAdjacentSetBits(int [,,]dp,
int n, int k,
int currentIndex,
int adjacentSetBits,
int lastBit)
{
/* Base Case when we form bit
string of length n */
if (currentIndex == n)
{
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
if (dp[currentIndex, adjacentSetBits,
lastBit] != -1)
{
return dp[currentIndex,
adjacentSetBits, lastBit];
}
int noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1)
{
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n,
k, currentIndex + 1,
adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n,
k, currentIndex + 1,
adjacentSetBits, 0);
}
else if (lastBit==0)
{
noOfWays += waysToKAdjacentSetBits(dp,
n, k, currentIndex + 1,
adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(dp,
n, k, currentIndex + 1,
adjacentSetBits, 0);
}
dp[currentIndex,adjacentSetBits,lastBit] = noOfWays;
return noOfWays;
}
// Driver Code
public static void Main(String []args)
{
int n = 5, k = 2;
/* dp[i,j,k] represents bit strings
of length i with f(bit string) = j
and last bit as k */
int [,,]dp = new int[MAX, MAX, 2];
// initialize the dp
for(int i = 0; i < MAX; i++)
for(int j = 0; j < MAX; j++)
for(int k1 = 0; k1 < 2; k1++)
dp[i, j, k1]=-1;
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
int totalWays = waysToKAdjacentSetBits(dp, n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(dp, n, k, 1, 0, 0);
Console.Write( "Number of ways = " + totalWays + "\n");
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// JavaScript program to find the number of Bit Strings
// of length N with K adjacent set bits
var MAX = 1000;
// Function to find the number of Bit Strings
// of length N with K adjacent set bits
function waysToKAdjacentSetBits(dp, n, k,
currentIndex, adjacentSetBits, lastBit)
{
/* Base Case when we form bit
string of length n */
if (currentIndex == n) {
// if f(bit string) = k, count this way
if (adjacentSetBits == k)
return 1;
return 0;
}
if (dp[currentIndex][adjacentSetBits][lastBit] != -1) {
return dp[currentIndex][adjacentSetBits][lastBit];
}
var noOfWays = 0;
/* Check if the last bit was set,
if it was set then call for
next index by incrementing the
adjacent bit count else just call
the next index with same value of
adjacent bit count and either set the
bit at current index or let it remain
unset */
if (lastBit == 1) {
// set the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n, k,
currentIndex + 1, adjacentSetBits + 1, 1);
// unset the bit at currentIndex
noOfWays += waysToKAdjacentSetBits(dp, n, k,
currentIndex + 1, adjacentSetBits, 0);
}
else if (!lastBit) {
noOfWays += waysToKAdjacentSetBits(dp, n, k,
currentIndex + 1, adjacentSetBits, 1);
noOfWays += waysToKAdjacentSetBits(dp, n, k,
currentIndex + 1, adjacentSetBits, 0);
}
dp[currentIndex][adjacentSetBits][lastBit] = noOfWays;
return noOfWays;
}
// Driver Code
var n = 5, k = 2;
/* dp[i][j][k] represents bit strings of length i
with f(bit string) = j and last bit as k */
var dp = Array.from(Array(MAX), ()=>Array(MAX));
for(var i =0; i<MAX; i++)
for(var j =0; j<MAX; j++)
dp[i][j] = new Array(2).fill(-1);
/* total ways = (ways by placing 1st bit as 1 +
ways by placing 1st bit as 0) */
var totalWays = waysToKAdjacentSetBits(dp, n, k, 1, 0, 1)
+ waysToKAdjacentSetBits(dp, n, k, 1, 0, 0);
document.write( "Number of ways = " + totalWays + "<br>");
</script>
Producción:
Number of ways = 6