Given 
and 
. The task is to find the number of binary strings of length n out of 2n such that they satisfy f(bit string) = k.
Where,
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
Examples:
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
Method 1 (Brute Force):
The simplest approach is to solve the problem recursively, by passing the current index, the value of f(bit string) formed till current index and the last bit we placed in the binary string formed till (current index – 1). If we reach the index where current index = n and the value of f(bit string) = k then we count this way else we dont.
Below is the implementation of the brute force approach:
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 bitsint 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 Codeint 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 bitsimport java.util.*;class solution{// Function to find the number of Bit Strings// of length N with K adjacent set bitsstatic 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 Codepublic 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 bitsdef 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 Coden = 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 bitsusing System;class GFG{// Function to find the number of Bit Strings// of length N with K adjacent set bitsstatic 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 Codepublic 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 |
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 bitsfunction 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> |
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 bitsfunction 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?> |
Output
Number of ways = 6
Method 2 (efficient): In method 1, there are overlapping subproblems to remove for which we can apply Dynamic Programming (Memoization).
To optimize method 1, we can apply memoization to the above recursive solution such that,
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
Below is the implementation of the efficient approach:
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 bitsint 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 Codeint 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 bitsclass solution{ static final int MAX=1000; // Function to find the number of Bit Strings// of length N with K adjacent set bitsstatic 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 Codepublic 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 bitsMAX = 1000 # Function to find the number of Bit Strings# of length N with K adjacent set bitsdef 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 noOfWaysn, k = 5, 2 """ dp[i][j][k] represents bit strings of length iwith 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 bitsclass GFG{static readonly int MAX=1000;// Function to find the number// of Bit Strings of length N// with K adjacent set bitsstatic 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 Codepublic 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 bitsvar MAX = 1000;// Function to find the number of Bit Strings// of length N with K adjacent set bitsfunction 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 Codevar 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> |
Output
Number of ways = 6
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a DP to store the solution of the subproblems and initialize it with 0.
- Initializing the base case, when length of bit string is 1
dp[1][0][1] = 1;
dp[1][0][0] = 1; - Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
- Now create a variable totalWays where totalWays = dp[n][k][0] + dp[n][k][1].
- Return the final solution stored in totalWays.
Implementation :
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 bitsint waysToKAdjacentSetBits(int n, int k) { // 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, 0, sizeof(dp)); // Initializing the base case // when length of bit string is 1 dp[1][0][1] = 1; dp[1][0][0] = 1; // Filling up the DP Table for(int i = 2; i <= n; i++) { for(int j = 0; j <= k; j++) { dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1]; if(j > 0) { dp[i][j][1] = dp[i-1][j-1][0]; } dp[i][j][1] += dp[i-1][j][1]; } } // Total number of ways int totalWays = dp[n][k][0] + dp[n][k][1]; return totalWays;}// Driver Codeint main() { int n = 5, k = 2; int totalWays = waysToKAdjacentSetBits(n, k); 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 bitsimport java.util.Arrays;public class Main {static final int MAX = 1000; // Function to find the number of Bit Strings// of length N with K adjacent set bitsstatic int waysToKAdjacentSetBits(int n, int k) { // 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]; for (int[][] array2D : dp) { for (int[] array1D : array2D) { Arrays.fill(array1D, 0); } } // Initializing the base case // when length of bit string is 1 dp[1][0][1] = 1; dp[1][0][0] = 1; // Filling up the DP Table for (int i = 2; i <= n; i++) { for (int j = 0; j <= k; j++) { dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1]; if (j > 0) { dp[i][j][1] = dp[i - 1][j - 1][0]; } dp[i][j][1] += dp[i - 1][j][1]; } } // Total number of ways int totalWays = dp[n][k][0] + dp[n][k][1]; return totalWays;}// Driver Codepublic static void main(String[] args) { int n = 5, k = 2; int totalWays = waysToKAdjacentSetBits(n, k); System.out.println("Number of ways = " + totalWays);}} |
Python3
# Function to find the number of Bit Strings# of length N with K adjacent set bitsdef waysToKAdjacentSetBits(n, k): # dp[i][j][k] represents bit strings of length i # with f(bit string) = j and last bit as k dp = [[[0 for i in range(2)] for j in range(k + 1)] for l in range(n + 1)] # Initializing the base case # when length of bit string is 1 dp[1][0][1] = 1 dp[1][0][0] = 1 # Filling up the DP Table for i in range(2, n + 1): for j in range(k + 1): dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1] if j > 0: dp[i][j][1] = dp[i - 1][j - 1][0] dp[i][j][1] += dp[i - 1][j][1] # Total number of ways totalWays = dp[n][k][0] + dp[n][k][1] return totalWays# Driver Codeif __name__ == '__main__': n = 5 k = 2 totalWays = waysToKAdjacentSetBits(n, k) print(f"Number of ways = {totalWays}") |
C#
using System;class MainClass { static int WaysToKAdjacentSetBits(int n, int k) { // dp[i][j][k] represents bit strings of length i // with f(bit string) = j and last bit as k int[,,] dp = new int[1000, 1000, 2]; // Initializing the base case // when length of bit string is 1 dp[1, 0, 1] = 1; dp[1, 0, 0] = 1; // Filling up the DP Table for(int i = 2; i <= n; i++) { for(int j = 0; j <= k; j++) { dp[i, j, 0] = dp[i-1, j, 0] + dp[i-1, j, 1]; if(j > 0) { dp[i, j, 1] = dp[i-1, j-1, 0]; } dp[i, j, 1] += dp[i-1, j, 1]; } } // Total number of ways int totalWays = dp[n, k, 0] + dp[n, k, 1]; return totalWays; } public static void Main() { int n = 5, k = 2; int totalWays = WaysToKAdjacentSetBits(n, k); Console.WriteLine("Number of ways = " + totalWays); }} |
Javascript
function waysToKAdjacentSetBits(n, k) { // dp[i][j][k] represents bit strings of length i // with f(bit string) = j and last bit as k const dp = new Array(n + 1).fill(0).map(() => new Array(k + 1).fill(0).map(() => new Array(2).fill(0))); // Initializing the base case // when length of bit string is 1 dp[1][0][1] = 1; dp[1][0][0] = 1; // Filling up the DP Table for (let i = 2; i <= n; i++) { for (let j = 0; j <= k; j++) { dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1]; if (j > 0) { dp[i][j][1] = dp[i - 1][j - 1][0]; } dp[i][j][1] += dp[i - 1][j][1]; } } // Total number of ways const totalWays = dp[n][k][0] + dp[n][k][1]; return totalWays;}// Driver Codeconst n = 5, k = 2;const totalWays = waysToKAdjacentSetBits(n, k);console.log("Number of ways =", totalWays); |
Output
Number of ways = 6
Time Complexity: O(n*k)
Auxiliary Space: O(n*k)
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