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Count binary strings with k times appearing adjacent two set bits in C++
We are given with integers N and K. We have binary strings of length N, containing 0’s and 1’s only. The goal is to find the number of such strings of length N that have K consecutive 1’s. I.e, if N=3, and K=2, then count all 3-digit binary strings possible that have 2 consecutive 1’s.
Example − 111, here adjacent 1’s appear twice ( K times ).
In 011 and 110 adjacent 1’s appeared once only.
We will solve this by storing the results of previous values.
Taking 3D array count[x][y][z]. Where x is N, y is K and z is last digit of string ( 0 or 1 )
- For N=1, strings will be “0” and “1”. Count of adjacent 1’s=0.
So for any K. If N=1, count=0.
count[1][K][0] = count[1][K][1] = 0.
- When last digit is 0, To make adjacent 1’s count as K
All digits of length N-1 with K ones + last 0 → new length = N
If the count of K adjacent 1’s was C then adding 0 at the end won’t change that count.
count[N][K][0] = count[N-1][K][0] + count[N-1][K][1]
- When last digit is 1, To make adjacent 1’s count as K
All digits of length N-1 ending with 0 with K ones + last 1 → new length = N,
count[N-1][K][0]
All digits of length N-1 ending with 1 with K-1 ones + last 1 → new length = N,
count[N-1][K-1][1]
count[N][K][1] = count[N-1][K][0] + count[N-1][K-1][1]
Total such strings= count[N][K][0] + count[N][K][1]
Input
N=4, K=2
Output
Count of strings: 2
Explanation − 1110, 0111 are the only strings of length 4 where adjacent 1’s appear twice only.
1110- “11 10”, then “1 11 0” ( splitting to show which adjacent 1’s are counted ) 0111- “0 11 1”, then “01 11”. K=2, adjacent 1’s= 2 times
Input
N=3, K=1
Output
Count of strings: 2
Explanation − 110, 011. are the only strings of length 3where adjacent 1’s appear once.
In 111 adjacent 1’s appear twice.
Approach used in the below program is as follows
Integers N and K store the length of strings and no. of times adjacent 1’s appear in each.
Function stringcount(int n, int k) takes n and k as parameters and returns the count of strings with K times adjacent 1’s.
Array count[i][j][0/1] is used to store the count of strings of length i with j adjacent 1’s ending with 0 or 1.
Initial condition is count[1][0][0] = 1; count[1][0][1] = 1;
Now start from 2 length strings ( i=2) to n length. For each such string check count of K times adjacent 1’s. j=0;j<=k, from previous counts. Update current count[i][j][0] and count[i][j][1].
If j-1>=0, the count of adjacent 1’s is greater than 1. Then update count of strings ending with 1 as count[i][j][1] + count[i - 1][j - 1][1];
At the end add count[n][k][0] and count[n][k][1]. Store it as a result.
- Return the result as desired count.
Example
#include <bits/stdc++.h>
using namespace std;
int stringcount(int n, int k){
//count of strings with length=n and adajcent 1's=k
int count[n + 1][k + 1][2]={0};
//count[n][k][0] -- strings with length=n and adajcent 1's=k last character is 0
//count[n][k][1] -- strings with length=n and adajcent 1's=k last character is 1
// If n = 1 and k = 0.
count[1][0][0] = 1;
count[1][0][1] = 1;
for (int i = 2; i <= n; i++) {
// number of adjacent 1's can not exceed i-1
for (int j = 0; j <= k; j++) {
count[i][j][0] = count[i - 1][j][0] + count[i - 1][j][1];
count[i][j][1] = count[i - 1][j][0];
if (j - 1 >= 0)
count[i][j][1] =count[i][j][1] + count[i - 1][j - 1][1];
}
}
int result=count[n][k][0]+count[n][k][1];
return result;
}
int main(){
int N = 6, K = 3;
cout <<"Strings of length 6 and 3 adjacent 1's in each :"<< stringcount(N,K);
return 0;
}Output
Strings of length 6 and 3 adjacent 1's in each :7
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