给定正整数N ,任务是对长度为N的所有可能的不同二进制字符串进行计数,以使没有连续的1。
例子:
Input: N = 5
Output: 5
Explanation:
The non-negative integers <= 5 with their corresponding binary representations are:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only 3 has two consecutive 1’s. Therefore required count = 5
Input: N = 12
Output: 8
动态编程方法:本文已经讨论了动态编程方法。
方法:在本文中,讨论了一种使用digit-dp概念的方法。
- 类似于digit-dp问题,此处创建了一个3维表来存储计算值。假定N <2 31 – 1,并且每个数字的范围仅为2(0或1)。因此,桌子的尺寸取为32 x 2 x 2 。
- 构造表后,给定的数字将转换为二进制字符串。
- 然后,对数字进行迭代。对于每次迭代:
- 检查前一位数字是0还是1。
- 如果为0,则当前数字可以为0或1。
- 但是,如果前一个数字为1,则当前数字必须为0,因为在二进制表示中不能有两个连续的1。
- 现在,该表已完全填充,就像digit-dp问题一样。
下面是上述方法的实现
C++
// C++ program to count number of
// binary strings without consecutive 1’s
#include
using namespace std;
// Table to store the solution of
// every sub problem
int memo[32][2][2];
// Function to fill the table
/* Here,
pos: keeps track of current position.
f1: is the flag to check if current
number is less than N or not.
pr: represents the previous digit
*/
int dp(int pos, int fl, int pr, string& bin)
{
// Base case
if (pos == bin.length())
return 1;
// Check if this subproblem
// has already been solved
if (memo[pos][fl][pr] != -1)
return memo[pos][fl][pr];
int val = 0;
// Placing 0 at the current position
// as it does not violate the condition
if (bin[pos] == '0')
val = val + dp(pos + 1, fl, 0, bin);
// Here flag will be 1 for the
// next recursive call
else if (bin[pos] == '1')
val = val + dp(pos + 1, 1, 0, bin);
// Placing 1 at this position only if
// the previously inserted number is 0
if (pr == 0) {
// If the number is smaller than N
if (fl == 1) {
val += dp(pos + 1, fl, 1, bin);
}
// If the digit at current position is 1
else if (bin[pos] == '1') {
val += dp(pos + 1, fl, 1, bin);
}
}
// Storing the solution to this subproblem
return memo[pos][fl][pr] = val;
}
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
int findIntegers(int num)
{
// Convert N to binary form
string bin;
// Loop to convert N
// from Decimal to binary
while (num > 0) {
if (num % 2)
bin += "1";
else
bin += "0";
num /= 2;
}
reverse(bin.begin(), bin.end());
// Initialising the table with -1.
memset(memo, -1, sizeof(memo));
// Calling the function
return dp(0, 0, 0, bin);
}
// Driver code
int main()
{
int N = 12;
cout << findIntegers(N);
return 0;
}
Java
// Java program to count number of
// binary Strings without consecutive 1’s
class GFG{
// Table to store the solution of
// every sub problem
static int [][][]memo = new int[32][2][2];
// Function to fill the table
/* Here,
pos: keeps track of current position.
f1: is the flag to check if current
number is less than N or not.
pr: represents the previous digit
*/
static int dp(int pos, int fl, int pr, String bin)
{
// Base case
if (pos == bin.length())
return 1;
// Check if this subproblem
// has already been solved
if (memo[pos][fl][pr] != -1)
return memo[pos][fl][pr];
int val = 0;
// Placing 0 at the current position
// as it does not violate the condition
if (bin.charAt(pos) == '0')
val = val + dp(pos + 1, fl, 0, bin);
// Here flag will be 1 for the
// next recursive call
else if (bin.charAt(pos) == '1')
val = val + dp(pos + 1, 1, 0, bin);
// Placing 1 at this position only if
// the previously inserted number is 0
if (pr == 0) {
// If the number is smaller than N
if (fl == 1) {
val += dp(pos + 1, fl, 1, bin);
}
// If the digit at current position is 1
else if (bin.charAt(pos) == '1') {
val += dp(pos + 1, fl, 1, bin);
}
}
// Storing the solution to this subproblem
return memo[pos][fl][pr] = val;
}
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
static int findIntegers(int num)
{
// Convert N to binary form
String bin = "";
// Loop to convert N
// from Decimal to binary
while (num > 0) {
if (num % 2 == 1)
bin += "1";
else
bin += "0";
num /= 2;
}
bin = reverse(bin);
// Initialising the table with -1.
for(int i = 0; i < 32; i++){
for(int j = 0; j < 2; j++){
for(int l = 0; l < 2; l++)
memo[i][j][l] = -1;
}
}
// Calling the function
return dp(0, 0, 0, bin);
}
static String reverse(String input) {
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver code
public static void main(String[] args)
{
int N = 12;
System.out.print(findIntegers(N));
}
}
// This code is contributed by sapnasingh4991
Python 3
# Python program to count number of
# binary strings without consecutive 1’s
# Table to store the solution of
# every sub problem
memo=[[[-1 for i in range(2)] for j in range(2)] for k in range(32)]
# Function to fill the table
''' Here,
pos: keeps track of current position.
f1: is the flag to check if current
number is less than N or not.
pr: represents the previous digit
'''
def dp(pos,fl,pr,bin):
# Base case
if (pos == len(bin)):
return 1;
# Check if this subproblem
# has already been solved
if (memo[pos][fl][pr] != -1):
return memo[pos][fl][pr];
val = 0
# Placing 0 at the current position
# as it does not violate the condition
if (bin[pos] == '0'):
val = val + dp(pos + 1, fl, 0, bin)
# Here flag will be 1 for the
# next recursive call
elif (bin[pos] == '1'):
val = val + dp(pos + 1, 1, 0, bin)
# Placing 1 at this position only if
# the previously inserted number is 0
if (pr == 0):
# If the number is smaller than N
if (fl == 1):
val += dp(pos + 1, fl, 1, bin)
# If the digit at current position is 1
elif (bin[pos] == '1'):
val += dp(pos + 1, fl, 1, bin)
# Storing the solution to this subproblem
memo[pos][fl][pr] = val
return val
# Function to find the number of integers
# less than or equal to N with no
# consecutive 1’s in binary representation
def findIntegers(num):
# Convert N to binary form
bin=""
# Loop to convert N
# from Decimal to binary
while (num > 0):
if (num % 2):
bin += "1"
else:
bin += "0"
num //= 2
bin=bin[::-1];
# Calling the function
return dp(0, 0, 0, bin)
# Driver code
if __name__ == "__main__":
N = 12
print(findIntegers(N))
C#
// C# program to count number of
// binary Strings without consecutive 1’s
using System;
public class GFG{
// Table to store the solution of
// every sub problem
static int [,,]memo = new int[32,2,2];
// Function to fill the table
/* Here,
pos: keeps track of current position.
f1: is the flag to check if current
number is less than N or not.
pr: represents the previous digit
*/
static int dp(int pos, int fl, int pr, String bin)
{
// Base case
if (pos == bin.Length)
return 1;
// Check if this subproblem
// has already been solved
if (memo[pos,fl,pr] != -1)
return memo[pos,fl,pr];
int val = 0;
// Placing 0 at the current position
// as it does not violate the condition
if (bin[pos] == '0')
val = val + dp(pos + 1, fl, 0, bin);
// Here flag will be 1 for the
// next recursive call
else if (bin[pos] == '1')
val = val + dp(pos + 1, 1, 0, bin);
// Placing 1 at this position only if
// the previously inserted number is 0
if (pr == 0) {
// If the number is smaller than N
if (fl == 1) {
val += dp(pos + 1, fl, 1, bin);
}
// If the digit at current position is 1
else if (bin[pos] == '1') {
val += dp(pos + 1, fl, 1, bin);
}
}
// Storing the solution to this subproblem
return memo[pos,fl,pr] = val;
}
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
static int findints(int num)
{
// Convert N to binary form
String bin = "";
// Loop to convert N
// from Decimal to binary
while (num > 0) {
if (num % 2 == 1)
bin += "1";
else
bin += "0";
num /= 2;
}
bin = reverse(bin);
// Initialising the table with -1.
for(int i = 0; i < 32; i++){
for(int j = 0; j < 2; j++){
for(int l = 0; l < 2; l++)
memo[i,j,l] = -1;
}
}
// Calling the function
return dp(0, 0, 0, bin);
}
static String reverse(String input) {
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver code
public static void Main(String[] args)
{
int N = 12;
Console.Write(findints(N));
}
}
// This code contributed by Rajput-Ji
输出:
8
时间复杂度: O(L * log(N))
- O(log(N))将数字从十进制转换为二进制。
- O(L)填充表格,其中L是二进制形式的长度。