长度为 N 且至少有 X 0 和 Y 1 的二进制字符串的计数

给定三个数字N、X和Y ，找出长度为N且至少包含X 0和Y 1的唯一二进制字符串的计数。

例子：

Input: N=5, X=1, Y=2
Output: 25

Input: N=3, X=1, Y=1
Output: 6
Explanation: There are 3 binary strings of length 3 with at least 1 0s and 1 1s, such as: 001, 010, 100, 011, 101, 110

天真的方法：生成所有长度为N的二进制字符串，然后计算至少具有X 0和Y 1的字符串的数量。
时间复杂度： O(2^N)
辅助空间： O(1)

更好的方法：这个问题也可以使用组合数学来解决。如果长度是N ，并且给定的是X 0s ，那么将有Y (=NX) 1s 。所以我们需要为此找到唯一组合的数量，可以得到(N)C(X)或(N)C(Y)。现在对于所有唯一的二进制字符串，我们需要找到[X, NY]范围内i的值的nCi并将其添加到变量中。在所有迭代之后这个总和的值将是所需的计数。

高效方法：上述方法可以在帕斯卡三角形的帮助下进一步优化以计算nCr 。 请按照以下步骤解决问题：

• 初始化二维数组p[][]以使用帕斯卡三角形进行计算。
• 将变量sum初始化为0以存储答案。
• 使用变量i遍历范围[x, ny]并执行以下任务：
  • 将值p[n][i]添加到变量sum 。
• 执行上述步骤后，打印sum的值作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
long long int p[31][31];
 
// Function to use pascal triangle
void pascalTriangle()
{
    p[0][0] = 1;
    p[1][0] = 1;
    p[1][1] = 1;
    for (int i = 2; i < 31; i++) {
        p[i][0] = 1;
        for (int j = 1; j < i; j++)
            p[i][j] = p[i - 1][j]
                      + p[i - 1][j - 1];
        p[i][i] = 1;
    }
}
 
// Function to count the total number of ways
long long int countWays(int n, int x, int y)
{
 
    // Store the answer
    long long int sum = 0;
 
    // Traverse
    for (long long int i = x; i <= n - y; i++) {
        sum += p[n][i];
    }
    return sum;
}
 
// Driver Code
int main()
{
    pascalTriangle();
 
    int N = 5, X = 1, Y = 2;
    cout << countWays(N, X, Y) << endl;
 
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG {
 
    static int[][] p = new int[31][31];
 
    // Function to use pascal triangle
    static void pascalTriangle()
    {
        p[0][0] = 1;
        p[1][0] = 1;
        p[1][1] = 1;
        for (int i = 2; i < 31; i++) {
            p[i][0] = 1;
            for (int j = 1; j < i; j++)
                p[i][j] = p[i - 1][j] + p[i - 1][j - 1];
            p[i][i] = 1;
        }
    }
 
    // Function to count the total number of ways
    static long countWays(int n, int x, int y)
    {
 
        // Store the answer
        long sum = 0;
 
        // Traverse
        for (int i = x; i <= n - y; i++) {
            sum += p[n][i];
        }
        return sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        pascalTriangle();
 
        int N = 5;
        int X = 1;
        int Y = 2;
 
        System.out.println(countWays(N, X, Y));
    }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python program for the above approach
p = [[0 for i in range(31)] for i in range(31)]
 
# Function to use pascal triangle
def pascalTriangle():
    p[0][0] = 1
    p[1][0] = 1
    p[1][1] = 1
    for i in range(2, 31):
        p[i][0] = 1
        for j in range(1, i):
            p[i][j] = p[i - 1][j] + p[i - 1][j - 1]
        p[i][i] = 1
 
# Function to count the total number of ways
def countWays(n, x, y):
 
    # Store the answer
    sum = 0
 
    # Traverse
    for i in range(x, n - y + 1):
        sum += p[n][i]
    return sum
 
# Driver Code
pascalTriangle()
 
N = 5
X = 1
Y = 2
print(countWays(N, X, Y))
 
# This code is contributed by gfgking

C#

// C# code for the above approach
using System;
 
public class GFG {
 
    static int[,] p = new int[31,31];
 
    // Function to use pascal triangle
    static void pascalTriangle()
    {
        p[0,0] = 1;
        p[1,0] = 1;
        p[1,1] = 1;
        for (int i = 2; i < 31; i++) {
            p[i,0] = 1;
            for (int j = 1; j < i; j++)
                p[i,j] = p[i - 1,j] + p[i - 1,j - 1];
            p[i,i] = 1;
        }
    }
 
    // Function to count the total number of ways
    static long countWays(int n, int x, int y)
    {
 
        // Store the answer
        long sum = 0;
 
        // Traverse
        for (int i = x; i <= n - y; i++) {
            sum += p[n,i];
        }
        return sum;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        pascalTriangle();
 
        int N = 5;
        int X = 1;
        int Y = 2;
 
        Console.WriteLine(countWays(N, X, Y));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

    // JavaScript program for the above approach
 
    let p = new Array(31).fill(0).map(() => new Array(31).fill(0));
 
    // Function to use pascal triangle
    const pascalTriangle = () => {
        p[0][0] = 1;
        p[1][0] = 1;
        p[1][1] = 1;
        for (let i = 2; i < 31; i++) {
            p[i][0] = 1;
            for (let j = 1; j < i; j++)
                p[i][j] = p[i - 1][j]
                    + p[i - 1][j - 1];
            p[i][i] = 1;
        }
    }
 
    // Function to count the total number of ways
    const countWays = (n, x, y) => {
 
        // Store the answer
        let sum = 0;
 
        // Traverse
        for (let i = x; i <= n - y; i++) {
            sum += p[n][i];
        }
        return sum;
    }
 
    // Driver Code
 
    pascalTriangle();
 
    let N = 5, X = 1, Y = 2;
    document.write(countWays(N, X, Y));
 
// This code is contributed by rakeshsahni
 

输出

25

时间复杂度： O(N)
辅助空间： O(1)