# Count of binary string of length N with at least X 0s and Y 1s

• Last Updated : 14 Dec, 2021

Given three numbers N, X, and Y, find the count of unique binary strings of length N having at least X 0s and Y 1s.

Examples:

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

Naive approach: Generate all binary strings of length N and then count the number of strings with at least X 0s and Y 1s.
Time Complexity: O(2^N)
Auxiliary Space: O(1)

Better Approach: This problem can also be solved using Combinatorics. If the length is N, and given is X 0s, then there will be Y (=N-X) 1s. So we need to find the number of unique combinations for this, which can be obtained as (N)C(X) or (N)C(Y). Now for all unique binary strings, we need to find the nCi for values of i in the range [X, N-Y] and add it to a variable. The value of this sum after all iterations will be the required count.

Efficient Approach: The above approach can further be optimized with the help of the Pascal triangle to calculate nCr. Follow the steps below to solve the problem:

• Initialize the 2-D array p[][] to calculate using the pascal triangle.
• Initialize the variable sum as 0 to store the answer.
• Iterate over the range [x, n-y] using the variable i and perform the following tasks:
• Add the value p[n][i] to the variable sum.
• After performing the above steps, print the value of sum as the answer.

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `long` `long` `int` `p;`   `// Function to use pascal triangle` `void` `pascalTriangle()` `{` `    ``p = 1;` `    ``p = 1;` `    ``p = 1;` `    ``for` `(``int` `i = 2; i < 31; i++) {` `        ``p[i] = 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

 ``

Output

`25`

Time Complexity: O(N)
Auxiliary Space: O(1)

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