Given two non-negative integers N and K, return the number of binary strings of length N with exactly K ones placed only on even positions
Example:
Input: N = 6, K = 2 Output: 3 Explanation: [ "101000", "100010", "001010" ]Notes: 2 * K <= N <= 30
The core challenge of this problem is to generate binary strings of a given length N such that exactly K ones are placed at even positions. This problem is significant in combinatorial generation and has applications in areas like coding theory and binary sequence analysis.
Potential pitfalls include misunderstanding the even positions (0-based indexing) and ensuring that the number of ones does not exceed the available even positions.
To solve this problem, we need to consider the following steps:
A naive solution would involve generating all possible binary strings of length N and counting those that meet the criteria. However, this is not optimal due to the exponential number of possible strings.
An optimized approach involves combinatorial mathematics. Specifically, we can use the binomial coefficient to count the number of ways to choose K positions out of the available even positions.
Here is a step-by-step breakdown of the algorithm:
Math.floor(N / 2)
.// Function to calculate binomial coefficient
function binomialCoefficient(n, k) {
if (k > n) return 0;
if (k === 0 || k === n) return 1;
let res = 1;
for (let i = 0; i < k; i++) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
// Main function to count binary strings with K ones on even positions
function countBinaryStrings(N, K) {
// Calculate the number of even positions
const evenPositions = Math.floor(N / 2);
// Calculate the number of ways to place K ones in these even positions
return binomialCoefficient(evenPositions, K);
}
// Example usage
const N = 6;
const K = 2;
console.log(countBinaryStrings(N, K)); // Output: 3
The time complexity of the binomial coefficient calculation is O(K)
, as it involves a loop that runs K times. The space complexity is O(1)
since we are using a constant amount of extra space.
Potential edge cases include:
Examples:
N = 4, K = 0 -> Output: 1 N = 4, K = 3 -> Output: 0
To test the solution comprehensively, consider the following test cases:
Example test cases:
console.log(countBinaryStrings(6, 2)); // Output: 3 console.log(countBinaryStrings(4, 1)); // Output: 2 console.log(countBinaryStrings(4, 0)); // Output: 1 console.log(countBinaryStrings(4, 3)); // Output: 0
When approaching such problems, consider the following tips:
In this blog post, we discussed how to count binary strings of length N with exactly K ones placed at even positions. We explored a combinatorial approach to solve the problem efficiently and provided a detailed explanation of the algorithm and its implementation in JavaScript.
Understanding and solving such problems is crucial for developing strong problem-solving skills and a deep understanding of combinatorial mathematics.
For further reading and practice, consider the following resources:
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