LeetCode 2450: Number of Distinct Binary Strings After Applying Operations Solution

Master LeetCode problem 2450 (Number of Distinct Binary Strings After Applying Operations), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2450. Number of Distinct Binary Strings After Applying Operations

Medium

Math String

Problem Explanation

Explanation:

Given a string of length n, we can apply two operations:

Swap: Choose any two indices i and j (0-indexed) and swap the characters at positions i and j.
Reverse: Reverse the string between indices i and j (inclusive).

Our goal is to find the number of distinct binary strings that can be obtained after applying a sequence of these operations.

To solve this problem, we can use the concept of permutation cycles. We can consider the binary string as a set of connected components where each component is a cycle. These cycles are formed by the positions of 1 in the binary string. By analyzing the cycles, we can determine the number of distinct binary strings that can be formed.

Algorithm:

Initialize a set to store the binary strings we have already encountered.
For each binary string, identify the cycles formed by the positions of 1.
Calculate the number of permutations for each cycle.
Multiply the number of permutations of all cycles to get the total number of distinct binary strings.
Return the count of distinct binary strings.

Time Complexity:

The time complexity of this algorithm is O(n) where n is the length of the binary string.

Space Complexity:

The space complexity is O(n) to store the set of encountered binary strings.

Solution Code

class Solution {
    public int numDifferentIntegers(String word) {
        Set<String> set = new HashSet<>();
        int n = word.length();
        
        for (int i = 0; i < n; i++) {
            if (Character.isDigit(word.charAt(i))) {
                int j = i;
                while (j < n && Character.isDigit(word.charAt(j))) {
                    j++;
                }
                while (i < j && word.charAt(i) == '0') {
                    i++;
                }
                set.add(word.substring(i, j));
                i = j;
            }
        }
        
        return set.size();
    }
}

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