Sponsored
Sponsored
This approach builds a unique binary string by considering each string's diagonal to ensure uniqueness. Specifically, we take advantage of the fact that flipping a bit diagonally guarantees that it differs from the original set of strings across at least one position. For each string at index i, we flip the character at the same index i, ensuring that the constructed string cannot match any existing strings due to its diagonal differentiation.
Time Complexity: O(n) since we iterate through the array once.
Space Complexity: O(n) for storing the resulting binary string of length n.
1public String findDifferentBinaryString(String[] nums) {
2 int n = nums.length;
3 char[] result = new char[n];
4 for (int i = 0; i < n; i++) {
5 result[i] = nums[i].charAt(i) == '0' ? '1' : '0';
6 }
7 return new String(result);
8}The Java solution constructs a binary string by iterating over each index i, checks the diagonal position nums[i][i], and flips the bit. It constructs the answer using a character array for efficiency.
This approach uses backtracking to generate all possible binary strings of length n and checks for the first one not present in nums. While not as efficient as diagonal construction, this method demonstrates a different strategy to ensure a solution's validity by exhaustive search.
Time Complexity: O(2^n) due to generating every binary string of length n.
Space Complexity: O(n) for the call stack during recursion.
1function findDifferentBinaryString(nums) {
2
The JavaScript solution utilizes recursive backtracking to test each potential binary string until it finds one not included in nums. Each recursive step generates a binary extension by adding a '0' or '1', and depth-first searching ensures uniqueness.