LeetCode 1980 - Find Unique Binary String

The problem is asking us to find a binary string of length n that does not exist in a given list nums of unique binary strings, where each string in nums also has length n.

leetcodemediumarrayhash-tablestringbacktracking

LeetCode Problem 1980

Difficulty: 🟡 Medium
Topics: Array, Hash Table, String, Backtracking

Solution

Problem Understanding

The problem is asking us to find a binary string of length n that does not exist in a given list nums of unique binary strings, where each string in nums also has length n. In simpler terms, we are given n strings that consist of 0s and 1s, and we need to generate another string of the same length that is not already in the list. The order of the strings does not matter, and multiple valid answers are acceptable.

The input nums is guaranteed to have n unique binary strings, so there are exactly n strings of length n already in the set. The number of all possible binary strings of length n is 2^n. Since n can be at most 16, the total number of possible strings is up to 65536, which is small enough to consider full enumeration if needed. However, the constraints also allow us to design an optimal solution using clever observations rather than brute-force enumeration.

Important edge cases include: n = 1 (the smallest possible strings), n = 16 (largest length for performance testing), and situations where nums contains strings that cover sequences at the extremes (like "000...0" or "111...1"). The problem guarantees uniqueness of input strings, so we do not need to handle duplicates.

Approaches

The brute-force approach is straightforward. One can generate all 2^n possible binary strings of length n and check each one against the set of strings in nums. As soon as we find a string not present in the set, we return it. This guarantees correctness because all possible strings are considered, but it is inefficient since generating all 2^n strings grows exponentially with n and would be unnecessary when n is large.

The optimal approach leverages the concept of a diagonalization technique, inspired by Cantor's diagonal argument. For a list of strings of length n, consider the i-th character of the i-th string. If we flip that character (0 to 1 or 1 to 0), the resulting string cannot exist in the original list because it differs from every string in at least one position. This approach guarantees uniqueness and runs in linear time with respect to n.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * 2^n) O(n * 2^n) Generate all possible strings and check membership
Optimal O(n) O(1) Use diagonalization to construct a guaranteed unique string

Algorithm Walkthrough

  1. Initialize an empty result string that will store the new binary string.
  2. Iterate over the list of strings using index i from 0 to n-1.
  3. For each string at index i, take the character at position i.
  4. Flip the character: if it is '0', append '1' to the result; if it is '1', append '0'.
  5. Continue this process for all n characters.
  6. After finishing the loop, return the result string.

Why it works: The diagonalization ensures that the resulting string differs from the i-th string at the i-th position for all i. Therefore, it cannot match any string in the original list. This guarantees that the string is unique.

Python Solution

from typing import List

class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:
        n = len(nums)
        result = []
        for i in range(n):
            # Flip the i-th character of the i-th string
            result.append('1' if nums[i][i] == '0' else '0')
        return ''.join(result)

In this implementation, we first determine the length n of the input list. We then iterate over each string, flipping the diagonal character to build a new string. The join function converts the list of characters into a final string. This directly implements the diagonalization algorithm.

Go Solution

func findDifferentBinaryString(nums []string) string {
    n := len(nums)
    result := make([]byte, n)
    for i := 0; i < n; i++ {
        if nums[i][i] == '0' {
            result[i] = '1'
        } else {
            result[i] = '0'
        }
    }
    return string(result)
}

In Go, we use a byte slice to build the resulting string. The diagonalization approach is identical, flipping the i-th character of the i-th string. Finally, we convert the byte slice to a string before returning.

Worked Examples

Example 1: nums = ["01","10"]

i nums[i][i] flipped result
0 '0' '1' "1"
1 '0' '1' "11"

Result: "11" is not in nums.

Example 2: nums = ["00","01"]

i nums[i][i] flipped result
0 '0' '1' "1"
1 '1' '0' "10"

Result: "10" is not in nums.

Example 3: nums = ["111","011","001"]

i nums[i][i] flipped result
0 '1' '0' "0"
1 '1' '0' "00"
2 '1' '0' "000"

Result: "000" is not in nums.

Complexity Analysis

Measure Complexity Explanation
Time O(n) We iterate once over all strings and flip one character per string
Space O(n) The result string of length n is constructed

The diagonalization method is optimal because it guarantees uniqueness while avoiding the exponential cost of generating all possible strings.

Test Cases

# Provided examples
assert Solution().findDifferentBinaryString(["01","10"]) in ["00", "11"]  # Example 1
assert Solution().findDifferentBinaryString(["00","01"]) in ["10", "11"]  # Example 2
assert Solution().findDifferentBinaryString(["111","011","001"]) in ["000","010","100","101","110"]  # Example 3

# Boundary cases
assert Solution().findDifferentBinaryString(["0"]) == "1"  # n = 1, single string
assert Solution().findDifferentBinaryString(["1"]) == "0"  # n = 1, single string

# Larger n
nums = [format(i, '016b') for i in range(16)]
assert Solution().findDifferentBinaryString(nums) not in nums  # n = 16
Test Why
["01","10"] Validates a small n=2 case
["00","01"] Checks alternate solution output
["111","011","001"] Validates n=3, multiple correct answers
["0"], ["1"] Minimum length edge cases
n = 16 sequential strings Tests larger input size, ensures algorithm scales

Edge Cases

Case 1: n = 1

When n is 1, there is only one character in the string. Flipping it ensures uniqueness. This case could break naive implementations that assume multiple characters.

Case 2: All zeros or all ones in nums

If nums contains "000" or "111", a naive approach may accidentally generate the same string. Diagonalization guarantees at least one bit is flipped, avoiding collisions.

Case 3: Maximum n (n = 16)

Even with the largest input, the algorithm runs efficiently because it only flips one character per string. This ensures O(n) time rather than attempting to enumerate 2^16 strings.

This approach handles all edge cases reliably.