Find Unique Binary String

Find Unique Binary String - Problem

Given an array of strings nums containing n unique binary strings each of length n, return a binary string of length n that does not appear in nums.

If there are multiple answers, you may return any of them.

Input & Output

Example 1 — Basic Case

$ Input: nums = ["01","10"]

› Output: "00"

💡 Note: Using diagonal approach: flip bits at positions [0,1] from strings ["01","10"] → flip '0' to '1' and '1' to '0' → "10". Or we can return "00" or "11" as they're not in the input.

Example 2 — Single Element

$ Input: nums = ["0"]

› Output: "1"

💡 Note: Only one string "0" exists, so we return the other possible string "1".

Example 3 — Three Strings

$ Input: nums = ["111","011","001"]

› Output: "101"

💡 Note: Using diagonal: flip bits at positions [0,1,2] from ["111","011","001"] → flip '1','1','1' → "000". Any missing string like "101" is valid.

Constraints

n == nums.length
1 ≤ n ≤ 16
nums[i].length == n
nums[i] is either '0' or '1'
All the strings of nums are unique

Visualization

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Asked in

G Google 12 f Facebook 8 a Amazon 6

The key insight is using Cantor's diagonal argument: construct a string that differs from each input string at position i by flipping the i-th bit of the i-th string. Best approach is diagonal flipping with Time: O(n), Space: O(n).

Common Approaches

✓ Greedy

⏱️ Time: N/A Space: N/A

Hash Map

⏱️ Time: N/A Space: N/A

Backtracking Construction

⏱️ Time: O(n²) Space: O(n)

Construct the binary string character by character. At each position, try '0' then '1'. If the current partial string could match any input string, backtrack and try the other option.

Cantor's Diagonal Argument

⏱️ Time: O(n) Space: O(n)

Use Cantor's diagonal argument: for each position i, look at the i-th character of the i-th string and flip it. This guarantees the result differs from every input string in at least one position.

Generate All Possible Strings

⏱️ Time: O(n × 2^n) Space: O(n)

Create all possible binary strings of length n by counting from 0 to 2^n-1 in binary. Check each generated string against the input array until we find one that doesn't exist.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    // Find intersection using array as hash for digits 0-9
    int hash1[10] = {0};
    for (int i = 0; i < nums1Size; i++) {
        hash1[nums1[i]] = 1;
    }
    
    int intersection[10];
    int intersectionSize = 0;
    for (int i = 0; i < nums2Size; i++) {
        if (hash1[nums2[i]]) {
            intersection[intersectionSize++] = nums2[i];
        }
    }
    
    // If common digit exists, return the smallest one
    if (intersectionSize > 0) {
        int minIntersection = intersection[0];
        for (int i = 1; i < intersectionSize; i++) {
            if (intersection[i] < minIntersection) {
                minIntersection = intersection[i];
            }
        }
        return minIntersection;
    }
    
    // No common digits - combine two smallest
    int min1 = nums1[0];
    for (int i = 1; i < nums1Size; i++) {
        if (nums1[i] < min1) {
            min1 = nums1[i];
        }
    }
    
    int min2 = nums2[0];
    for (int i = 1; i < nums2Size; i++) {
        if (nums2[i] < min2) {
            min2 = nums2[i];
        }
    }
    
    int option1 = min1 * 10 + min2;
    int option2 = min2 * 10 + min1;
    return (option1 < option2) ? option1 : option2;
}

int main() {
    char line1[1000], line2[1000];
    int nums1[100], nums2[100];
    int nums1Size = 0, nums2Size = 0;
    
    // Parse first array
    fgets(line1, sizeof(line1), stdin);
    char* ptr = strchr(line1, '[') + 1;
    char* token = strtok(ptr, ",]");
    while (token != NULL) {
        nums1[nums1Size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse second array
    fgets(line2, sizeof(line2), stdin);
    ptr = strchr(line2, '[') + 1;
    token = strtok(ptr, ",]");
    while (token != NULL) {
        nums2[nums2Size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(nums1, nums1Size, nums2, nums2Size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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