Find Kth Bit in Nth Binary String

Find Kth Bit in Nth Binary String - Problem

Given two positive integers n and k, the binary string Sn is formed as follows:

S1 = "0"

Si = Si-1 + "1" + reverse(invert(Si-1)) for i > 1

Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).

For example, the first four strings in the above sequence are:

S1 = "0"
S2 = "011"
S3 = "0111001"
S4 = "011100110110001"

Return the kth bit in Sn. It is guaranteed that k is valid for the given n.

Input & Output

Example 1 — Basic Case

$ Input: n = 3, k = 1

› Output: 0

💡 Note: S3 = "0111001". The 1st bit is "0"

Example 2 — Middle Position

$ Input: n = 4, k = 11

› Output: 1

💡 Note: S4 = "011100110110001". The 11th bit is "1"

Example 3 — Second Half

$ Input: n = 2, k = 3

› Output: 1

💡 Note: S2 = "011". The 3rd bit is "1"

Constraints

1 ≤ n ≤ 20
1 ≤ k ≤ 2ⁿ - 1

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is recognizing the recursive structure: each Sn follows the pattern Sn-1 + "1" + reverse(invert(Sn-1)). Instead of building massive strings, we can use this pattern to directly navigate to the kth position. The optimal approach uses recursion to identify which section contains our target and applies position mapping with inversion when needed. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Build Complete String

⏱️ Time: O(2ⁿ) Space: O(2ⁿ)

Start with S1 = "0" and iteratively build each Si by concatenating Si-1 + "1" + reverse(invert(Si-1)) until we reach Sn. Then return the character at position k-1.

Recursive Pattern Recognition

⏱️ Time: O(n) Space: O(n)

Each string Si has a predictable structure: Si-1 + '1' + reverse(invert(Si-1)). The length of Si is 2^i - 1, with the middle character always being '1'. We can recursively determine which part contains our target position.

Brute Force - Build Complete String — Algorithm Steps

Start with S1 = "0"
For each i from 2 to n, build Si = Si-1 + "1" + reverse(invert(Si-1))
Return the kth character (index k-1) from Sn

Visualization

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Step-by-Step Walkthrough

Start with S1

S1 = "0"

Build S2

S2 = "0" + "1" + reverse(invert("0")) = "011"

Build S3

S3 = "011" + "1" + reverse(invert("011")) = "0111001"

Code -

solution.c — C

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

char* solution(int n, int k) {
    char* s = malloc(2000000);
    strcpy(s, "0");
    
    for (int i = 2; i <= n; i++) {
        int len = strlen(s);
        char* inverted = malloc(len + 1);
        
        // Invert
        for (int j = 0; j < len; j++) {
            inverted[j] = (s[j] == '0') ? '1' : '0';
        }
        inverted[len] = '\0';
        
        // Reverse
        for (int j = 0; j < len / 2; j++) {
            char temp = inverted[j];
            inverted[j] = inverted[len - 1 - j];
            inverted[len - 1 - j] = temp;
        }
        
        // Concatenate
        strcat(s, "1");
        strcat(s, inverted);
        free(inverted);
    }
    
    char* result = malloc(2);
    result[0] = s[k - 1];
    result[1] = '\0';
    free(s);
    return result;
}

int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    char* result = solution(n, k);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ)

Each string Si has length 2^i - 1, so building Sn requires O(2^n) operations

✓ Linear Growth

Space Complexity

O(2ⁿ)

Store the complete string Sn which has length 2^n - 1

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Build Complete String — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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