Binary String With Substrings Representing 1 To N

Binary String With Substrings Representing 1 To N - Problem

Given a binary string s and a positive integer n, return true if the binary representation of all the integers in the range [1, n] are substrings of s, or false otherwise.

A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — All Found

$ Input: s = "0110", n = 3

› Output: true

💡 Note: Binary representations: 1="1", 2="10", 3="11". All are substrings of "0110": "1" at index 1, "10" at index 1, "11" at index 2.

Example 2 — Missing Number

$ Input: s = "0110", n = 4

› Output: false

💡 Note: Binary of 4 is "100". The string "0110" does not contain "100" as a substring, so return false.

Example 3 — Single Digit

$ Input: s = "1", n = 1

› Output: true

💡 Note: Only need to check if binary of 1 ("1") exists in "1", which it does.

Constraints

1 ≤ s.length ≤ 1000
s[i] is either '0' or '1'
1 ≤ n ≤ 10⁹

Visualization

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Asked in

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The key insight is to check if binary representations of all numbers 1 to n exist as substrings. The optimized approach checks from larger numbers first for potential early termination. Time: O(n × m), Space: O(log n).

Common Approaches

✓ Brute Force Search

⏱️ Time: O(n × m) Space: O(log n)

For each number from 1 to n, convert it to binary representation and check if this binary string exists as a substring in s. This approach checks every number sequentially.

Reverse Order Search

⏱️ Time: O(n × m) Space: O(log n)

Instead of checking from 1 to n, check from n down to 1. Larger numbers have longer binary representations and are more likely to be missing, allowing for earlier termination.

Brute Force Search — Algorithm Steps

Convert each number i (1 to n) to binary string
Check if binary string exists as substring in s
Return false if any number's binary is not found

Visualization

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Step-by-Step Walkthrough

Convert to Binary

Transform each number 1 to n into binary

Search String

Check if binary exists as substring

Result

Return false if any binary not found

Code -

solution.c — C

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

void toBinary(int num, char* binary) {
    int index = 0;
    if (num == 0) {
        binary[0] = '0';
        binary[1] = '\0';
        return;
    }
    
    char temp[32];
    int tempIndex = 0;
    while (num > 0) {
        temp[tempIndex++] = (num % 2) + '0';
        num /= 2;
    }
    
    for (int i = tempIndex - 1; i >= 0; i--) {
        binary[index++] = temp[i];
    }
    binary[index] = '\0';
}

int solution(char* s, int n) {
    for (int i = 1; i <= n; i++) {
        char binaryRep[32];
        toBinary(i, binaryRep);
        if (strstr(s, binaryRep) == NULL) {
            return 0;
        }
    }
    return 1;
}

int main() {
    char s[1000];
    int n;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &n);
    int result = solution(s, n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

For each of n numbers, we search in string of length m

✓ Linear Growth

Space Complexity

O(log n)

Space for binary representation of largest number

⚡ Linearithmic Space

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Related Problems

Common Approaches

Brute Force Search — Algorithm Steps

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Code -

Time & Space Complexity

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