Binary Prefix Divisible By 5 - Practice Coding Problems

Binary Prefix Divisible By 5 - Problem

You are given a binary array nums (0-indexed). We define x_i as the number whose binary representation is the subarray nums[0..i] (from most-significant-bit to least-significant-bit).

For example, if nums = [1,0,1], then x₀ = 1, x₁ = 2, and x₂ = 5.

Return an array of booleans answer where answer[i] is true if x_i is divisible by 5.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,0,1]

› Output: [false,false,true]

💡 Note: Binary prefixes: [1]→1 (not divisible by 5), [1,0]→2 (not divisible by 5), [1,0,1]→5 (divisible by 5)

Example 2 — All Zeros

$ Input: nums = [0,1,1]

› Output: [true,false,false]

💡 Note: Binary prefixes: [0]→0 (divisible by 5), [0,1]→1 (not divisible by 5), [0,1,1]→3 (not divisible by 5)

Example 3 — Longer Sequence

$ Input: nums = [1,1,1,0,1]

› Output: [false,false,false,false,false]

💡 Note: Binary prefixes: [1]→1, [1,1]→3, [1,1,1]→7, [1,1,1,0]→14, [1,1,1,0,1]→29. None are divisible by 5

Constraints

1 ≤ nums.length ≤ 3 × 10⁴
nums[i] is either 0 or 1

Visualization

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Understanding the Visualization

Input

Binary array [1,0,1]

Process

Check each prefix: [1]→1, [1,0]→2, [1,0,1]→5

Output

Boolean array showing divisibility by 5

Key Takeaway

🎯 Key Insight: Use modular arithmetic to track remainder % 5 instead of full decimal values

Asked in

G Google 15 a Amazon 12 f Facebook 8

The key insight is to use modular arithmetic to track only the remainder when divided by 5, avoiding integer overflow. Best approach uses modular arithmetic with formula: remainder = (remainder * 2 + bit) % 5. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Convert Each Prefix	O(n²)	O(1)	Convert each binary prefix to decimal and check divisibility by 5
Modular Arithmetic - Track Remainder Only	O(n)	O(1)	Use modular arithmetic to avoid integer overflow by only tracking remainder mod 5

Brute Force - Convert Each Prefix — Algorithm Steps

Iterate through each position i
Convert binary prefix nums[0..i] to decimal
Check if decimal value is divisible by 5

Visualization

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Step-by-Step Walkthrough

Binary Prefix

Extract prefix from start to current position

Convert to Decimal

Calculate decimal value of binary prefix

Check Divisibility

Test if decimal value is divisible by 5

Code -

solution.c — C

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

bool* solution(int* nums, int numsSize, int* returnSize) {
    bool* result = (bool*)malloc(numsSize * sizeof(bool));
    *returnSize = numsSize;
    
    for (int i = 0; i < numsSize; i++) {
        long long decimalValue = 0;
        for (int j = 0; j <= i; j++) {
            decimalValue = decimalValue * 2 + nums[j];
        }
        result[i] = (decimalValue % 5 == 0);
    }
    return result;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[100000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int returnSize;
    bool* result = solution(nums, numsSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf(result[i] ? "true" : "false");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each position i, we convert prefix of length i+1 to decimal

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to store current decimal value

✓ Linear Space

28.5K Views

Medium Frequency

~15 min Avg. Time

845 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Convert Each Prefix — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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