Partitioning Into Minimum Number Of Deci-Binary Numbers

Partitioning Into Minimum Number Of Deci-Binary Numbers - Problem

String Greedy Medium

A deci-binary number is a special decimal number where each digit is either 0 or 1 (no leading zeros allowed). Think of it as a binary-looking number that we treat as decimal.

Examples of deci-binary numbers: 101, 1100, 1, 10

NOT deci-binary: 112 (contains digit 2), 3001 (contains digit 3), 0101 (leading zero)

🎯 Your Challenge: Given a string n representing a positive decimal integer, find the minimum number of positive deci-binary numbers that sum up to n.

Goal: Return the smallest count of deci-binary numbers needed to create the target sum.

Input & Output

example_1.py — Basic Case

$ Input: n = "32"

› Output: 3

💡 Note: We can partition 32 into three deci-binary numbers: 10 + 11 + 11 = 32. The maximum digit is 3, so we need exactly 3 deci-binary numbers.

example_2.py — Larger Number

$ Input: n = "82734"

› Output: 8

💡 Note: The maximum digit in 82734 is 8. We need exactly 8 deci-binary numbers. For example: 11111 + 11111 + 11111 + 11111 + 11111 + 11111 + 10111 + 10101 = 82734.

example_3.py — Edge Case

$ Input: n = "27346209830709182346"

› Output: 9

💡 Note: Even for very large numbers, we only need to find the maximum digit. Here it's 9, so the answer is 9.

Constraints

1 ≤ n.length ≤ 10⁵
n consists of only digits.
n does not contain any leading zeros and represents a positive integer.

Visualization

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Understanding the Visualization

Identify target heights

Each digit in the input represents the target height for that position's tower.

Understand block constraints

Each deci-binary 'block' can only add 0 or 1 height to any tower position.

Find the bottleneck

The tallest tower determines how many blocks we need minimum.

Prove optimality

We can always construct exactly max-digit blocks to achieve any target configuration.

Key Takeaway

🎯 Key Insight: The maximum digit is always the answer because it represents the bottleneck position that requires the most deci-binary contributions.

Asked in

G Google 45 a Amazon 38 f Meta 29 ⊞ Microsoft 22 Apple 15

This problem has an elegant greedy solution: the answer is simply the maximum digit in the input string. Since each deci-binary number can contribute at most 1 to any digit position, we need exactly as many deci-binary numbers as the largest digit requires.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(2^(10^n))	O(2^n)	Generate all possible deci-binary combinations until we find the minimum count
Greedy Maximum Digit (Optimal)	O(n)	O(1)	The answer is simply the maximum digit in the string

Brute Force (Try All Combinations) — Algorithm Steps

Generate all possible deci-binary numbers up to the length of input
Use backtracking to try all combinations
For each combination, check if sum equals target
Keep track of minimum count found
Return the minimum count

Visualization

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Step-by-Step Walkthrough

Generate deci-binary numbers

Create all possible deci-binary numbers: 1, 10, 11, 100, 101, 110, 111...

Try all combinations

Use backtracking to try every possible sum combination

Track minimum

Keep the smallest count that achieves the target sum

Code -

solution.c — C

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

int isDeciBinary(int num) {
    while (num > 0) {
        int digit = num % 10;
        if (digit != 0 && digit != 1) return 0;
        num /= 10;
    }
    return 1;
}

int backtrack(int remaining, int count, int start) {
    if (remaining == 0) return count;
    
    int minCount = INT_MAX;
    for (int num = start; num <= remaining; num++) {
        if (isDeciBinary(num)) {
            int result = backtrack(remaining - num, count + 1, num);
            if (result < minCount) {
                minCount = result;
            }
        }
    }
    return minCount;
}

int minPartitions(char* n) {
    int target = atoi(n);
    return backtrack(target, 0, 1);
}

int main() {
    char n[100];
    scanf("%s", n);
    printf("%d\n", minPartitions(n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^(10^n))

Exponential - we generate all possible deci-binary numbers and try all combinations

✓ Linear Growth

Space Complexity

O(2^n)

Space needed to store all generated combinations

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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