LeetCode #1689 — MEDIUM

Partitioning Into Minimum Number Of Deci-Binary Numbers

Move from brute-force thinking to an efficient approach using greedy strategy.

The Problem

Problem Statement

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.

Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Example 1:

Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32

Example 2:

Input: n = "82734"
Output: 8

Example 3:

Input: n = "27346209830709182346"
Output: 9

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not. Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"32"

Example 2

"82734"

Example 3

"27346209830709182346"

Step 02

Core Insight

What unlocks the optimal approach

Think about if the input was only one digit. Then you need to add up as many ones as the value of this digit.
If the input has multiple digits, then you can solve for each digit independently, and merge the answers to form numbers that add up to that input.
Thus the answer is equal to the max digit.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1689: Partitioning Into Minimum Number Of Deci-Binary Numbers
class Solution {
    public int minPartitions(String n) {
        int ans = 0;
        for (int i = 0; i < n.length(); ++i) {
            ans = Math.max(ans, n.charAt(i) - '0');
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1689: Partitioning Into Minimum Number Of Deci-Binary Numbers
func minPartitions(n string) (ans int) {
	for _, c := range n {
		ans = max(ans, int(c-'0'))
	}
	return
}

# Accepted solution for LeetCode #1689: Partitioning Into Minimum Number Of Deci-Binary Numbers
class Solution:
    def minPartitions(self, n: str) -> int:
        return int(max(n))

// Accepted solution for LeetCode #1689: Partitioning Into Minimum Number Of Deci-Binary Numbers
impl Solution {
    pub fn min_partitions(n: String) -> i32 {
        n.as_bytes().iter().fold(0, |ans, &c| ans.max((c - b'0') as i32))
    }
}

// Accepted solution for LeetCode #1689: Partitioning Into Minimum Number Of Deci-Binary Numbers
function minPartitions(n: string): number {
    return Math.max(...n.split('').map(Number));
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.