LeetCode 1689 — Partitioning Into Minimum Number of Deci-Binary Numbers (C++) (Day-01)

When I first saw this problem, it looked like a math + greedy problem , but the solution turns out to be surprisingly simple. Let’s break it down. 📌 Problem A deci-binary number is a number where each digit is either 0 or 1 . Examples: 101 1100 10001 Invalid examples: 112 3021 You are given a decimal string n . Your task is to return the minimum number of deci-binary numbers needed so their sum equals n . 💡 Key Observation Each deci-binary number can contribute at most 1 to any digit position . Example: n = "32" We can build it like this: 10 11 11 Sum: 10 + 11 + 11 = 32 We needed 3 numbers . Why? Because the largest digit in 32 is 3 . 🔑 Core Insight The minimum number of deci-binary numbers required = the maximum digit in the string . Why this works: If a digit in n is 8 , we need at least 8 numbers , since each deci-binary number can contribute only 1 at that position . Example: n = "82734" Digits: 8 2 7 3 4 Maximum digit: 8 So the answer is: 8 🧠 Algorithm Traverse the string. Convert