daily-2026-03-01-1689-partitioning-into-minimum-number-of-deci-binary-numbers

problems/1689-partitioning-into-minimum-number-of-deci-binary-numbers/analysis.md

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+# 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers
+
+[LeetCode Link](https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/)
+
+Difficulty: Medium
+Topics: String, Greedy
+Acceptance Rate: 89.0%
+
+## Hints
+
+### Hint 1
+
+Think about what a deci-binary number can contribute to each digit position. Since each digit in a deci-binary number is either 0 or 1, what does that tell you about how many deci-binary numbers you need to "build up" a particular digit?
+
+### Hint 2
+
+Consider each digit of `n` independently. If a digit is `d`, you need at least `d` deci-binary numbers that have a `1` in that position. Can you think about the problem as: what is the bottleneck digit that requires the most deci-binary numbers?
+
+### Hint 3
+
+The answer is simply the maximum digit in `n`. Each deci-binary number contributes at most 1 to each digit position, so the digit with the largest value dictates the minimum count. And you can always construct exactly that many deci-binary numbers to cover all positions — just "peel off" one layer at a time by placing a 1 wherever the remaining digit is still positive.
+
+## Approach
+
+The key insight is that each deci-binary number has digits that are either 0 or 1. When we sum multiple deci-binary numbers, each digit position accumulates independently (no carries needed if we think of it digit-by-digit).
+
+For a digit `d` at some position in `n`, we need exactly `d` deci-binary numbers to have a `1` in that position. The remaining deci-binary numbers will have a `0` there.
+
+Since every digit position must be satisfied, the total number of deci-binary numbers we need is determined by the position that demands the most — i.e., the maximum digit in `n`.
+
+**Example walkthrough with n = "32":**
+- Digit `3` requires 3 deci-binary numbers with a `1` in the tens place.
+- Digit `2` requires 2 deci-binary numbers with a `1` in the ones place.
+- Maximum digit is `3`, so we need 3 deci-binary numbers.
+- Construction: `10 + 11 + 11 = 32` — the first number has a `1` only in the tens place, while the second and third have `1`s in both places.
+
+**Algorithm:**
+1. Iterate through each character in the string `n`.
+2. Track the maximum digit value seen.
+3. Return the maximum digit value.
+
+## Complexity Analysis
+
+Time Complexity: O(n) — single pass through the string where n is the number of digits.
+Space Complexity: O(1) — only a single variable to track the maximum.
+
+## Edge Cases
+
+- **Single digit (e.g., "1"):** The answer is that digit itself. The simplest case.
+- **All same digits (e.g., "111"):** The answer equals that repeated digit. Every deci-binary number used contributes a `1` everywhere.
+- **Contains a 9:** The answer is 9. This is the theoretical maximum since digits range from 0–9.
+- **Very long string:** The algorithm handles strings up to 10^5 characters efficiently in O(n) time.
+- **String like "10":** The max digit is 1, so only one deci-binary number is needed — the number itself.

problems/1689-partitioning-into-minimum-number-of-deci-binary-numbers/problem.md

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+---
+number: "1689"
+frontend_id: "1689"
+title: "Partitioning Into Minimum Number Of Deci-Binary Numbers"
+slug: "partitioning-into-minimum-number-of-deci-binary-numbers"
+difficulty: "Medium"
+topics:
+  - "String"
+  - "Greedy"
+acceptance_rate: 8901.3
+is_premium: false
+created_at: "2026-03-01T03:27:02.039910+00:00"
+fetched_at: "2026-03-01T03:27:02.039910+00:00"
+link: "https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/"
+date: "2026-03-01"
+---
+
+# 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers
+
+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.

problems/1689-partitioning-into-minimum-number-of-deci-binary-numbers/solution_daily_20260301.go

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+package main
+
+// Greedy: the minimum number of deci-binary numbers needed equals the
+// maximum digit in n, since each deci-binary number contributes at most
+// 1 to each digit position.
+func minPartitions(n string) int {
+	maxDigit := 0
+	for _, ch := range n {
+		d := int(ch - '0')
+		if d > maxDigit {
+			maxDigit = d
+		}
+		if maxDigit == 9 {
+			break
+		}
+	}
+	return maxDigit
+}

problems/1689-partitioning-into-minimum-number-of-deci-binary-numbers/solution_daily_20260301_test.go

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+package main
+
+import "testing"
+
+func TestMinPartitions(t *testing.T) {
+	tests := []struct {
+		name     string
+		input    string
+		expected int
+	}{
+		{"example 1: two digits summing to 32", "32", 3},
+		{"example 2: five digits with max 8", "82734", 8},
+		{"example 3: large number with max digit 9", "27346209830709182346", 9},
+		{"edge case: single digit 1", "1", 1},
+		{"edge case: single digit 9", "9", 9},
+		{"edge case: all ones", "111", 1},
+		{"edge case: number is itself deci-binary", "10", 1},
+		{"edge case: all nines", "999", 9},
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			result := minPartitions(tt.input)
+			if result != tt.expected {
+				t.Errorf("minPartitions(%q) = %d, want %d", tt.input, result, tt.expected)
+			}
+		})
+	}
+}