LeetCode 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Solutions

Solution 1: Quick Thinking

The problem is equivalent to finding the maximum number in the string.

The time complexity is O(n), where n is the length of the string. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustC

class Solution: def minPartitions(self, n: str) -> int: return int(max(n))(code-box)

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; } }(code-box)

class Solution { public: int minPartitions(string n) { int ans = 0; for (char& c : n) { ans = max(ans, c - '0'); } return ans; } };(code-box)

func minPartitions(n string) (ans int) { for _, c := range n { ans = max(ans, int(c-'0')) } return }(code-box)

function minPartitions(n: string): number { return Math.max(...n.split('').map(Number)); }(code-box)

impl Solution { pub fn min_partitions(n: String) -> i32 { n.as_bytes().iter().fold(0, |ans, &c| ans.max((c - b'0') as i32)) } }(code-box)

int minPartitions(char* n) { int ans = 0; for (int i = 0; n[i]; i++) { int v = n[i] - '0'; if (v > ans) { ans = v; } } return ans; }(code-box)