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Question
Formatted question description: https://leetcode.ca/all/1689.html
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.
Algorithm
Find the largest single digit… More of math proof about it…
Code
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class Solution { public int minPartitions(String n) { int max = 0; int length = n.length(); for (int i = 0; i < length; i++) { int digit = n.charAt(i) - '0'; max = Math.max(max, digit); } return max; } } ############ 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; } }
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// OJ: https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/ // Time: O(N) // Space: O(1) class Solution { public: int minPartitions(string n) { return *max_element(begin(n), end(n)) - '0'; } };
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class Solution: def minPartitions(self, n: str) -> int: return int(max(n)) ############ # 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers # https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/ class Solution: def minPartitions(self, n: str) -> int: res = float('-inf') for c in n: res = max(int(c), res) return res
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func minPartitions(n string) (ans int) { for _, c := range n { if t := int(c - '0'); ans < t { ans = t } } return }
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function minPartitions(n: string): number { let nums = n.split('').map(d => parseInt(d)); let ans = Math.max(...nums); return ans; }
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impl Solution { pub fn min_partitions(n: String) -> i32 { let mut ans = 0; for c in n.as_bytes() { ans = ans.max((c - b'0') as i32); } ans } }