Question

1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

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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 $O(1)$.

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

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;
  }
}

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

func minPartitions(n string) (ans int) {
  for _, c := range n {
    if t := int(c - '0'); ans < t {
      ans = t
    }
  }
  return
}

function minPartitions(n: string): number {
  return Math.max(...n.split('').map(d => parseInt(d)));
}

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
  }
}

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;
}