1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

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 $O(1)$.

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

Java Code
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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;
    }
}

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

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

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

Rust Code
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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
    }
}

C Code
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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;
}