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.

Constraints

• 1 <= n.length <= 105

• n consists of only digits.

• n does not contain any leading zeros and represents a positive integer.

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: n = "32"

Output: 3

Explanation: 10 + 11 + 11 = 32

Example 2

Input: n = "82734"

Output: 8

Explanation:

Example 3

Input: n = "27346209830709182346"

Output: 9

Solutions

Solution 1

/**
 * Time complexity : O(M), since we need to iterate over string n to 
 *    find out the maximum digit.
 * Space complexity : O(1), since no extra data structure is required. 
 *    Note that here we do not take the space of input n into account.
 */

class Solution {
    public int minPartitions(String n) {
        if(n == null || n.length() == 0) {
            return 0;
        }
        int count = 0;
        for(int i = 0; i < n.length(); i++) {
            count = Math.max(count, n.charAt(i)-'0');
        }
        return count;
    }
}

Follow up