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

  • Binarysearch

  • GeeksforGeeks

  • ProgramCreek

  • YouTube

Examples

Input: n = "32"

Output: 3

Explanation: 10 + 11 + 11 = 32

Solutions

/**
 * 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

Last updated

Was this helpful?