1689. Partitioning Into Minimum Number Of Deci-Binary Numbers
阿新 • • 發佈:2020-12-13
package LeetCode_1689 /** * 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers * https://leetcode.com/problems/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. 1 <= n.length <= 105 2. n consists of only digits. 3. n does not contain any leading zeros and represents a positive integer. **/ class Solution { /* * solution: greedy, get the max digit, Time:O(L), Space:O(1); * for example: 135: * init 5 deci-binary number with length is 3: * 000 * 000 * 000 * 000 * 000 * then we can fill like below: * 111 * 011 * 011 * 001 * 001 * 111+11+11+1+1 = 135 **/ fun minPartitions(n: String): Int { var max = 0 for (c in n) { max = Math.max(max, c - '0') } return max } }