# [LeetCode]#1694. Reformat Phone Number

• 2 digits: A single block of length 2.
• 3 digits: A single block of length 3.
• 4 digits: Two blocks of length 2 each.
`Input: number = "1-23-45 6"Output: "123-456"Explanation: The digits are "123456".Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".Step 2: There are 3 digits remaining, so put them in a single block of length 3. The 2nd block is "456".Joining the blocks gives "123-456".`
`Input: number = "123 4-567"Output: "123-45-67"Explanation: The digits are "1234567".Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".Step 2: There are 4 digits left, so split them into two blocks of length 2. The blocks are "45" and "67".Joining the blocks gives "123-45-67".`
`Input: number = "123 4-5678"Output: "123-456-78"Explanation: The digits are "12345678".Step 1: The 1st block is "123".Step 2: The 2nd block is "456".Step 3: There are 2 digits left, so put them in a single block of length 2. The 3rd block is "78".Joining the blocks gives "123-456-78".`
`Input: number = "12"Output: "12`
1. number=”123 4–5678"
2. replace “-” with “”
3. replace “ “ with “”
4. if n size >3, 123-. left number is 45678
5. if n size >3, 123–456-. Left number is 78
6. Return 123–456–78
`class Solution:    def reformatNumber(self, number):        n = "".join(number.split("-"))        n="".join(n.split(" "))        ans = ''        while len(n) > 4:            ans += n[:3] + '-'            n = n[3:]        if len(n) == 4:            ans += n[:2] + '-' + n[2:]        else:            ans += n        return ans`

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