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Program to find reformatted phone number in Python
Suppose we have a phone number as string. The phone number number consists of digits, spaces and/or dashes '-'. We want to reformat the phone number in a certain manner. There are few rules −
Remove all spaces and dashes at beginning
Group the digits from left side to right side into blocks of length 3 until there are 4 or less digits are left.
The final digits are then grouped like −
For 2 digits: A single block of length 2.
For 3 digits: A single block of length 3.
For 4 digits: Two more blocks of length 2 each.
These blocks are then clubbed by dashes. We have to find the reformatted phone number.
So, if the input is like s = "9-6-84102-4 7-8", then the output will be “968-410-24-78”
To solve this, we will follow these steps −
digits := a blank string
blk := a blank string
for each character i in s, do
if i is numeric, then
blk := blk concatenate i
if size of blk is same as 3, then
digits := digits concatenate blk concatenate dash("-")
blk := a blank string
if size of blk is same as 0, then
return substring of digits from index 0 to size of digits-1]
otherwise when size of blk is same as 1, then
return substring of digits from index 0 to size of digits-2] concatenate dash("-") concatenate second last character of digits concatenate blk
otherwise when size of blk is same as 2, then
return digits concatenate blk
Example (Python)
Let us see the following implementation to get better understanding −
def solve(s): digits = "" blk = "" for i in s: if i.isnumeric(): blk += i if len(blk) == 3: digits += blk+"-" blk = "" if len(blk) == 0: return digits[:-1] elif len(blk) == 1: return digits[:-2]+"-"+digits[-2]+blk elif len(blk) == 2: return digits+blk s = "9-6-84102-4 7-8" print(solve(s))
Input
"9-6-84102-4 7-8"
Output
968-410-24-78
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