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Program to find all possible IP address after restoration in C++
Suppose we have a string with only digits, we have to restore it by forming all possible valid IP address combinations. We know that a valid IP address consists of exactly four integers (each integer is in range 0 to 255) separated by single period symbol.
So, if the input is like ip = "25525511136", then the output will be ["254.25.40.123", "254.254.0.123"]
To solve this, we will follow these steps −
- Define a function convertToNum(), this will take s, start, end,
- num := 0
- for initialize i := start, when i <= end, update (increase i by 1), do −
- num := (num * 10) + (s[i] - ASCII of '0')
- if num > 255, then −
- return 10000
- return num
- Define a function addDots(), this will take positions,
- res := blank string
- x := 0, posIndex := 0
- for initialize i := 0, when i < size of positions, update (increase i by 1), do −
- num := positions[i]
- create one string str1
- temp := num as string
- res := res + temp
- if i < size of positions, then −
- res := res concatenate "."
- return res
- Define a function solve(), this will take s, one string array result, an array positions, dotCount, this is initialize it with 3, startIndex, this is initialize it with 0,
- if not dotCount is non-zero and ((size of s - 1) - startIndex + 1) 1, then −
- temp := convertToNum(s, startIndex, size of s)
- if temp >= 0 and temp <= 255, then −
- insert temp at the end of positions
- res := addDots(positions)
- if size of res is same as size of s, then −
- insert res at the end of result
- return
- for initialize i := startIndex, when i < size of s, update (increase i by 1), do −
- temp := convertToNum(s, startIndex, i)
- if temp >= 0 and temp <= 255, then −
- insert temp at the end of positions
- solve(s, result, positions, dotCount - 1, i + 1)
- delete last element from positions
- Define one function genIp this will take a string s
- Define an array result
- Define an array position
- solve(s, result, position)
- return result
- From the main method call genIp(A)
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
typedef long long int lli;
class Solution {
public:
lli convertToNum(string s,int start, int end){
lli num = 0;
for (int i = start; i <= end; i++) {
num = (num * 10) + (s[i] - '0');
if (num > 255)
return 10000;
}
return num;
}
string addDots(vector <int> positions){
string res = "";
int x = 0;
int posIndex = 0;
for (int i = 0; i < positions.size(); i++) {
int num = positions[i];
ostringstream str1;
str1 << num;
string temp = str1.str();
res += temp;
if (i < positions.size() - 1)
res += ".";
}
return res;
}
void solve(string s, vector <string> &result,vector <int> positions, int dotCount = 3, int startIndex = 0){
if (!dotCount && ((s.size() - 1) - startIndex + 1) >= 1) {
int temp = convertToNum(s, startIndex, s.size() - 1);
if (temp >= 0 && temp <= 255) {
positions.push_back(temp);
string res = addDots(positions);
if (res.size() - 3 == s.size()) {
result.push_back(res);
}
}
return;
}
for (int i = startIndex; i < s.size(); i++) {
int temp = convertToNum(s, startIndex, i);
if (temp >= 0 && temp <= 255) {
positions.push_back(temp);
solve(s, result, positions, dotCount - 1, i + 1);
positions.pop_back();
}
}
}
vector<string> genIp(string s){
vector<string> result;
vector<int> position;
solve(s, result, position);
return result;
}
vector<string> get_ip(string A) {
return genIp(A);
}};
main(){
Solution ob;
string ip = "25525511136";
print_vector(ob.get_ip(ip));
}Input
25525511136
Output
[255.255.11.136, 255.255.111.36, ]
Updated on: 12-Dec-2020
216 Views
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