Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Before and After Puzzle in C++
Suppose we have a list of phrases, generate a list of Before and After puzzles. Here a phrase is a string that consists of lowercase letters and spaces only. No space will be there at the starting and ending positions. There are no consecutive spaces in a phrase.
The before and after puzzles are phrases that are formed by merging two phrases where the last word of the first phrase is the same as the first word of the second phrase. We have to find the Before and After puzzles that can be formed by every two phrases phrases[i] and phrases[j] where I!= j. Note that the order of matching two phrases matters, we want to consider both orders.
We should find a list of distinct strings sorted lexicographically. So if the input is like phrases = ["mission statement", "a quick bite to eat", "a chip off the old block", "chocolate bar", "mission impossible", "a man on a mission", "block party", "eat my words", "bar of soap"], then the output will be: ["a chip off the old block party", "a man on a mission impossible", "a man on a mission statement", "a quick bite to eat my words", "chocolate bar of soap"].
To solve this, we will follow these steps −
Define an array of strings ret, sort the phrases array
define a map m, n := size of phrases array
for I in range 0 to n – 1
s := phrases[i], rspace := index of blank spaces from the right side
insert I into the list placed at m[when rspace is null, then s, otherwise find substring of s up to rspace + 1]
for I in range 0 to n – 1
s := phrases[i] lspace := index of blank spaces from the left side
x := when lspace is null, then s, otherwise find substring of s from 0 to lspace]
if m has x as key
v := m[x]
for j in range 0 to size of v
if v[j] is not I, then
insert phrases[v[j]] + substring of s(up to size of x) into ret
sort ret
delete unique of ret and return ret
Example (C++)
Let us see the following implementation to get a 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;
}
class Solution {
public:
vector<string> beforeAndAfterPuzzles(vector<string>& phrases) {
vector <string> ret;
sort(phrases.begin(), phrases.end());
unordered_map <string, vector <int> > m;
int n = phrases.size();
for(int i = 0; i < n; i++){
string s = phrases[i];
auto rspace = s.rfind(' ');
m[rspace == string::npos ? s : s.substr(rspace + 1)].push_back(i);
}
for(int i = 0; i < n; i++){
string s = phrases[i];
auto lspace = s.find(' ');
string x = (lspace == string::npos? s : s.substr(0, lspace));
if(m.count(x)){
vector <int>& v = m[x];
for(int j = 0; j < v.size(); j++){
if(v[j] != i){
ret.push_back(phrases[v[j]] + s.substr(x.size()));
}
}
}
}
sort(ret.begin(), ret.end());
ret.erase(unique(ret.begin(), ret.end()), ret.end());
return ret;
}
};
main(){
vector<string> v = {"mission statement","a quick bite to eat","a chip off the old block","chocolate bar","mission impossible","a man on a mission","block party","eat my words","bar of soap"};
Solution ob;
print_vector(ob.beforeAndAfterPuzzles(v));
}Input
["mission statement","a quick bite to eat","a chip off the old block","chocolate bar","mission impossible","a man on a mission","block party","eat my words","bar of soap"]
Output
[a chip off the old block party, a man on a mission impossible, a man on a mission statement, a quick bite to eat my words, chocolate bar of soap, ]
81 Views
