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Iterator for Combination in C++
Suppose we have to design an Iterator class, that consists of few operations −
- Define a constructor that takes a string characters of sorted distinct lowercase English letters and a number combinationLength as parameter.
- Define a function next() that returns the next combination of length combinationLength in alphabetic order.
- Define another function hasNext() that returns True if and only if there exists a next combination.
So if the input is like −
CombinationIterator iterator = new CombinationIterator("xyz", 2);
iterator.next(); // returns "xy"
iterator.hasNext(); // returns true
iterator.next(); // returns "xz"
iterator.hasNext(); // returns true
iterator.next(); // returns "yz"
iterator.hasNext(); // returns falseTo solve this, we will follow these steps −
- Create an array comb of strings, and index idx
- Define a method makeCombs(), this will take string s, integer variable l, a temp string, that is initially empty, and start, that is initially 0. This is defined as follows −
- if size of temp = l, then insert temp into comb and return
- for i in range start to size of s
- makeCombs(s, l, temp + s[i], i + 1)
- printVector() method will take the array of string as input, this will display the elements of that array
- Define the constructor as follows: this will take string c and cl,
- call makeCombs(c, cl), set idx := 0
- The next() method will increase idx and return comb[idx - 1]
- the hasNext() method will return true if idx is not same as size of comb, otherwise false.
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class CombinationIterator {
public:
vector <string> combs;
int idx;
void makeCombs(string s, int l, string temp ="", int start = 0){
if(temp.size() == l){
combs.push_back(temp);
return;
}
for(int i = start; i < s.size(); i++){
makeCombs(s, l, temp + s[i], i + 1);
}
}
void printVector(vector <string> v){
for(int i = 0; i < v.size(); i++){
cout << v[i] << "\n";
}
cout << endl;
}
CombinationIterator(string c, int cl) {
makeCombs(c, cl);
idx = 0;
}
string next() {
idx++;
return combs[idx - 1];
}
bool hasNext() {
return !(idx == combs.size());
}
};
main(){
CombinationIterator ob("xyz", 2);
cout << (ob.next()) << endl;
cout << (ob.hasNext()) << endl;
cout << (ob.next()) << endl;
cout << (ob.hasNext()) << endl;
cout << (ob.next()) << endl;
cout << (ob.hasNext()) << endl;
}Input
Initialize with “xyz” and 2, then call next() and hasNext() multiple times
Output
xy 1 xz 1 yz 0
Updated on: 30-Apr-2020
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