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Expression Add Operators in C++
Suppose we have a string that holds only digits from 0 to 9. And one target value is given. We have to return all possibilities to add binary operators +, - and * into the digits to get the target values. So if the input is like “232” and the target is 8, then the answer will be [“2*3+2”, “2+3*2”]
To solve this, we will follow these steps −
Define a method called solve(), this will take index, s, curr, target, temp, mult −
if idx >= size of s, then,
if target is same as curr, then,
insert temp at the end of ret
return
aux := empty string
for initializing i := idx, when i < size of s, increase i by 1 do −
aux = aux + s[i]
if aux[0] is same as '0' and size of aux> 1, then,
Skip to the next iteration, ignore following part
if idx is same as 0, then,
call solve(i + 1, s, aux as integer, target, aux, aux as integer)
Otherwise
call solve(i + 1, s, curr + aux as integer, target, temp + " + " + aux, aux as integer)
call solve(i + 1, s, curr - aux as integer, target, temp + " - " + aux, - aux as integer)
call solve(i + 1, s, curr - mult + mult * aux as integer, target, temp + " * " + aux, mult * aux as integer)
From the main method call solve(0, num, 0, target, empty string, 0)
return ret
Example
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;
}
typedef long long int lli;
class Solution {
public:
vector <string> ret;
void solve(int idx, string s, lli curr, lli target, string temp, lli mult){
//cout << temp << " " << curr << endl;
if(idx >= s.size()){
if(target == curr){
ret.push_back(temp);
}
return;
}
string aux = "";
for(int i = idx; i < s.size(); i++){
aux += s[i];
if(aux[0] == '0' && aux.size() > 1) continue;
if(idx == 0){
solve(i + 1, s, stol(aux), target, aux, stol(aux));
} else {
solve(i + 1, s, curr + stol(aux), target, temp + "+" + aux, stol(aux));
solve(i + 1, s, curr - stol(aux), target, temp + "-" + aux, -stol(aux));
solve(i + 1, s, curr - mult + mult * stol(aux), target, temp + "*" + aux, mult * stol(aux));
}
}
}
vector<string> addOperators(string num, int target) {
solve(0, num, 0, target, "", 0);
return ret;
}
};
main(){
Solution ob;
print_vector(ob.addOperators("232", 8));
}Input
"232", 8
Output
[2+3*2, 2*3+2, ]
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