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Maximum Length of a Concatenated String with Unique Characters in C++
Suppose we have an array of strings arr. The string s is a concatenation of a sub-sequence of arr which have unique characters. Find the maximum possible length of s. If the input is like [“cha”, “r”, “act”, “ers”], then the output will be 6, possible solutions are “chaers” and “acters”.
To solve this, we will follow these steps −
- make one method called ok() and this will take strings s and t. This will act like below
- make one map x
- for i in range 0 to size of s
- increase x[s[i]] by 1
- if x[s[i]] > 1, then return false
- for i in range 0 to size of t
- increase x[t[i]] by 1
- if x[t[i]] > 1, then return false
- return true
- the actual method will be look like below −
- make an array of strings called v, and ans := 0, insert one empty string into v
- for i in range 0 to size of arr
- n := size of v
- for j in range 0 to n – 1
- if ok(v[j], arr[i]) is true, then
- t := v[j] + arr[i]
- insert t into v
- ans := max of ans and size of t
- if ok(v[j], arr[i]) is true, then
- return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool ok(string s, string t){
map <char, int > x;
for(int i = 0; i < s.size(); i++){
x[s[i]]++;
if(x[s[i]] >1)return false;
}
for(int i = 0; i < t.size(); i++){
x[t[i]]++;
if(x[t[i]]>1)return false;
}
return true;
}
int maxLength(vector<string>& arr) {
vector <string> v;
int ans = 0;
v.push_back("");
for(int i = 0; i < arr.size(); i++){
int n = v.size();
for(int j = 0; j < n; j++){
if(ok(v[j],arr[i])){
string t = v[j]+arr[i];
v.push_back(t);
ans = max(ans,(int)t.size());
}
}
}
return ans;
}
};
main(){
vector<string> v = {"cha","r","act","ers"};
Solution ob;
cout << (ob.maxLength(v));
}Input
["cha","r","act","ers"]
Output
6
Updated on: 02-May-2020
330 Views
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