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Partition Labels in C++
Suppose we have a string S of lowercase letters is given. We will partition this string into as many parts as possible so that each letter appears in at most one part, finally return a list of integers representing the size of these parts. So if the string is like “ababcbacadefegdehijhklij”, output is [9,7,8], because the partitions are “ababcbaca”, “defegde”, “hijhklij”. So this is a partition so that each letter occurs in at most one part. A partition like "ababcbacadefegde", "hijhklij" is not correct, because it splits S into less parts.
To solve this, we will follow these steps −
- define one map called cnt
- for i in range 0 to s, cnt[s[i]] := i
- j := 0, start := 0 and i := 0 and n := size of s
- define one array ans
- while i < n
- j := max of j and cnt[s[i]]
- if i = j, then insert i – start + 1 into ans and start := i + 1
- increase i by 1
- return ans
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;
}
class Solution {
public:
vector<int> partitionLabels(string s) {
map <char, int> cnt;
for(int i = 0; i < s.size(); i++)cnt[s[i]] = i;
int j = 0, start = 0;
int i = 0;
int n = s.size();
vector <int> ans;
while(i < n){
j = max(j, cnt[s[i]]);
if( i == j){
ans.push_back(i-start+ 1);
start = i + 1;
}
i++;
}
return ans;
}
};
main(){
Solution ob;
print_vector(ob.partitionLabels("ababcbacadefegdehijhklij"));
}Input
"ababcbacadefegdehijhklij"
Output
[9,7,8]
Updated on: 29-Apr-2020
258 Views
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