Optimal Partition of String LeetCode Solution

Problem – Optimal Partition of String LeetCode Solution

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = "abacaba"
Output: 4
Explanation:
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.

Example 2:

Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").

Constraints:

  • 1 <= s.length <= 105
  • s consists of only English lowercase letters.

Optimal Partition of String LeetCode Solution in C++

class Solution {
public:
    bool checkBit(int &flag, int &n){
        return flag & (1<<n);
    }
    void setBit(int &flag, int &n){
        flag = flag | (1<<n);
    }
    int partitionString(string s) {
        int flag = 0;
        int i = 0, ans = 1;
        while(i < s.size()){
            int n = s[i] - 'a';
            if( checkBit(flag, n) ) {
                flag = 0;
                ans++;
            }
            setBit(flag, n);
            i++;
        }
        return ans;
    }
};

Optimal Partition of String LeetCode Solution in Java

class Solution {
    public int partitionString(String s) {
        int ans = 1;
        HashSet<Character> st = new HashSet<>();
        for(int i=0;i<s.length();i++){
 		  // Insert Till we find duplicate element.
            if(!st.contains(s.charAt(i))){
                st.add(s.charAt(i));
            }
            else{
			 // If we found duplicate char then increment count and clear set and start with new set.
                ans++;
                st.clear();
                st.add(s.charAt(i));
            }
        }
        return ans;
    }
}

Optimal Partition of String LeetCode Solution in Python

class Solution:
    def partitionString(self, s: str) -> int:

        def isUnique(s):
            return len(set(s)) == len(s)
			
		left = 0
        res = 0
        for right in range(1,len(s)+1):
            if not isUnique(s[left:right]):
                res += 1
                left = right-1
                
        return res+1 # don't forget there is still one partition we didn't add into res
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