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Formatted question description: https://leetcode.ca/all/2405.html

2405. Optimal Partition of String

  • Difficulty: Medium.
  • Related Topics: .
  • Similar Questions: Longest Substring Without Repeating Characters, Longest Substring with At Least K Repeating Characters, Partition Labels, Partition Array into Disjoint Intervals.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public int partitionString(String s) {
        int count = (s.isEmpty()) ? 0 : 1;
        HashSet<Character> letters = new HashSet<Character>(); // creating a set to find unique letters in substring
        for (int i = 0; i < s.length(); i++) {
            if (letters.contains(s.charAt(i))) { // if we encounter a duplicate
                letters.clear(); // remove all letters in set
                count++; // increment count
            }
            letters.add(s.charAt(i)); 
        }
        return count;
    }
}

############

class Solution {
    public int partitionString(String s) {
        int v = 0;
        int ans = 1;
        for (char c : s.toCharArray()) {
            int i = c - 'a';
            if (((v >> i) & 1) == 1) {
                v = 0;
                ++ans;
            }
            v |= 1 << i;
        }
        return ans;
    }
}

  • class Solution:
    def partitionString(self, s: str) -> int:
        ans, v = 1, 0
        for c in s:
            i = ord(c) - ord('a')
            if (v >> i) & 1:
                v = 0
                ans += 1
            v |= 1 << i
        return ans

############

# 2405. Optimal Partition of String
# https://leetcode.com/problems/optimal-partition-of-string/

class Solution:
    def partitionString(self, s: str) -> int:
        n = len(s)
        res = 1
        seen = set()
        
        for j, x in enumerate(s):
            if x in seen:
                seen.clear()
                res += 1
                
            seen.add(x)
        
        return res


  • class Solution {
public:
    int partitionString(string s) {
        int ans = 1;
        int v = 0;
        for (char c : s) {
            int i = c - 'a';
            if ((v >> i) & 1) {
                v = 0;
                ++ans;
            }
            v |= 1 << i;
        }
        return ans;
    }
};

  • func partitionString(s string) int {
	ans, v := 1, 0
	for _, c := range s {
		i := int(c - 'a')
		if v>>i&1 == 1 {
			v = 0
			ans++
		}
		v |= 1 << i
	}
	return ans
}

  • function partitionString(s: string): number {
    const set = new Set();
    let res = 1;
    for (const c of s) {
        if (set.has(c)) {
            res++;
            set.clear();
        }
        set.add(c);
    }
    return res;
}


  • use std::collections::HashSet;
impl Solution {
    pub fn partition_string(s: String) -> i32 {
        let mut set = HashSet::new();
        let mut res = 1;
        for c in s.as_bytes().iter() {
            if set.contains(c) {
                res += 1;
                set.clear();
            }
            set.insert(c);
        }
        res
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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