README_EN.md

763. Partition Labels

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Description

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part.

Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s.

Return a list of integers representing the size of these parts.

 

Example 1:

Input: s = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits s into less parts.

Example 2:

Input: s = "eccbbbbdec"
Output: [10]

 

Constraints:

• 1 <= s.length <= 500
• s consists of lowercase English letters.

Solutions

Python3

class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        last = [0] * 26
        for i, c in enumerate(s):
            last[ord(c) - ord('a')] = i
        ans = []
        left = right = 0
        for i, c in enumerate(s):
            right = max(right, last[ord(c) - ord('a')])
            if i == right:
                ans.append(right - left + 1)
                left = right + 1
        return ans

Java

class Solution {
    public List<Integer> partitionLabels(String s) {
        int[] last = new int[26];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            last[s.charAt(i) - 'a'] = i;
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0, left = 0, right = 0; i < n; ++i) {
            right = Math.max(right, last[s.charAt(i) - 'a']);
            if (i == right) {
                ans.add(right - left + 1);
                left = right + 1;
            }
        }
        return ans;
    }
}

TypeScript

function partitionLabels(s: string): number[] {
    const n = s.length;
    let last = new Array(26);
    for (let i = 0; i < n; i++) {
        last[s.charCodeAt(i) - 'a'.charCodeAt(0)] = i;
    }
    let ans = [];
    let left = 0,
        right = 0;
    for (let i = 0; i < n; i++) {
        right = Math.max(right, last[s.charCodeAt(i) - 'a'.charCodeAt(0)]);
        if (i == right) {
            ans.push(right - left + 1);
            left = right + 1;
        }
    }
    return ans;
}

C++

class Solution {
public:
    vector<int> partitionLabels(string s) {
        vector<int> last(26);
        int n = s.size();
        for (int i = 0; i < n; ++i) last[s[i] - 'a'] = i;
        vector<int> ans;
        for (int i = 0, left = 0, right = 0; i < n; ++i)
        {
            right = max(right, last[s[i] - 'a']);
            if (i == right)
            {
                ans.push_back(right - left + 1);
                left = right + 1;
            }
        }
        return ans;
    }
};

Go

func partitionLabels(s string) []int {
	last := make([]int, 26)
	n := len(s)
	for i := 0; i < n; i++ {
		last[s[i]-'a'] = i
	}
	var ans []int
	for i, left, right := 0, 0, 0; i < n; i++ {
		right = max(right, last[s[i]-'a'])
		if i == right {
			ans = append(ans, right-left+1)
			left = right + 1
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Rust

impl Solution {
    pub fn partition_labels(s: String) -> Vec<i32> {
        let n = s.len();
        let bytes = s.as_bytes();
        let mut inx_arr = [0; 26];
        for i in 0..n {
            inx_arr[(bytes[i] - b'a') as usize] = i;
        }
        let mut res = vec![];
        let mut left = 0;
        let mut right = 0;
        for i in 0..n {
            right = right.max(inx_arr[(bytes[i] - b'a') as usize]);
            if right == i {
                res.push((right - left + 1) as i32);
                left = i + 1;
            }
        }
        res
    }
}

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