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763. Partition Labels

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

  • 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<>();
        int mx = 0, j = 0;
        for (int i = 0; i < n; ++i) {
            mx = Math.max(mx, last[s.charAt(i) - 'a']);
            if (mx == i) {
                ans.add(i - j + 1);
                j = i + 1;
            }
        }
        return ans;
    }
}

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

  • class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        last = {c: i for i, c in enumerate(s)}
        mx = j = 0
        ans = []
        for i, c in enumerate(s):
            mx = max(mx, last[c])
            if mx == i:
                ans.append(i - j + 1)
                j = i + 1
        return ans


  • func partitionLabels(s string) (ans []int) {
	last := [26]int{}
	for i, c := range s {
		last[c-'a'] = i
	}
	var mx, j int
	for i, c := range s {
		mx = max(mx, last[c-'a'])
		if mx == i {
			ans = append(ans, i-j+1)
			j = i + 1
		}
	}
	return
}

  • function partitionLabels(s: string): number[] {
    const last: number[] = Array(26).fill(0);
    const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    const n = s.length;
    for (let i = 0; i < n; ++i) {
        last[idx(s[i])] = i;
    }
    const ans: number[] = [];
    for (let i = 0, j = 0, mx = 0; i < n; ++i) {
        mx = Math.max(mx, last[idx(s[i])]);
        if (mx === i) {
            ans.push(i - j + 1);
            j = i + 1;
        }
    }
    return ans;
}


  • /**
 * @param {string} s
 * @return {number[]}
 */
var partitionLabels = function (s) {
    const last = new Array(26).fill(0);
    const idx = c => c.charCodeAt() - 'a'.charCodeAt();
    const n = s.length;
    for (let i = 0; i < n; ++i) {
        last[idx(s[i])] = i;
    }
    const ans = [];
    for (let i = 0, j = 0, mx = 0; i < n; ++i) {
        mx = Math.max(mx, last[idx(s[i])]);
        if (mx === i) {
            ans.push(i - j + 1);
            j = i + 1;
        }
    }
    return ans;
};


  • public class Solution {
    public IList<int> PartitionLabels(string s) {
        int[] last = new int[26];
        int n = s.Length;
        for (int i = 0; i < n; i++) {
            last[s[i] - 'a'] = i;
        }
        IList<int> ans = new List<int>();
        for (int i = 0, j = 0, mx = 0; i < n; ++i) {
            mx = Math.Max(mx, last[s[i] - 'a']);
            if (mx == i) {
                ans.Add(i - j + 1);
                j = i + 1;
            }
        }
        return ans;
    }
}

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

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