LeetCode – Partition Labels (Java)

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A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.

For example:

Input: S = “ababfeefhijkh”
Output: [4,4,5]

Explanation:
The partition is “abab”, “feef”, “hijkh”. This is a partition so that each letter appears in at most one part.

Java Solution

public List<Integer> partitionLabels(String S) {
    LinkedHashMap<Character, int[]> map = new LinkedHashMap<>();
 
    for (int i = 0; i < S.length(); i++) {
        char c = S.charAt(i);
        int[] range = map.getOrDefault(c, new int[]{-1, -1});
        if (range[0] == -1) {
            range[0] = i;
            range[1] = i;
        } else {
            range[1] = i;
        }
        map.put(c, range);
    }
 
    int[] temp = null;
    List<Integer> result = new ArrayList<>();
    for (Map.Entry<Character, int[]> entry : map.entrySet()) {
        int[] curr = entry.getValue();
        if (temp == null) {
            temp = curr;
        } else {
            if (curr[0] < temp[1]) {
                temp = new int[]{temp[0], Math.max(curr[1], temp[1])};
            } else {
                result.add(temp[1] - temp[0] + 1);
                temp = curr;
            }
        }
    }
 
    result.add(temp[1] - temp[0] + 1);
 
    return result;
}

2 thoughts on “LeetCode – Partition Labels (Java)”

  2. O(n) solution

    class Solution {
    public:
    vector partitionLabels(string s) {
    unordered_map mp;
    for(int i = 0; i < s.size(); i++) mp[s[i]] = i;
    vector partitionLastIndex;
    partitionLastIndex.push_back(0);
    for(int i = 0; i < s.size(); i++) {
    if(partitionLastIndex[partitionLastIndex.size() - 1] 0; i--) {
    partitionLastIndex[i] -= partitionLastIndex[i - 1];
    }
    partitionLastIndex[0]++;
    /* handle the corner empty string case */
    if(s.size() == 0) partitionLastIndex[0] = 0;
    return partitionLastIndex;
    }
    };

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