Partition Labes
Post
Cancel

# LeetCode #763: Partition Labes (C/C++).

medium

source: https://leetcode.com/problems/partition-labels
C/C++ Solution to LeetCode problem 763. Partition Labes.

## Problem

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.

## Examples

### Example 1:

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.

## Solution

Using Hash Table to track last index of each character, and two pointers to determine where a new partition starts and where it ends.

• We track the last location of each character using the hash table.
• We iterate again through the string, and we determine if we are in the last position for this character.
• If so, then we determine its size (position of the char minus last position for previous label)
• We add that size to the result and we set this as the new last position.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 class Solution { public: vector<int> partitionLabels(string s) { vector<int> result; unordered_map<char, int> positions; for (int i = 0; i < s.size(); i += 1) positions[s[i]] = i; int maxP = 0; int lastP = -1; for (int i = 0; i < s.size(); i += 1) { maxP = max(maxP, positions[s[i]]); if (maxP == i) { result.push_back(maxP - lastP); lastP = maxP; } } return result; } };