We use a hash map to record the last occurrence of each character. Then we iterate through the string and maintain a boundary which is the furthest last occurrence of any character seen in the current partition.

1. last = {char: i for i, char in enumerate(s)}.
2. start = 0, end = 0, res = [].
3. For i, char in enumerate(s):
   • end = max(end, last[char]).
   • If i == end:
     • Partition found! Append end - start + 1 to res.
     • start = i + 1.
4. Return res.

• Time Complexity: O(N).
• Space Complexity: O(1) (size of alphabet).

def partition_labels(s):
    last = {c: i for i, c in enumerate(s)}
    start = end = 0
    res = []
    
    for i, c in enumerate(s):
        end = max(end, last[c])
        if i == end:
            res.append(end - start + 1)
            start = i + 1
            
    return res