Optimal Partition of String

Optimal Partition of String

What is this problem about?

The Optimal Partition of String problem asks you to partition a string into the minimum number of substrings such that each substring has no repeated characters. This coding problem uses a greedy approach: extend the current substring until a repeated character forces a new partition. The hash table, string, and greedy interview pattern is directly demonstrated.

Why is this asked in interviews?

Microsoft, Amazon, Google, and Bloomberg ask this because it tests greedy partitioning with a character-seen tracking set. The greedy choice — always extend the current partition as far as possible before creating a new one — minimizes the number of partitions.

Algorithmic pattern used

Greedy with character set. Maintain a set of characters in the current partition. Scan each character: if it's already in the set, start a new partition (clear the set, increment count). Add the current character to the set. Return the total count (initialized to 1 for the first partition).

Example explanation

s="abacaba".

• 'a': set={a}. count=1.
• 'b': set={a,b}.
• 'a': 'a' in set → new partition! count=2, set={a}.
• 'c': set={a,c}.
• 'a': 'a' in set → new partition! count=3, set={a}.
• 'b': set={a,b}.
• 'a': 'a' in set → count=4, set={a}. Answer = 4.

Common mistakes candidates make

• Starting count at 0 (should be 1 — the first partition always exists).
• Not clearing the character set when starting a new partition.
• Including the repeated character in the old partition before starting the new one (it goes into the new partition).
• Using a frequency map instead of a set (frequency isn't needed, just presence).

Interview preparation tip

Greedy partitioning problems where "extend until constraint is violated, then start new group" are common. The character set approach is O(n) and O(26) space. After mastering this, practice "minimum number of operations to make string k-distinct" and similar greedy partitioning variants. The greedy insight — always extend as far as possible before creating a new group — minimizes the partition count.