Find the odd one out among the four letter-strings by analyzing letter repetition and pattern regularity: USTO, OOTU, TTOU, SSTO.

Introduction / Context:
Odd-one-out problems require isolating the unique item that breaks a rule the other items share. The most reliable approach is to test simple, consistent properties such as repeated letters, alphabetical position, vowel–consonant structure, or arrangement symmetry. Here we compare four short letter-strings: USTO, OOTU, TTOU, and SSTO.

Given Data / Assumptions:

• All options are 4-letter sequences containing only the letters O, S, T, U in different orders.
• “Odd” should be determined by a crisp, objective property that three items share and one item lacks (or vice versa).

Concept / Approach:
Check for repetition. In OOTU, the letter O occurs twice; in TTOU, the letter T occurs twice; in SSTO, the letter S occurs twice. By contrast, USTO does not repeat any letter—each character U, S, T, O appears exactly once. Therefore, USTO is the lone string without duplication, making it the odd one out under the “has a repeated letter” rule shared by the others.

Step-by-Step Solution:
1) Scan each string for repeated letters.2) Mark OOTU (O repeated), TTOU (T repeated), SSTO (S repeated) as having duplication.3) Observe USTO has all distinct letters; pick it as the odd one.

Verification / Alternative check:
Other plausible checks (vowel count, alphabetic order) do not separate the set as cleanly as the duplication property. The repetition test provides a unique, unambiguous selection.

Why Other Options Are Wrong:
OOTU, TTOU, and SSTO all share the property “one letter appears twice,” hence they form a group and are not odd.

Common Pitfalls:
Overcomplicating patterns or hunting for multiple rules can mislead. Start from the simplest measurable property—in this case, letter repetition.

Final Answer:
USTO