How to Solve Minimum Additions to Make Valid String

To solve Minimum Additions to Make Valid String, track the expected sequence 'abc' while scanning the word. Insert missing letters whenever the current character does not match the expected one. Using a state transition dynamic programming approach minimizes redundant operations and guarantees the fewest insertions.

Problem Statement

You are given a string word containing only the letters 'a', 'b', and 'c'. You may insert any number of letters 'a', 'b', or 'c' anywhere in the string. Return the minimum number of insertions needed to transform word into a valid string.

A string is valid if it can be represented as one or more concatenations of the substring 'abc'. For example, 'abc', 'abcabc', and 'abcabcabc' are valid. Example: if word = 'b', you need 2 insertions: an 'a' before and a 'c' after to form 'abc'.

Examples

Example 1

Input: word = "b"

Output: 2

Insert the letter "a" right before "b", and the letter "c" right next to "b" to obtain the valid string "abc".

Example 2

Input: word = "aaa"

Output: 6

Insert letters "b" and "c" next to each "a" to obtain the valid string "abcabcabc".

Example 3

Input: word = "abc"

Output: 0

word is already valid. No modifications are needed.

Constraints

• 1 <= word.length <= 50
• word consists of letters "a", "b" and "c" only.

Solution Approach

State Tracking with Expected Pointer

Maintain a pointer for the current position in the word and a pointer cycling through 'abc'. Compare characters and insert missing letters to match the expected sequence. Count insertions each time a mismatch occurs.

Greedy Scan with Modular Index

Iterate through the string while tracking the expected character using modulo arithmetic. Whenever the current character does not match, increment the insertion counter and advance the expected pointer accordingly. This handles consecutive repeated letters efficiently.

Dynamic Programming for Accumulated Inserts

Define dp[i] as the minimum insertions required for the first i characters to form valid segments of 'abc'. Update dp using transitions based on the expected next character. This captures all sequences while avoiding unnecessary insertions.

Complexity Analysis

Time complexity is O(n) for scanning the string with a pointer, or O(n) using DP for state transitions over word length n. Space complexity is O(1) for greedy approach or O(n) for DP array storing minimal insertions.

What Interviewers Usually Probe

• Check if candidates can maintain the cyclic 'abc' state efficiently.
• Observe whether they identify insertion points greedily or use dynamic programming.
• Look for handling of consecutive missing characters without over-inserting.

Common Pitfalls or Variants

Common pitfalls

• Failing to cycle through 'abc' correctly when characters repeat.
• Overcounting insertions by not considering the expected sequence state.
• Using brute-force insertion checking instead of a DP or pointer-based approach.

Follow-up variants

• Allow additional letters beyond 'a', 'b', 'c' and compute minimal deletions to restore validity.
• Find the minimal number of deletions instead of insertions to achieve valid repeated 'abc' segments.
• Compute minimal insertions when word must form 'xyz' repeated patterns instead of 'abc'.

How GhostInterview Helps

• Provides step-by-step guidance to track the expected 'abc' sequence while scanning the word.
• Highlights insertion points and counts dynamically, reducing wasted operations and errors.
• Offers examples illustrating DP and greedy approaches specific to this cyclic pattern problem.

Topic Pages

• String
• Dynamic Programming
• Stack
• Greedy

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FAQ

What is the core idea behind Minimum Additions to Make Valid String?

The core idea is to maintain the expected sequence 'abc' while scanning the word and insert missing letters at mismatches to minimize insertions.

Can this problem be solved without dynamic programming?

Yes, a greedy pointer-based approach cycling through 'abc' can achieve optimal results with O(1) extra space.

How do repeated letters affect the insertion count?

Repeated letters may require multiple consecutive insertions to restore the 'abc' pattern, which is why tracking the expected state is essential.

What is the time complexity for a DP solution?

The DP solution runs in O(n) time, where n is the length of the input word, because each character updates a fixed number of state transitions.

Why is this problem considered a state transition dynamic programming pattern?

Because each character depends on the expected next character in the sequence, and DP tracks minimal insertions across these transitions efficiently.