Move from brute-force thinking to an efficient approach using dynamic programming strategy.
Given a string word to which you can insert letters "a", "b" or "c" anywhere and any number of times, return the minimum number of letters that must be inserted so that word becomes valid.
A string is called valid if it can be formed by concatenating the string "abc" several times.
Example 1:
Input: word = "b" Output: 2 Explanation: 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 Explanation: Insert letters "b" and "c" next to each "a" to obtain the valid string "abcabcabc".
Example 3:
Input: word = "abc" Output: 0 Explanation: word is already valid. No modifications are needed.
Constraints:
1 <= word.length <= 50word consists of letters "a", "b" and "c" only. Problem summary: Given a string word to which you can insert letters "a", "b" or "c" anywhere and any number of times, return the minimum number of letters that must be inserted so that word becomes valid. A string is called valid if it can be formed by concatenating the string "abc" several times.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Stack · Greedy
"b"
"aaa"
"abc"
merge-strings-alternately)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2645: Minimum Additions to Make Valid String
class Solution {
public int addMinimum(String word) {
String s = "abc";
int ans = 0, n = word.length();
for (int i = 0, j = 0; j < n; i = (i + 1) % 3) {
if (word.charAt(j) != s.charAt(i)) {
++ans;
} else {
++j;
}
}
if (word.charAt(n - 1) != 'c') {
ans += word.charAt(n - 1) == 'b' ? 1 : 2;
}
return ans;
}
}
// Accepted solution for LeetCode #2645: Minimum Additions to Make Valid String
func addMinimum(word string) (ans int) {
s := "abc"
n := len(word)
for i, j := 0, 0; j < n; i = (i + 1) % 3 {
if word[j] != s[i] {
ans++
} else {
j++
}
}
if word[n-1] == 'b' {
ans++
} else if word[n-1] == 'a' {
ans += 2
}
return
}
# Accepted solution for LeetCode #2645: Minimum Additions to Make Valid String
class Solution:
def addMinimum(self, word: str) -> int:
s = 'abc'
ans, n = 0, len(word)
i = j = 0
while j < n:
if word[j] != s[i]:
ans += 1
else:
j += 1
i = (i + 1) % 3
if word[-1] != 'c':
ans += 1 if word[-1] == 'b' else 2
return ans
// Accepted solution for LeetCode #2645: Minimum Additions to Make Valid String
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2645: Minimum Additions to Make Valid String
// class Solution {
// public int addMinimum(String word) {
// String s = "abc";
// int ans = 0, n = word.length();
// for (int i = 0, j = 0; j < n; i = (i + 1) % 3) {
// if (word.charAt(j) != s.charAt(i)) {
// ++ans;
// } else {
// ++j;
// }
// }
// if (word.charAt(n - 1) != 'c') {
// ans += word.charAt(n - 1) == 'b' ? 1 : 2;
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #2645: Minimum Additions to Make Valid String
function addMinimum(word: string): number {
const s: string = 'abc';
let ans: number = 0;
const n: number = word.length;
for (let i = 0, j = 0; j < n; i = (i + 1) % 3) {
if (word[j] !== s[i]) {
++ans;
} else {
++j;
}
}
if (word[n - 1] === 'b') {
++ans;
} else if (word[n - 1] === 'a') {
ans += 2;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.