LeetCode #3675 — MEDIUM

Minimum Operations to Transform String

Move from brute-force thinking to an efficient approach using greedy strategy.

The Problem

Problem Statement

You are given a string s consisting only of lowercase English letters.

You can perform the following operation any number of times (including zero):

Choose any character c in the string and replace every occurrence of c with the next lowercase letter in the English alphabet.

Return the minimum number of operations required to transform s into a string consisting of only 'a' characters.

Note: Consider the alphabet as circular, thus 'a' comes after 'z'.

Example 1:

Input: s = "yz"

Output: 2

Explanation:

Change 'y' to 'z' to get "zz".
Change 'z' to 'a' to get "aa".
Thus, the answer is 2.

Example 2:

Input: s = "a"

Output: 0

Explanation:

The string "a" only consists of 'a' characters. Thus, the answer is 0.

Constraints:

1 <= s.length <= 5 * 10⁵
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting only of lowercase English letters. You can perform the following operation any number of times (including zero): Choose any character c in the string and replace every occurrence of c with the next lowercase letter in the English alphabet. Return the minimum number of operations required to transform s into a string consisting of only 'a' characters. Note: Consider the alphabet as circular, thus 'a' comes after 'z'.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"yz"

Example 2

"a"

Step 02

Core Insight

What unlocks the optimal approach

Each operation shifts every occurrence of a chosen character forward by one in the alphabet (with wrap-around).
For any character <code>c</code>, the number of moves required to turn it into <code>'a'</code> is <code>(26 - (ord(c) - ord('a'))) % 26</code>.
You can plan operations so characters that need more shifts are advanced first and cause merges that don't increase the total number of moves; therefore the minimum number of moves equals the maximum, over characters appearing in <code>s</code>, of <code>(26 - (ord(c) - ord('a'))) % 26</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3675: Minimum Operations to Transform String
class Solution {
    public int minOperations(String s) {
        int ans = 0;
        for (char c : s.toCharArray()) {
            if (c != 'a') {
                ans = Math.max(ans, 26 - (c - 'a'));
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3675: Minimum Operations to Transform String
func minOperations(s string) (ans int) {
	for _, c := range s {
		if c != 'a' {
			ans = max(ans, 26-int(c-'a'))
		}
	}
	return
}

# Accepted solution for LeetCode #3675: Minimum Operations to Transform String
class Solution:
    def minOperations(self, s: str) -> int:
        return max((26 - (ord(c) - 97) for c in s if c != "a"), default=0)

// Accepted solution for LeetCode #3675: Minimum Operations to Transform 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 #3675: Minimum Operations to Transform String
// class Solution {
//     public int minOperations(String s) {
//         int ans = 0;
//         for (char c : s.toCharArray()) {
//             if (c != 'a') {
//                 ans = Math.max(ans, 26 - (c - 'a'));
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3675: Minimum Operations to Transform String
function minOperations(s: string): number {
    let ans = 0;
    for (const c of s) {
        if (c !== 'a') {
            ans = Math.max(ans, 26 - (c.charCodeAt(0) - 97));
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

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.