Is Minimum Rotations to Dial a Number II easy or hard?

It is rated Medium with a 53.1% acceptance rate on FleetCode. The hard part is recognizing that the shorter arc in either direction is always the right local choice. Once you see the circular distance formula, the implementation is only a few lines.

How to solve Minimum Rotations to Dial a Number II in O(n)?

Start with the current position at the starting dial value and iterate through the target string. For each character, compute the absolute difference from the current position, take the smaller of that difference and 26 minus that difference, add it to a running total, and update the current position to the character. Return the total after the loop.

Minimum Rotations to Dial a Number II Python/Java solution?

In Python, loop over the string, convert each character to its index with ord(c) - ord('a'), compute d = abs(target - current), and add min(d, 26 - d) to the total. The Java version uses the same logic with charAt and Math.min. Both run in O(n) time and O(1) space.

What is the best approach for Minimum Rotations to Dial a Number II?

The best approach is a greedy single pass that adds the shorter circular arc between each consecutive pair of characters. For each character, compute the raw difference and take the minimum of that difference and the complement on the dial. This runs in O(n) time with O(1) extra space.

Is Minimum Rotations to Dial a Number II asked at Google/Amazon/Meta?

This problem is tagged General on FleetCode and does not have confirmed company tags in our data. It tests greedy reasoning and modular distance on a circle, which appear in interviews at many large tech companies. Check company-specific question lists on LeetCode for current reports.

What data structure is used in Minimum Rotations to Dial a Number II?

The solution uses only a few integer variables: one for the current position and one for the running total. No hash map, array, or stack is required. The string is read once, so the input itself is the only data structure involved.

What is the time complexity of Minimum Rotations to Dial a Number II?

The optimal solution runs in O(n) time, where n is the length of the target string, because each character is processed once with constant work. A step-by-step simulation that rotates notch by notch runs in O(26n), which is still linear in practice but does more work per character.

4071. Minimum Rotations to Dial a Number II - Practice Online

You are given an integer n and a string s of length n consisting of digits.

The dial contains the digits 0 through 9 in order and is circular, so 0 and 9 are adjacent. The pointer initially points to 0.

To dial each digit of s in order, rotate the pointer until it points to that digit. Each rotation moves the pointer to an adjacent digit, and you may rotate in either direction. Dialing a digit that the pointer already points to requires no rotations.

Before dialing, you may perform the following operation at most once:

Choose an index k such that 0 <= k < n and reverse the suffix s[k..n - 1].

Return the minimum total number of rotations needed to dial the string after optimally choosing whether to perform the operation and which suffix to reverse.

Example 1:

Input: n = 4, s = "1502"

Output: 9

Explanation:

Reverse the suffix starting at k = 1 to obtain "1205", then dial it.

Step	From	To	Rotations
1	0	1	1
2	1	2	1
3	2	0	2
4	0	5	5

The total is 1 + 1 + 2 + 5 = 9, which is the minimum total number of rotations.

Example 2:

Input: n = 4, s = "2916"

Output: 12

Explanation:

Choose not to reverse a suffix and dial "2916".

Step	From	To	Rotations
1	0	2	2
2	2	9	3
3	9	1	2
4	1	6	5

The total is 2 + 3 + 2 + 5 = 12, which is the minimum total number of rotations.

Example 3:

Input: n = 4, s = "4219"

Output: 6

Explanation:

Reverse the suffix starting at k = 0, which reverses the entire string, to obtain "9124", then dial it.

Step	From	To	Rotations
1	0	9	1
2	9	1	2
3	1	2	1
4	2	4	2

The total is 1 + 2 + 1 + 2 = 6, which is the minimum total number of rotations.

Constraints:

1 <= n == s.length <= 10⁵
s consists only of digits '0' to '9'