0917MediumArrayHash TableDynamic ProgrammingGreedyEnumerationPrefix Sum

Most Dials Reading k After One Nudge

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Trains the technique from

LeetCode 3434Maximum Frequency After Subarray Operation

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Dial readings are given as nums.

Choose exactly one contiguous stretch of the readings and add the same whole number to every reading in it. That number may be negative, and it may be zero.

Return the greatest number of readings that can end up equal to k.

Examples

Example 1

Input: nums = [3, 7, 3, 9, 7, 3, 7], k = 7
Output: 4

Three readings already show 7. Adding four to the stretch covering positions 0 to 2 turns two threes into sevens while pushing the single 7 inside it off, for a net gain of one, giving four in all.

Example 2

Input: nums = [1, 1, 1, 1, 1], k = 2
Output: 5

Adding one to the whole list turns every reading into 2.

Example 3

Input: nums = [7, 7, 7, 7], k = 3
Output: 4

Every reading is the same, so one shift moves as many of them onto 3 as the stretch covers, and covering all four is best.

Constraints

1 <= nums.length <= 10^5
1 <= nums[i] <= 50
1 <= k <= 50

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_frequency(nums: list[int], k: int) -> int:

Java

public int maxFrequency(int[] nums, int k)

The intended approach

Time O(n * V) for n readings over a reading range V of fifty.Space O(1).

A single shift applied to one stretch can only ever turn one value into `k`, whichever value the shift maps there. So fix that value `v` and ask what the best stretch is: inside the stretch, every reading equal to `v` becomes `k` and is gained, while every reading already equal to `k` is pushed off it and is lost. Readings outside the stretch keep whatever they were. Score each position `+1` when it holds `v`, `-1` when it holds `k`, and nothing otherwise; the best stretch is then the largest running total over any stretch, which one pass finds by carrying a running sum and dropping it whenever it falls below zero. Adding that gain to the readings already equal to `k` gives the answer for that `v`, and the best over the fifty possible values wins. Allowing the empty stretch, or equivalently a shift of zero, is what keeps the answer from falling below the count already there.

Pythonentry point: max_frequency

def max_frequency(nums, k):
    already = 0
    for v in nums:
        if v == k:
            already += 1
    gain = 0
    for v in range(1, 51):
        if v == k:
            continue
        running = 0
        best = 0
        for x in nums:
            if x == v:
                running += 1
            elif x == k:
                running -= 1
            if running < 0:
                running = 0
            if running > best:
                best = running
        if best > gain:
            gain = best
    return already + gain

The obvious version

Try every stretch and every shift that could matter, apply it to a copy of the readings, and count how many equal `k`. Cubic in the reading count and it never argues about which value the shift should target.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_frequency(nums, k):
    n = len(nums)
    best = 0
    for left in range(n):
        for right in range(left, n):
            for shift in range(-60, 61):
                count = 0
                for i in range(n):
                    value = nums[i]
                    if left <= i <= right:
                        value += shift
                    if value == k:
                        count += 1
                if count > best:
                    best = count
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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