
Problem Statement
Given a 0-indexed array nums and a non-negative integer k, the task is to modify the elements of the array through a series of permissible operations to maximize its beauty. The beauty of an array is defined as the length of the longest subsequence where all elements are equal. Each element, denoted by nums[i], can be replaced once in an operation by any integer in the inclusive range of [nums[i] - k, nums[i] + k]. The goal is to perform these operations judiciously to maximize the beauty of the resulting array.
Examples
Example 1
Input:
nums = [4,6,1,2], k = 2
Output:
3
Explanation:
In this example, we apply the following operations: - Choose index 1, replace it with 4 (from range [4,8]), nums = [4,4,1,2]. - Choose index 3, replace it with 4 (from range [0,4]), nums = [4,4,1,4]. After the applied operations, the beauty of the array nums is 3 (subsequence consisting of indices 0, 1, and 3). It can be proven that 3 is the maximum possible length we can achieve.
Example 2
Input:
nums = [1,1,1,1], k = 10
Output:
4
Explanation:
In this example we don't have to apply any operations. The beauty of the array nums is 4 (whole array).
Constraints
1 <= nums.length <= 1050 <= nums[i], k <= 105
Approach and Intuition
To achieve the maximum beauty of the array, which is the length of the longest subsequence of identical elements, we can strategically use the flexibility provided by the range of values [nums[i] - k, nums[i] + k] for each element. By understanding how these ranges overlap, we can determine the most frequent element that can be attained through permitted modifications.
Here is a detailed plan:
Track Possible Values:
First, create a map to count potential values that each element innumscan be altered to fall into (considering the range defined byk). This requires checking each element's possible value range and counting these possibilities.Strategy for Overlapping Ranges:
For each elementnums[i], if its modification range overlaps significantly with another, it might be beneficial to transform both to the same value. This maximizes the number of same elements in subsequence for maximum beauty.Simulation of Changes:
Intuitively, you might want to convert numbers to the most frequent potential number within their allowed range. For example, if many numbers can potentially be turned into the number10, and10is within their k-range, transforming all such numbers to10would increase the subsequence of10s, thereby increasing beauty.Utilize Frequency Counts:
Using a frequency count of potential values or a histogram can assist in determining the most popular target number to transform others into. This count can dictate which number within the[nums[i] - k, nums[i] + k]range should be targeted to transform the currentnum[i]into, based on the potential to align it with as many other numbers as possible.Final Calculation:
Once the optimal value to converge multiple array elements to has been determined, count the maximum number of elements that can be legally modified to this value or other closely competitive values. This count forms the maximum beauty of the array. This is achieved by a simple collection and summary of how many numbers can reach each possible maximum value.
By understanding and exploiting the flexibility provided by k, and focusing transformations to unify as many elements of nums as possible, we maximize the array's beauty. This involves recognizing cluster points where multiple ranges overlap and aiming for transformation towards these cluster points.
Solutions
- C++
- Java
- Python
class Solution {
public:
int maxBeautyValue(vector<int>& elements, int m) {
if (elements.size() == 1) return 1;
int maxBeauty = 0;
int highestValue = 0;
for (int el : elements) highestValue = max(highestValue, el);
vector<int> rangeCount(highestValue + 1, 0);
for (int el : elements) {
rangeCount[max(el - m, 0)]++;
if (el + m + 1 <= highestValue)
rangeCount[el + m + 1]--;
}
int runningSum = 0;
for (int i : rangeCount) {
runningSum += i;
maxBeauty = max(maxBeauty, runningSum);
}
return maxBeauty;
}
};
The provided solution in C++ is designed to solve the problem of determining the maximum beauty of an array after applying a specific operation. The operation in consideration allows you to increase the beauty of each element by a certain number of times, up to a maximum limit defined by m. The function maxBeautyValue aims to compute the highest possible beauty of the entire array after these operations are applied optimally.
Here's the logic breakdown:
Begin by handling an edge case where the length of the vector
elementsis 1. In such a case, the maximum beauty value is1.Initialize two variables,
maxBeautyto track the maximum beauty value calculated andhighestValueto store the highest number found in theelements.Traverse the
elementsarray to determine thehighestValue.Create a vector
rangeCountwith a size ofhighestValue + 1, initialized to zero. This vector will help in tracking the range of beauties feasible by adjusting every element in theelementswith up tomincrements or decrements.Again traverse through
elements. For each elementel, increment the position representingel - mby 1 inrangeCountand decrement the position representingel + m + 1(if it's within bounds) to mark the end of the range affectingel.Now, compute the running sum of values in the
rangeCountby iterating over it. This sum helps in determining the potentials of various beauty levels as enhanced by the operations. UpdatemaxBeautyeach time if the current running sum exceeds it.Finally, return the value of
maxBeautywhich represents the highest beauty achievable using the described operation.
This function efficiently computes the desired output using a range overlap technique without modifying the array in place, ensuring an optimized solution leveraging mathematical calculations and logical reasoning.
class Solution {
public int calculateMaxBeauty(int[] flowers, int range) {
if (flowers.length == 1) return 1;
int highestBeauty = 0;
int largestValue = 0;
for (int flower : flowers) largestValue = Math.max(largestValue, flower);
int[] frequency = new int[largestValue + 1];
for (int flower : flowers) {
frequency[Math.max(flower - range, 0)]++;
if (flower + range + 1 <= largestValue) {
frequency[flower + range + 1]--;
}
}
int currentFrequency = 0;
for (int freq : frequency) {
currentFrequency += freq;
highestBeauty = Math.max(highestBeauty, currentFrequency);
}
return highestBeauty;
}
}
The given Java code solves a problem where you must calculate the maximum beauty of an array after applying a certain operation. The operation adds a beauty value to each element within a specified range of range units. This summary breaks down the steps and the logic used to achieve the result in the provided Java code.
Initialize Variables: First, the program checks if the array contains only one flower. If so, it returns a beauty of 1 as, with one flower, the beauty can't be less than that. Otherwise, two variables are initialized:
highestBeautyto keep track of the maximum beauty observed, andlargestValueto store the maximum value in theflowersarray.Determine the Range of Values: Loop through the
flowersarray to determine the largest value, which is necessary for setting up the frequency array that later stages of the algorithm depend on.Set Up Frequency Changes Within Range: A frequency array is populated based on the values in
flowersarray adjusted by the allowed range. If a flower’s value,flower, is within range [flower - range, flower + range], the effect is registered in the frequency array:- Frequency increment at
flower - range(or 0 if the result is negative). - Frequency decrement at
flower + range + 1if it remains within the bounds of the largest value observed.
- Frequency increment at
Calculate Maximum/Current Beauty: A loop goes through the frequency array to continuously add to a
currentFrequencycounter. This counter keeps track of the accumulated frequency at each point. ThehighestBeautyis updated whenevercurrentFrequencyexceeds the previously knownhighestBeauty.Return Highest Beauty: The highest value of
highestBeautythrough the iterations is then returned as the maximum potential beauty of the array after the defined operation.
This approach efficiently calculates the maximum beauty of an array with the application of the range-based operation by leveraging optimized frequency computation within specified ranges.
class Solution:
def calcMaxBeauty(self, items: list[int], limit: int) -> int:
# Return 1 for single element scenarios
if len(items) == 1:
return 1
highest_value = max(items) # Determine highest value in the list
tracker = [0] * (highest_value + 1) # Store changes in this tracker
# Adjust the tracker values based on the permissible range
for item in items:
lower_bound = max(item - limit, 0) # Lower bound of modification
tracker[lower_bound] += 1 # Increment at the beginning of the range
if item + limit + 1 <= highest_value:
tracker[item + limit + 1] -= 1 # Decrement right after the range
aggregate_beauty = 0
accumulated = 0 # Cumulative count
# Compute the cumulative sums and find maximum beauty
for i in tracker:
accumulated += i
aggregate_beauty = max(aggregate_beauty, accumulated)
return aggregate_beauty
For the given problem, the Python solution revolves around calculating the maximum beauty of an array after a specified operation that involves adjusting the array's elements within a given limit. The code uses an approach similar to the "prefix sum" technique to determine the largest cumulative increase that can be reached when modifying each element within the specified range constraints.
Here’s how the code functions:
- Initially, it checks if the array contains only one element. In this scenario, it returns a beauty of
1as there's no room for modification that could change the array. - It identifies the highest value in the array to set up a tracker array. This tracker will note modifications across possible values up to this maximum.
- For each item in the array, the code calculates possible modifications within the limit. It increases the starting point of modification and decreases the point just beyond the range of modification.
- The tracker array accumulates these modifications to reflect the total adjustments possible at each point.
- Finally, it iterates over the accumulated tracker values to find the maximum, representing the maximum beauty.
This approach ensures efficient calculation of possible values by maintaining a record of incremental modifications rather than recomputing possible values for each array item repeatedly.