Finding the perfect balance in a dataset often requires trimming the outliers that skew your results. This problem challenges you to find the most efficient way to prune an array so that the largest value doesn't dwarf the smallest one by more than a specific factor.
Problem Summary
You're given:
- An array of integers called
nums. - An integer
krepresenting the maximum allowed ratio between the largest and smallest elements.
Your goal:
- Calculate the minimum number of elements you need to remove to make the array balanced. An array is balanced if .
Intuition: The Power of Sorting
To satisfy a condition involving the minimum and maximum elements, our first instinct should be to sort the array. Once the array is sorted, any contiguous subarray (a slice of the array) will have its first element as the minimum and its last element as the maximum.
The logic follows a greedy "two-pointer" or "sliding window" style of thinking. If we fix a certain element as our minimum, we want to include as many subsequent elements as possible that satisfy the condition.
The provided solution uses a very clever, condensed approach. It iterates through the sorted array and keeps track of a "valid window" starting from an index i. If the current element a exceeds the allowed limit based on the element at A[i], we effectively "remove" an element by incrementing our count. Because we want the minimum removals, we are essentially looking for the maximum number of elements we can keep.
Walkthrough: Understanding the Examples
Let's look at Example 2: nums = [1, 6, 2, 9], k = 3.
-
Sort the array:
[1, 2, 6, 9]. -
Initialize:
i = 0. -
Check 1: Current element is 1. is true.
iremains 0. (Valid window:[1]) -
Check 2: Current element is 2. is true.
iremains 0. (Valid window:[1, 2]) -
Check 3: Current element is 6. is true. This element is too large for our current minimum. We increment
ito 1. -
Check 4: Current element is 9. We check against the new minimum at index 1 (which is 2). is true. We increment
ito 2. -
Result: The final value of
iis 2. This means we removed 2 elements to keep the array balanced.
Code Implementation
C++
class Solution {
public:
int minRemoval(vector<int>& A, int k) {
// Sort to easily identify min and max in any range
sort(A.begin(), A.end());
int i = 0;
for (int a : A) {
// If current element is too large for the current min at A[i]
// we increment i, effectively shrinking the 'kept' count
if (a > 1LL * A[i] * k) {
i++;
}
}
return i;
}
};
Python
class Solution:
def minRemoval(self, nums: List[int], k: int) -> int:
# Sort the numbers to maintain a sliding window of valid elements
nums.sort()
i = 0
for a in nums:
# If the current max (a) exceeds min (nums[i]) * k,
# we must remove an element
if a > nums[i] * k:
i += 1
return i
JavaScript
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var minRemoval = function(nums, k) {
// Sort numerically as JS default sort is lexicographical
nums.sort((a, b) => a - b);
let i = 0;
for (let a of nums) {
// Check if the current element violates the balance condition
if (a > nums[i] * k) {
i++;
}
}
return i;
};
Key Takeaways
- Sorting as Preprocessing: When a problem mentions "minimum" and "maximum" constraints, sorting often simplifies the search space from to .
-
Sliding Window Logic: By maintaining a pointer
i, we can track the start of a valid range and determine how many elements fall outside that range. -
Integer Overflow: In C++, multiplying two large integers can exceed the standard
intlimit. Using1LL(Long Long) ensures the calculation is handled correctly.
Final Thoughts
This problem is a fantastic example of how a "balanced" state is defined in real-world systems. For instance, in load balancing for servers or financial portfolio risk management, we often need to ensure that no single entity is disproportionately larger than the others to prevent system failure or volatility. Mastering this logic helps you build systems that stay within safe operating parameters.
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