Determine the Minimum Sum of a k-avoiding Array
MedPrompt
You are given two integers, n and k.
An array of distinct positive integers is called a k-avoiding array if there does not exist any pair of distinct elements that sum to k.
Return the minimum possible sum of a k-avoiding array of length n.
Example 1:
Input: n = 5, k = 4
Output: 18
Explanation: Consider the k-avoiding array [1,2,4,5,6], which has a sum of 18.
It can be proven that there is no k-avoiding array with a sum less than 18.Example 2:
Input: n = 2, k = 6
Output: 3
Explanation: We can construct the array [1,2], which has a sum of 3.
It can be proven that there is no k-avoiding array with a sum less than 3.
Constraints:
1 <= n, k <= 50
Approaches
2 approaches with complexity analysis and trade-offs.
This approach greedily builds the k-avoiding array by iterating through positive integers 1, 2, 3, ... and adding an integer num to our array if it doesn't violate the k-avoiding condition. To do this efficiently, we maintain a set of the numbers already added to the array. For each candidate number num, we check if k - num is already in the set. If it's not, we add num to the set and to our total sum. We repeat this process until the array has n elements.
Algorithm
- Initialize a
HashSet<Integer>namedseen. - Initialize
sum = 0L,count = 0, andnum = 1. - Start a
whileloop that continues as long ascount < n. - Inside the loop, check if
seencontains the valuek - num. - If it does not, add
numtoseen, addnumtosum, and incrementcount. - Increment
numin every iteration. - After the loop terminates, return
sum.
Walkthrough
To find the minimum sum, we should always try to include the smallest possible positive integers. This suggests a greedy strategy.
We iterate through integers starting from 1. Let's call the current integer num.
We maintain a HashSet to store the elements we've chosen for our array. This allows for quick O(1) average time lookups.
For each num, we check if adding it would violate the condition. An element x violates the condition if there's already an element y in our set such that x + y = k. So, for our candidate num, we check if k - num exists in the HashSet.
If k - num is not in the set, it's safe to add num. We add it to the set, add its value to the running sum, and increment the count of elements we've found.
If k - num is in the set, we must skip num and try the next integer, num + 1.
We continue this process until we have collected n numbers.
Here is a code snippet demonstrating the logic:
import java.util.HashSet;import java.util.Set; class Solution { public long minimumSum(int n, int k) { Set<Integer> seen = new HashSet<>(); long sum = 0; int count = 0; int num = 1; while (count < n) { if (!seen.contains(k - num)) { seen.add(num); sum += num; count++; } num++; } return sum; }}Complexity
Time
O(n + k). In the worst-case scenario, to find `n` numbers, we might have to iterate up to `num` being roughly `n + k/2`. For example, if `k` is small, we skip many numbers. Since `n, k <= 50`, the number of iterations is small. Each iteration involves a hash set operation, which is O(1) on average.
Space
O(n). The `HashSet` stores at most `n` elements that are added to the k-avoiding array.
Trade-offs
Pros
Simple to understand and implement as it directly simulates the greedy decision-making process.
Correct and works for the given constraints.
Cons
Slightly less efficient than a purely mathematical approach.
Uses extra space to store the seen numbers.
By analyzing the pattern of the greedy approach, we can derive a direct mathematical formula to calculate the sum without any iteration. The key observation is that for any pair of integers (i, k-i), we can include at most one in our array. To minimize the sum, we always choose the smaller one, i. This leads to a clear division of which numbers are chosen first.
Algorithm
- Calculate
m = k / 2. - Check if
n <= m. - If true, return the sum of the first
nintegers:(long)n * (n + 1) / 2. - If false, it means
n > m. Calculate the sum in two parts:- Part 1: Sum of the first
mintegers:sum1 = (long)m * (m + 1) / 2. - Part 2: We need
rem_count = n - mmore numbers. These numbers will bek, k+1, .... Calculate the sum of this arithmetic series:sum2 = rem_count * (2L * k + rem_count - 1) / 2.
- Part 1: Sum of the first
- Return
sum1 + sum2.
Walkthrough
The greedy strategy always picks the smallest available number. This means we will pick 1, then 2, and so on, as long as the k-avoiding property is maintained.
For any integer i < k/2, the pair is (i, k-i). The greedy choice is to pick i. This means we will pick all integers from 1 up to m = k/2 (integer division), provided we need that many numbers.
This observation splits the problem into two cases:
Case 1: n <= k / 2
If we need n numbers and n is less than or equal to k/2, we can simply pick the first n positive integers: {1, 2, ..., n}. The maximum sum of any two distinct elements in this set is n + (n-1) = 2n-1. Since n <= k/2, 2n <= k, which means 2n-1 < k. Thus, this set is k-avoiding. The minimum sum is the sum of the first n integers, which is n * (n+1) / 2.
Case 2: n > k / 2
Let m = k / 2. We first pick the smallest m integers that don't conflict with each other: {1, 2, ..., m}. The sum is m * (m+1) / 2.
By picking {1, ..., m}, we are forbidden from picking {k-1, ..., k-m}. The smallest number not yet picked or forbidden is k.
We have already picked m numbers and need n - m more. We pick the next n - m available numbers, which are {k, k+1, ..., k + (n-m) - 1}.
The sum of these additional numbers can be calculated using the arithmetic series formula.
The total sum is the sum of the first part (1 to m) and the second part (k onwards).
class Solution { public long minimumSum(int n, int k) { long m = k / 2; if (n <= m) { return (long)n * (n + 1) / 2; } else { long sum_first_part = m * (m + 1) / 2; long remaining_count = n - m; // The next numbers start from k. This is an arithmetic series: // k, k+1, ..., k + remaining_count - 1 // Sum = count * (first + last) / 2 // Sum = remaining_count * (k + k + remaining_count - 1) / 2 long sum_second_part = remaining_count * (2L * k + remaining_count - 1) / 2; return sum_first_part + sum_second_part; } }}Complexity
Time
O(1). The solution involves a fixed number of arithmetic operations, regardless of the input values of `n` and `k`.
Space
O(1). The solution uses only a few variables to store intermediate calculations, not dependent on the input size.
Trade-offs
Pros
Extremely efficient, with constant time complexity.
No extra space is required.
Provides a direct solution without simulation.
Cons
Requires mathematical insight to derive the formula, which might be less intuitive than the simulation.
Solutions
Solution
class Solution {public int minimumSum(int n, int k) { int s = 0, i = 1; boolean[] vis = new boolean[k + n * n + 1]; while (n-- > 0) { while (vis[i]) { ++i; } vis[i] = true; if (k >= i) { vis[k - i] = true; } s += i; } return s; }}class Solution {public: int minimumSum(int n, int k) { int s = 0, i = 1; bool vis[k + n * n + 1]; memset(vis, false, sizeof(vis)); while (n--) { while (vis[i]) { ++i; } vis[i] = true; if (k >= i) { vis[k - i] = true; } s += i; } return s; }};class Solution: def minimumSum(self, n: int, k: int) -> int: s, i = 0, 1 vis = set() for _ in range(n): while i in vis: i += 1 vis . add(i) vis . add(k - i) s += i return sVideo walkthrough
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