2829. Determine the Minimum Sum of a k-avoiding Array
Problem Description
In this problem, we are working with a very specific type of array called a "k-avoiding" array. An array is considered to be k-avoiding if there are no two distinct elements within it that add up to the number k. Our objective is to construct a k-avoiding array that has a length of n, where n is a given integer, and we want to make sure that the sum of the elements of this array is as small as possible. We're asked to find and return this minimum possible sum.
Intuition
The key to solving this problem is to avoid any pairs of numbers that sum up to k. A straightforward approach is to start from the smallest positive integer and keep adding distinct numbers to the array, making sure that none of these integer pairs sum to k.
We can do this by maintaining a set of integers that should not be included in the array (to avoid summing to k). As we iterate and add a new element i to the array, we also add k - i to the set of integers to avoid because if k - i was ever included later, it would sum with i to make k, violating the k-avoiding property.
If we encounter a number that is already in the set of numbers to avoid, we simply increment i to find the next suitable candidate.
This approach will ensure that we are always adding the smallest possible integers to our array, hence achieving the minimum possible sum for a k-avoiding array of length n.
How many ways can you arrange the three letters A, B and C?
Solution Approach
The solution involves a greedy approach and a set to keep track of visited numbers and their complementary numbers that sum to k. Here is a step-by-step breakdown of the algorithm using the provided solution code.
-
Initialize a sum
sto0, which will hold the sum of the k-avoiding array values, and an indexito1, which is the smallest positive integer we can add to our array. -
Initialize an empty set
visto keep track of numbers that are already either used in the array or that should be avoided because their corresponding pairs would sum up tok. -
Iterate
ntimes, since we need to addndistinct integers to our k-avoiding array. For each iteration:- Increment
iuntil you find a value that is not present in thevisset. This is the next smallest number that can be added to the array without risking a sum ofkwith any other number already in the array. - Add
ito thevisset because it is now part of our k-avoiding array. - Add
k - ito thevisset to avoid any future numbers from creating a pair withithat sums tok. - Add
ito our running totals, asiis now a confirmed element of our k-avoiding array.
- Increment
-
After we have added
nnumbers to the array, we returns, which now holds the minimum possible sum of the k-avoiding array.
The two key data structures and patterns used in this solution are:
- Greedy Algorithm: By always choosing the next smallest integer not yet visited or restricted, we ensure that we are minimizing the sum at each step, leading to a global minimum sum.
- Set: By using a set for
vis, we can quickly check if a number has already been used or is off-limits and thus efficiently manage our pool of potential array elements.
This algorithm is efficient because it operates on a simple principle of avoidance and uses sets for fast lookup, adding up to an overall time complexity of O(n) since we perform a constant amount of work for each of the n iterations to construct the k-avoiding array.
Which of the following is a good use case for backtracking?
Example Walkthrough
Let's consider constructing a k-avoiding array with k = 7 and length n = 5. We want to end up with an array which has no two distinct elements that add up to 7, and with the smallest possible sum.
-
We initialize the sum
s = 0and indexi = 1which is our starting point for choosing array elements. Our setvisis also initialized to an empty set. -
Now, we will add numbers to our k-avoiding array, while also updating the set
viswith each new number and the number that when combined with it adds up to our target k (7). Our loop will runntimes. -
On the first iteration,
i = 1is not invis, so we can add it to our array. We then updatevisto include1and7 - 1 = 6(to avoid later). The sumsnow becomess = 1. -
On the second iteration, we try
i = 2. Since2is not invis, we can use it. We updatevisby adding2and7 - 2 = 5. The sumsbecomess = 1 + 2 = 3. -
On the third iteration,
i = 3is not invis, and its pair that would sum to7is4which is not invisyet either, so we can add3safely. We updateviswith3and4. The sumsis nows = 3 + 3 = 6. -
The fourth iteration, however, is interesting—
iis incremented to4, but4is invis, indicating a sum with3would equalk = 7, so we cannot use it. We incrementito5, which is also invis. We keep incrementing until we reachi = 8, which is safe to use. We add8and its pair7 - 8 = -1(although negative numbers are not relevant for our case, we're working with positive integers) tovis, and update our sum tos = 6 + 8 = 14. -
On our final iteration,
i = 9is not invis, so we can use it safely in our k-avoiding array. We add9and its pair7 - 9 = -2tovis, even though those negative numbers won't impact our positive integer set. The final sumsis14 + 9 = 23.
Our final k-avoiding array might look like [1, 2, 3, 8, 9] which has no elements that add up to 7, and the sum of its elements is minimum possible, which is 23.
The steps effectively demonstrate a greedy approach, continuously choosing the next smallest integer not in the avoidance set, and efficiently constructing a k-avoiding array with the smallest possible sum.
Solution Implementation
1class Solution:
2 def minimumSum(self, num_elements: int, target: int) -> int:
3 # Initialize the sum and the starting integer
4 current_sum, current_int = 0, 1
5
6 # This set will keep track of the visited or used integers
7 visited = set()
8
9 # Iterate for the number of elements required
10 for _ in range(num_elements):
11 # Avoid using integers already in the visited set
12 while current_int in visited:
13 current_int += 1
14
15 # Add the current integer to the visited set
16 visited.add(current_int)
17
18 # Add the complement of current_int with respect to the target
19 visited.add(target - current_int)
20
21 # Add the current integer to the sum
22 current_sum += current_int
23
24 # Return the calculated sum
25 return current_sum
26
27# Example usage:
28# solution = Solution()
29# print(solution.minimumSum(4, 10)) # Example call to the method
301class Solution {
2 public int minimumSum(int numElements, int cancellationFactor) {
3 int sum = 0; // Variable to keep track of the sum of the chosen elements
4 int smallestEligible = 1; // Variable to store the smallest eligible number that can be used
5 boolean[] visited = new boolean[numElements * numElements + cancellationFactor + 1]; // Array to mark visited numbers
6
7 // Repeat until all 'numElements' elements have been added
8 while (numElements-- > 0) {
9 // Loop to find the next unvisited number starting from smallestEligible
10 while (visited[smallestEligible]) {
11 ++smallestEligible;
12 }
13
14 // Mark this number as visited
15 visited[smallestEligible] = true;
16
17 // If the number is large enough, mark its complementary number (relative to 'cancellationFactor') as visited
18 if (cancellationFactor >= smallestEligible) {
19 visited[cancellationFactor - smallestEligible] = true;
20 }
21
22 // Add the number to the sum
23 sum += smallestEligible;
24 }
25
26 // Return the computed sum
27 return sum;
28 }
29}
301#include <vector>
2#include <cstring>
3
4class Solution {
5public:
6 int minimumSum(int numElements, int delta) {
7 int sum = 0; // Used to store the sum of chosen elements
8 int currentElement = 1; // Starting element for selection
9 // Vector to keep track of visited elements, initialized to false
10 std::vector<bool> visited(numElements * numElements + delta + 1, false);
11
12 // Process each element, decrementing numElements as we go
13 while (numElements--) {
14 // Find the first unvisited element starting from currentElement
15 while (visited[currentElement]) {
16 ++currentElement;
17 }
18 // Mark this element as visited
19 visited[currentElement] = true;
20 // If delta is sufficient, mark the corresponding element as visited
21 if (delta >= currentElement) {
22 visited[delta - currentElement] = true;
23 }
24 // Add the current element to the sum
25 sum += currentElement;
26 }
27 return sum; // Return the computed sum
28 }
29};
301function minimumSum(numElements: number, offset: number): number {
2 let sum = 0; // Will hold the sum of the minimum elements
3 let currentElement = 1; // Start with the first natural number
4 // An array to keep track of visited numbers; initialized with 'false'
5 const visited = Array<boolean>(numElements * numElements + offset + 1).fill(false);
6
7 // Iterate until we have selected 'numElements' elements
8 while (numElements--) {
9 // Find the next current element that has not been visited
10 while (visited[currentElement]) {
11 ++currentElement;
12 }
13
14 // Mark the current element as visited
15 visited[currentElement] = true;
16
17 // If the offset minus the current element is non-negative, mark it as visited
18 if (offset >= currentElement) {
19 visited[offset - currentElement] = true;
20 }
21
22 // Accumulate the sum of selected elements
23 sum += currentElement;
24 }
25
26 // Return the sum of the minimum elements
27 return sum;
28}
29Depth first search can be used to find whether two components in a graph are connected.
Time and Space Complexity
The given Python code aims to find the minimum sum of a sequence of n unique numbers, whereby for each selected number i, the number k - i is not selected. It initializes a sum s to 0, an index i to 1, and a set vis to track visited numbers.
Time Complexity
The time complexity of the code depends on the loop that runs n times – once for each number we need to find. Inside this loop, there is a while i in vis loop that keeps incrementing i until it finds a non-visited number. In the worst-case scenario, this inner while loop can run for each number from 1 to n if all are previously marked as visited. However, in practice, the loop will run considerably fewer times than n for each outer loop iteration, because it only needs to find the next unvisited number.
Assuming U(n) to be the average number of iterations for finding the next unvisited number over all n iterations, the time complexity can be approximated as O(n * U(n)). Since it's difficult to precisely define U(n), we can consider it as a factor that is less than n. However, for the sake of upper-bound estimation, let's consider U(n) in the worst case to be n, which would give us a time complexity of O(n^2) in the worst case.
Space Complexity
The space complexity is easier to analyze. The code only uses a set to store the visited numbers. At most, this set will contain 2n elements (each element and its complement with respect to k), so the space complexity is O(n).
Learn more about how to find time and space complexity quickly using problem constraints.
How would you design a stack which has a function min that returns the minimum element in the stack, in addition to push and pop? All push, pop, min should have running time O(1).
Recommended Readings
Greedy Introduction iframe width 560 height 315 src https www youtube com embed WTslqPbj7I title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture web share allowfullscreen iframe When do we use greedy Greedy algorithms tend to solve optimization problems Typically they will ask you
Math for Technical Interviews How much math do I need to know for technical interviews The short answer is about high school level math Computer science is often associated with math and some universities even place their computer science department under the math faculty However the reality is that you
Top Patterns to Conquer the Technical Coding Interview Should the written word bore you fear not A delightful video alternative awaits iframe width 560 height 315 src https www youtube com embed LW8Io6IPYHw title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture
Got a question? Ask the Teaching Assistant anything you don't understand.
Still not clear? Ask in the Forum, Discord or Submit the part you don't understand to our editors.