2834. Find the Minimum Possible Sum of a Beautiful Array
Problem Description
The problem presents us with two positive integers, n
and target
. We are asked to find an array nums
meeting specific criteria, being defined as beautiful:
- The array
nums
should have a length equal ton
. - All elements in
nums
must be distinct positive integers. - There must not be any two distinct indices
i
andj
within the range[0, n - 1]
for whichnums[i] + nums[j]
equalstarget
.
The goal is to determine the minimum possible sum of a beautiful array, with the added detail that the result should be given modulo 10^9 + 7
. This modulus operation ensures that the numbers stay within a reasonable range, given the constraints of large number handling in most programming languages.
Intuition
The solution relies on constructing the beautiful array iteratively while maintaining two key insights:
- Since we want to minimize the sum of
nums
, we should start adding the smallest positive integers available, which are naturally consecutive starting from 1. - To prevent any two numbers from being able to sum up to
target
, when we add a new numberi
tonums
, we must make sure thattarget - i
is not possible to be added in future steps. This is because if we havei
in the array, and later addtarget - i
, we would violate the condition that no two numbers can sum totarget
.
We use a set, vis
, to keep track of numbers that cannot be a part of nums
to satisfy the condition above. This is so we can quickly check if a number is disallowed before adding it to our array. Each iteration, we look for the smallest unvisited number starting from 1, add it to the sum, mark its complement with respect to target
in vis
, and move to the next smallest number.
This iterative process is repeated n
times to fill the nums
array while ensuring all conditions for a beautiful array are met and that the sum remains as small as possible.
Solution Approach
The implementation of the solution is straightforward and methodical. Let's break down the steps in the solution code:
- Initialize an empty set
vis
which will store the integers that cannot be included in thenums
array, to prevent summing up to thetarget
. ans
will hold the running sum of thenums
array as we find the correct integers to include.- We start iterating through numbers to include in the
nums
array starting withi = 1
, which ensures we start with the smallest positive integer.
Now we go into a loop that runs n
times – once for each number that we need to add to our nums
array:
- Check
vis
for the next available number: Since no two numbers should add up totarget
, before considering the integeri
to add toans
, we check if it's already in the setvis
. If it is, it means its complementary number (that would sum totarget
) is already part of thenums
array, so we incrementi
to the next number and check again. - Update the sum and set: Once we find a number that is not in
vis
, it means we can safely add it toans
without "violating" thetarget
sum condition. We addi
toans
, then we addtarget - i
tovis
. Addingtarget - i
ensures that in the subsequent iterations, we don't pick a number that could combine with our currenti
to sum totarget
. - Iterate: We increment
i
to consider the next integer in the following iteration.
Finally, we return the total sum ans
modulo 10^9 + 7
. This modulus ensures that our final answer fits within the limits for integer values as prescribed by the problem and is a common practice in competitive programming to prevent integer overflow.
In this implementation, we use a greedy algorithm starting with the smallest possible integer and moving up. The set vis
ensures constant time complexity O(1)
checks and insertions, providing us with an efficient way to track and prevent selecting numbers that could pair up to form target
.
The overall time complexity of the solution is O(n)
since we iterate over n
elements, and the space complexity is also O(n)
due to potentially storing up to n
integers in the vis
set.
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Start EvaluatorExample Walkthrough
Let's use a small example to illustrate the solution approach. Suppose we have the following input:
n = 5
target = 10
Following the steps outlined in the solution approach:
- Initialize a set and sum variable: We start with an empty set
vis = {}
and an integer for our running sumans = 0
. - Iterate through numbers starting with
i = 1
: Our goal is to iterate 5 times, asn = 5
.
Now let's walk through each iteration:
-
First Iteration (
i = 1
): 4. Checkvis
:1
is not invis
, so we can consider it. 5. Updateans
andvis
:ans = 0 + 1 = 1
,vis = {10 - 1 = 9}
. 6. Iterate to the next number:i = 2
. -
Second Iteration (
i = 2
): 4. Checkvis
:2
is not invis
, so it's safe to add. 5. Updateans
andvis
:ans = 1 + 2 = 3
,vis = {9, 10 - 2 = 8}
. 6. Iterate to the next number:i = 3
. -
Third Iteration (
i = 3
): 4. Checkvis
:3
is not invis
, so we take it. 5. Updateans
andvis
:ans = 3 + 3 = 6
,vis = {9, 8, 10 - 3 = 7}
. 6. Iterate to the next number:i = 4
. -
Fourth Iteration (
i = 4
): 4. Checkvis
:4
is not invis
, so we can use it. 5. Updateans
andvis
:ans = 6 + 4 = 10
,vis = {9, 8, 7, 10 - 4 = 6}
. 6. Iterate to the next number:i = 5
. -
Fifth Iteration (
i = 5
): 4. Checkvis
:5
is not invis
, but adding it we need to consider that its target complement would be5
(since10 - 5 = 5
), and we're trying to add5
now, so it's okay to choose it as5
will not be added again. 5. Updateans
andvis
: Since5
can be added toans
, it's updated toans = 10 + 5 = 15
, andvis
would include its target complement, but since it's the same, no new entry is added tovis
. 6. There are no more iterations, as we've reachedn
additions.
Finally, we return ans % (10^9 + 7)
, which in this case is 15 % (10^9 + 7) = 15
, since 15
is already less than 10^9 + 7
.
The final beautiful array that satisfies all conditions could be [1, 2, 3, 4, 5]
with the minimum possible sum being 15
. The set vis
helped us to avoid including the numbers which would sum up to the target with any other number already in the array.
Remember, in scenarios where n
and target
are much larger, the modulo operation would assure that the output fits within the 32-bit or 64-bit signed integer range commonly used in programming languages.
Solution Implementation
1class Solution:
2 def minimum_possible_sum(self, n: int, target: int) -> int:
3 # Initialize a set to keep track of visited numbers.
4 visited = set()
5
6 # Initialize the answer (sum) to zero.
7 answer = 0
8
9 # Initialize the current integer we are going to add to the sum.
10 current_int = 1
11
12 # Loop 'n' times to find 'n' unique numbers to add to the sum.
13 for _ in range(n):
14
15 # Find the next unvisited integer to add to the sum.
16 while current_int in visited:
17 current_int += 1
18
19 # Add the current integer to the sum.
20 answer += current_int
21
22 # Mark the counterpart (target - current_int) as visited.
23 visited.add(target - current_int)
24
25 # Move to the next integer.
26 current_int += 1
27
28 # Return the computed sum after 'n' additions.
29 return answer
30
1class Solution {
2 public long minimumPossibleSum(int n, int target) {
3 // Create an array to keep track of visited numbers
4 boolean[] visited = new boolean[n + target];
5 // Initialize the answer (sum) to 0
6 long sum = 0;
7
8 // Loop over the numbers starting from 1 up to n
9 for (int i = 1; n > 0; --n, ++i) {
10 // If the current number has been visited, skip to the next one
11 while (visited[i]) {
12 ++i;
13 }
14 // Add the smallest unvisited number to the sum
15 sum += i;
16
17 // If the target is greater than or equal to the current number, mark the corresponding number as visited
18 if (target >= i && (target - i) < visited.length) {
19 visited[target - i] = true;
20 }
21 }
22 // Return the final sum which is the minimum possible sum
23 return sum;
24 }
25}
26
1#include <cstring>
2
3class Solution {
4public:
5 // Function to calculate the minimum possible sum of 'n' unique positive integers
6 // such that no pair of integers adds up to 'target'.
7 long long minimumPossibleSum(int n, int target) {
8 // Create an array to keep track of the numbers that should not be used
9 // because they would add up to the target with a number already in use.
10 bool visited[n + target];
11 // Initialize the 'visited' array to false, indicating no numbers have been used yet.
12 memset(visited, false, sizeof(visited));
13
14 // Initialize the sum to 0, the sum will be accrued over the iteration.
15 long long sum = 0;
16 // Iterating over the potential numbers starting from 1.
17 for (int i = 1; n > 0; ++i) {
18 // Skip the numbers that we can't use because they have a pair already in use.
19 while (visited[i]) {
20 ++i;
21 }
22 // Add the current number to our sum.
23 sum += i;
24
25 // Check if the counterpart of the current number (target - i) can potentially
26 // be used in the future, and mark it as visited to avoid using it.
27 if (target >= i) {
28 visited[target - i] = true;
29 }
30
31 // Decrease the count of remaining numbers to add to our sum.
32 --n;
33 }
34 return sum; // Return the final sum.
35 }
36};
37
1/**
2 * Calculates the minimum possible sum of 'n' distinct integers where
3 * each integer 'i' in the array does not equal 'target - i'.
4 *
5 * @param {number} n - The count of distinct integers to sum
6 * @param {number} target - The target value that must not be met by the expression 'i' + 'array[i]'
7 * @returns {number} - The minimum possible sum of 'n' distinct integers.
8 */
9function minimumPossibleSum(n: number, target: number): number {
10 // Initialize a boolean array with false values to track visited numbers
11 const visited: boolean[] = Array(n + target).fill(false);
12 let sum = 0; // Initialize the sum of integers.
13
14 // Iterate over the range of possible values until 'n' distinct integers are found
15 for (let i = 1; n > 0; ++i, --n) {
16 // Skip over the numbers that have already been visited
17 while (visited[i]) {
18 ++i;
19 }
20 // Add the current integer to the sum
21 sum += i;
22
23 // Mark the corresponding pair value as visited if it falls within the array bounds
24 if (target >= i && (target - i) < visited.length) {
25 visited[target - i] = true;
26 }
27 }
28
29 // Return the minimum sum of 'n' distinct integers
30 return sum;
31}
32
Time and Space Complexity
The provided Python code snippet finds the minimum possible sum of a sequence of n
integers such that each value and its complement with respect to target
are unique in the sequence.
Time Complexity
The time complexity of the code is O(n)
.
Here is the breakdown:
- There is a
for
loop that goesn
times, which isO(n)
. - Inside the loop, there is a
while
loop that continues untili
is not invis
. Since thewhile
loop incrementsi
each time a collision withvis
is detected and the number of possible collisions is limited byn
, the amortized time complexity due to thewhile
loop isO(1)
. - Inserting and checking the presence of an item in a
set
in Python isO(1)
on average, asset
is implemented as a hash table.
Therefore, the time complexity for the complete for loop is essentially O(n)
.
Space Complexity
The space complexity of the code is O(n)
.
Here is the breakdown:
- A
set
namedvis
is used to keep track of visited numbers, which would at most storen
elements because for every element we add, we loopn
times. - Other than
vis
, only a few variables are used (ans
,i
) with constant space requirement.
Thus, the set
vis
dictates the space complexity, which is O(n)
.
Learn more about how to find time and space complexity quickly using problem constraints.
Consider the classic dynamic programming of fibonacci numbers, what is the recurrence relation?
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