grid47 Exploring patterns and algorithms
Sep 1, 2024
5 min read
Solution to LeetCode 667: Beautiful Arrangement II Problem
Given two integers n and k, construct a list of n different positive integers from 1 to n such that the absolute differences between consecutive elements contain exactly k distinct integers. Return any valid solution.
Problem
Approach
Steps
Complexity
Input: The input consists of two integers n and k, where n is the size of the list and k is the number of distinct differences to be present.
Example: n = 4, k = 1
Constraints:
• 1 <= k < n <= 10^4
Output: The output should be a list of n distinct integers between 1 and n that satisfies the condition of having exactly k distinct differences.
Example: [1, 2, 3, 4]
Constraints:
• The answer list must contain distinct integers.
Goal: The goal is to construct the list such that the differences between consecutive elements have exactly k distinct values.
Steps:
• 1. Start by choosing numbers from 1 to n.
• 2. Arrange these numbers in a way that the absolute differences between consecutive numbers contain exactly k distinct values.
• 3. Adjust the placement of numbers to ensure exactly k distinct differences.
Goal: The problem has constraints on the size of the input list and the number of distinct differences.
Steps:
• 1 <= k < n <= 10^4
Assumptions:
• The integers in the output list must be distinct.
• There can be multiple valid answers.
• Input: n = 4, k = 1
• Explanation: The differences between consecutive elements in the list [1, 2, 3, 4] are all 1, so there is exactly 1 distinct difference.
• Input: n = 5, k = 2
• Explanation: The list [1, 5, 2, 4, 3] has differences of [4, 3, 2, 1], with exactly 2 distinct values: 1 and 3.
Approach: The approach for constructing the list involves systematically selecting elements such that the absolute differences between consecutive numbers meet the requirements.
Observations:
• We need to control the number of distinct absolute differences between consecutive elements.
• The arrangement of the numbers must balance the number of unique differences.
• By alternating between the smallest and largest remaining numbers, we can ensure a controlled difference distribution.
Steps:
• 1. Start with the smallest and largest numbers.
• 2. Alternately place these numbers in the list while tracking the distinct absolute differences.
• 3. Continue until the list is filled with n numbers.
Empty Inputs:
• The input will always have valid n and k, so no empty cases need to be handled.
Large Inputs:
• For large values of n and k, ensure that the algorithm is efficient enough to handle up to 10^4 elements.
Special Values:
• If k = 1, the list will have identical differences, which means the sequence must be in increasing or decreasing order.
Constraints:
• Ensure that the list contains distinct integers between 1 and n.
vector<int> constructArray(int n, int k) {
vector<int> res;
for(int i =1, j = n; i <= j; ) {
if(k >0) {
res.push_back(k--%2? i++: j--);
} else res.push_back(i++);
}
return res;
}
1 : Function Definition
vector<int> constructArray(int n, int k) {
This is the function definition for `constructArray`, which takes two parameters: `n` (the size of the array) and `k` (the number of elements with the largest absolute difference between consecutive numbers). It returns a vector of integers.
2 : Result Initialization
vector<int> res;
Initialize an empty vector `res` that will hold the result of the constructed array.
3 : Main Loop
for(int i =1, j = n; i <= j; ) {
Start a `for` loop with two pointers: `i` initialized to 1 (the first number) and `j` initialized to `n` (the last number). The loop continues until `i` exceeds `j`.
4 : Condition Check
if(k >0) {
Check if `k` is greater than 0, indicating that we should alternate between the smallest and largest numbers to maximize the difference.
5 : Push Number with Alternating Direction
res.push_back(k--%2? i++: j--);
If `k > 0`, decrement `k` and decide whether to push the next smallest (`i++`) or largest (`j--`) number, alternating between them. This maximizes the difference between consecutive numbers for the first `k` elements.
6 : Push Incremented Number
} else res.push_back(i++);
If `k` is 0, just push the next smallest number (`i++`) into the result vector. This ensures that the rest of the array is filled in increasing order.
7 : Return Result
return res;
Return the constructed result vector `res`, which contains the integers arranged with the maximum difference for the first `k` elements and the rest in increasing order.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) since we are constructing a list of size n and performing constant-time operations for each element.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) because we store the list of size n.
