LeetCode 2294: Partition Array Such That Maximum Difference Is K Solution
Master LeetCode problem 2294 (Partition Array Such That Maximum Difference Is K), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation
To solve this problem, we can use a greedy approach. We will sort the input array nums, and then iteratively create subsequences by adding elements to the current subsequence until the difference between the maximum and minimum values in that subsequence exceeds k. At that point, we start a new subsequence and continue the process. The minimum number of subsequences will be equal to the number of times we start a new subsequence.
The steps are as follows:
- Sort the input array
nums. - Initialize a variable
resto keep track of the number of subsequences. - Initialize a variable
startto store the starting index of the current subsequence. - Iterate through the sorted
numsarray:- If the difference between the current element and the element at index
startexceedsk, incrementresand setstartto the current index.
- If the difference between the current element and the element at index
- Return the value of
res.
The time complexity of this solution is O(nlogn) due to the sorting step, where n is the number of elements in the input array. The space complexity is O(1) as we are using a constant amount of extra space.
Solution Code
import java.util.Arrays;
class Solution {
public int partitionArray(int[] nums, int k) {
Arrays.sort(nums);
int res = 1;
int start = 0;
for (int i = 1; i < nums.length; i++) {
if (nums[i] - nums[start] > k) {
res++;
start = i;
}
}
return res;
}
}Try It Yourself
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