From d164977cebec7be669ff3c0beb9703b89efbb6b2 Mon Sep 17 00:00:00 2001 From: Jack Sangdahl Date: Wed, 18 Jun 2025 20:48:11 -0600 Subject: 2294. partition array such that maximum difference is k --- partition_arr_max_diff_k.py | 34 ++++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) create mode 100644 partition_arr_max_diff_k.py (limited to 'partition_arr_max_diff_k.py') diff --git a/partition_arr_max_diff_k.py b/partition_arr_max_diff_k.py new file mode 100644 index 0000000..555cdf9 --- /dev/null +++ b/partition_arr_max_diff_k.py @@ -0,0 +1,34 @@ +# 2294. Partition Array Such That Maximum Difference Is K + +""" +you are given an integer array 'nums' and an integer 'k'. you may partition +'nums' into one or more subsequences such that each element in 'nums' appears +in exactly one of the subsequences. return the minimum number of subsequences +needed such that the difference between the maximum and minimum values in +each subsequence is at most 'k'. a subsequence is a sequence that can be +derived from another sequence by deleting some or no elements without +changing the order of the remaining elements. +""" + + +class Solution(object): + def partitionArray(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: int + """ + nums.sort() + ans, j = 1, 0 + for i in range(len(nums)): + if nums[i] - nums[j] > k: + ans += 1 + j = i + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.partitionArray(nums=[3, 6, 1, 2, 5], k=2)) + print(obj.partitionArray(nums=[1, 2, 3], k=1)) + print(obj.partitionArray(nums=[2, 2, 4, 5], k=0)) -- cgit v1.2.3