Package g2101_2200.s2163_minimum_difference_in_sums_after_removal_of_elements

Class Solution

  • All Implemented Interfaces:
    
public final class Solution

    2163 - Minimum Difference in Sums After Removal of Elements.

    Hard

    You are given a 0-indexed integer array nums consisting of 3 * n elements.

    You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:

    • The first n elements belonging to the first part and their sum is <code>sum<sub>first</sub></code>.

    • The next n elements belonging to the second part and their sum is <code>sum<sub>second</sub></code>.

    The difference in sums of the two parts is denoted as <code>sum<sub>first</sub> - sum<sub>second</sub></code>.

    • For example, if <code>sum<sub>first</sub> = 3</code> and <code>sum<sub>second</sub> = 2</code>, their difference is 1.

    • Similarly, if <code>sum<sub>first</sub> = 2</code> and <code>sum<sub>second</sub> = 3</code>, their difference is -1.

    Return the minimum difference possible between the sums of the two parts after the removal of n elements.

    Example 1:

    Input: nums = 3,1,2

    Output: -1

    Explanation: Here, nums has 3 elements, so n = 1.

    Thus we have to remove 1 element from nums and divide the array into two equal parts.

    • If we remove nums0 = 3, the array will be 1,2. The difference in sums of the two parts will be 1 - 2 = -1.

    • If we remove nums1 = 1, the array will be 3,2. The difference in sums of the two parts will be 3 - 2 = 1.

    • If we remove nums2 = 2, the array will be 3,1. The difference in sums of the two parts will be 3 - 1 = 2.

    The minimum difference between sums of the two parts is min(-1,1,2) = -1.

    Example 2:

    Input: nums = 7,9,5,8,1,3

    Output: 1

    Explanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.

    If we remove nums2 = 5 and nums3 = 8, the resultant array will be 7,9,1,3. The difference in sums will be (7+9) - (1+3) = 12.

    To obtain the minimum difference, we should remove nums1 = 9 and nums4 = 1. The resultant array becomes 7,5,8,3. The difference in sums of the two parts is (7+5) - (8+3) = 1.

    It can be shown that it is not possible to obtain a difference smaller than 1.

    Constraints:

    • nums.length == 3 * n

    • <code>1 <= n <= 10<sup>5</sup></code>

    • <code>1 <= numsi<= 10<sup>5</sup></code>

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description

    • Field Summary

      Fields 
      Modifier and Type Field Description

    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()

    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description

    • Method Summary

      Modifier and Type Method Description
      final Long minimumDifference(IntArray nums)

      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait