Leetcode #2163: Minimum Difference in Sums After Removal of Elements
In this guide, we solve Leetcode #2163 Minimum Difference in Sums After Removal of Elements in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming, Heap (Priority Queue)
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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 nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.
- If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.
- If we remove nums[2] = 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.
Python Solution
class Solution:
def minimumDifference(self, nums: List[int]) -> int:
m = len(nums)
n = m // 3
s = 0
pre = [0] * (m + 1)
q1 = []
for i, x in enumerate(nums[: n * 2], 1):
s += x
heappush(q1, -x)
if len(q1) > n:
s -= -heappop(q1)
pre[i] = s
s = 0
suf = [0] * (m + 1)
q2 = []
for i in range(m, n, -1):
x = nums[i - 1]
s += x
heappush(q2, x)
if len(q2) > n:
s -= heappop(q2)
suf[i] = s
return min(pre[i] - suf[i + 1] for i in range(n, n * 2 + 1))
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.