LeetCode #2163 — HARD

Minimum Difference in Sums After Removal of Elements

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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. 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 sum_first.
The next n elements belonging to the second part and their sum is sum_second.

The difference in sums of the two parts is denoted as sum_first - sum_second.

For example, if sum_first = 3 and sum_second = 2, their difference is 1.
Similarly, if sum_first = 2 and sum_second = 3, 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 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.

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 nums[2] = 5 and nums[3] = 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 nums[1] = 9 and nums[4] = 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
1 <= n <= 10⁵
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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 sumfirst. The next n elements belonging to the second part and their sum is sumsecond. The difference in sums of the two parts is denoted as sumfirst - sumsecond. For example, if sumfirst = 3 and sumsecond = 2, their difference is 1. Similarly, if sumfirst = 2 and sumsecond = 3, their difference is -1. Return the minimum difference possible between the sums of the two parts after the removal of n elements.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[3,1,2]

Example 2

[7,9,5,8,1,3]

Core Insight

What unlocks the optimal approach

The lowest possible difference can be obtained when the sum of the first n elements in the resultant array is minimum, and the sum of the next n elements is maximum.
For every index i, think about how you can find the minimum possible sum of n elements with indices lesser or equal to i, if possible.
Similarly, for every index i, try to find the maximum possible sum of n elements with indices greater or equal to i, if possible.
Now for all indices, check if we can consider it as the partitioning index and hence find the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2163: Minimum Difference in Sums After Removal of Elements
class Solution {
    public long minimumDifference(int[] nums) {
        int m = nums.length;
        int n = m / 3;
        long s = 0;
        long[] pre = new long[m + 1];
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        for (int i = 1; i <= n * 2; ++i) {
            int x = nums[i - 1];
            s += x;
            pq.offer(x);
            if (pq.size() > n) {
                s -= pq.poll();
            }
            pre[i] = s;
        }
        s = 0;
        long[] suf = new long[m + 1];
        pq = new PriorityQueue<>();
        for (int i = m; i > n; --i) {
            int x = nums[i - 1];
            s += x;
            pq.offer(x);
            if (pq.size() > n) {
                s -= pq.poll();
            }
            suf[i] = s;
        }
        long ans = 1L << 60;
        for (int i = n; i <= n * 2; ++i) {
            ans = Math.min(ans, pre[i] - suf[i + 1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2163: Minimum Difference in Sums After Removal of Elements
func minimumDifference(nums []int) int64 {
	m := len(nums)
	n := m / 3
	s := 0
	pre := make([]int, m+1)
	q1 := hp{}
	for i := 1; i <= n*2; i++ {
		x := nums[i-1]
		s += x
		heap.Push(&q1, -x)
		if q1.Len() > n {
			s -= -heap.Pop(&q1).(int)
		}
		pre[i] = s
	}
	s = 0
	suf := make([]int, m+1)
	q2 := hp{}
	for i := m; i > n; i-- {
		x := nums[i-1]
		s += x
		heap.Push(&q2, x)
		if q2.Len() > n {
			s -= heap.Pop(&q2).(int)
		}
		suf[i] = s
	}
	ans := int64(1e18)
	for i := n; i <= n*2; i++ {
		ans = min(ans, int64(pre[i]-suf[i+1]))
	}
	return ans
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

# Accepted solution for LeetCode #2163: Minimum Difference in Sums After Removal of Elements
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))

// Accepted solution for LeetCode #2163: Minimum Difference in Sums After Removal of Elements
use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn minimum_difference(nums: Vec<i32>) -> i64 {
        let m = nums.len();
        let n = m / 3;
        let mut s = 0i64;
        let mut pre = vec![0i64; m + 1];
        let mut pq = BinaryHeap::new(); // max-heap

        for i in 1..=2 * n {
            let x = nums[i - 1] as i64;
            s += x;
            pq.push(x);
            if pq.len() > n {
                if let Some(top) = pq.pop() {
                    s -= top;
                }
            }
            pre[i] = s;
        }

        s = 0;
        let mut suf = vec![0i64; m + 1];
        let mut pq = BinaryHeap::new();

        for i in (n + 1..=m).rev() {
            let x = nums[i - 1] as i64;
            s += x;
            pq.push(Reverse(x));
            if pq.len() > n {
                if let Some(Reverse(top)) = pq.pop() {
                    s -= top;
                }
            }
            suf[i] = s;
        }

        let mut ans = i64::MAX;
        for i in n..=2 * n {
            ans = ans.min(pre[i] - suf[i + 1]);
        }

        ans
    }
}

// Accepted solution for LeetCode #2163: Minimum Difference in Sums After Removal of Elements
function minimumDifference(nums: number[]): number {
    const m = nums.length;
    const n = Math.floor(m / 3);
    let s = 0;
    const pre: number[] = Array(m + 1);
    const q1 = new MaxPriorityQueue<number>();
    for (let i = 1; i <= n * 2; ++i) {
        const x = nums[i - 1];
        s += x;
        q1.enqueue(x);
        if (q1.size() > n) {
            s -= q1.dequeue();
        }
        pre[i] = s;
    }
    s = 0;
    const suf: number[] = Array(m + 1);
    const q2 = new MinPriorityQueue<number>();
    for (let i = m; i > n; --i) {
        const x = nums[i - 1];
        s += x;
        q2.enqueue(x);
        if (q2.size() > n) {
            s -= q2.dequeue();
        }
        suf[i] = s;
    }
    let ans = Number.MAX_SAFE_INTEGER;
    for (let i = n; i <= n * 2; ++i) {
        ans = Math.min(ans, pre[i] - suf[i + 1]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.