LeetCode #1818 — MEDIUM

Minimum Absolute Sum Difference

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two positive integer arrays nums1 and nums2, both of length n.

The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed).

You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference.

Return the minimum absolute sum difference after replacing at most one element in the array nums1. Since the answer may be large, return it modulo 10⁹ + 7.

|x| is defined as:

x if x >= 0, or
-x if x < 0.

Example 1:

Input: nums1 = [1,7,5], nums2 = [2,3,5]
Output: 3
Explanation: There are two possible optimal solutions:
- Replace the second element with the first: [1,7,5] => [1,1,5], or
- Replace the second element with the third: [1,7,5] => [1,5,5].
Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3.

Example 2:

Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
Output: 0
Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an 
absolute sum difference of 0.

Example 3:

Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
Output: 20
Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20

Constraints:

n == nums1.length
n == nums2.length
1 <= n <= 10⁵
1 <= nums1[i], nums2[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given two positive integer arrays nums1 and nums2, both of length n. The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed). You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference. Return the minimum absolute sum difference after replacing at most one element in the array nums1. Since the answer may be large, return it modulo 109 + 7. |x| is defined as: x if x >= 0, or -x if x < 0.

Baseline thinking

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

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[1,7,5]
[2,3,5]

Example 2

[2,4,6,8,10]
[2,4,6,8,10]

Example 3

[1,10,4,4,2,7]
[9,3,5,1,7,4]

Core Insight

What unlocks the optimal approach

Go through each element and test the optimal replacements.
There are only 2 possible replacements for each element (higher and lower) that are optimal.

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

Upper-end input sizes

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 #1818: Minimum Absolute Sum Difference
class Solution {
    public int minAbsoluteSumDiff(int[] nums1, int[] nums2) {
        final int mod = (int) 1e9 + 7;
        int[] nums = nums1.clone();
        Arrays.sort(nums);
        int s = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
            s = (s + Math.abs(nums1[i] - nums2[i])) % mod;
        }
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            int d1 = Math.abs(nums1[i] - nums2[i]);
            int d2 = 1 << 30;
            int j = search(nums, nums2[i]);
            if (j < n) {
                d2 = Math.min(d2, Math.abs(nums[j] - nums2[i]));
            }
            if (j > 0) {
                d2 = Math.min(d2, Math.abs(nums[j - 1] - nums2[i]));
            }
            mx = Math.max(mx, d1 - d2);
        }
        return (s - mx + mod) % mod;
    }

    private int search(int[] nums, int x) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #1818: Minimum Absolute Sum Difference
func minAbsoluteSumDiff(nums1 []int, nums2 []int) int {
	n := len(nums1)
	nums := make([]int, n)
	copy(nums, nums1)
	sort.Ints(nums)
	s, mx := 0, 0
	const mod int = 1e9 + 7
	for i, a := range nums1 {
		b := nums2[i]
		s = (s + abs(a-b)) % mod
	}
	for i, a := range nums1 {
		b := nums2[i]
		d1, d2 := abs(a-b), 1<<30
		j := sort.SearchInts(nums, b)
		if j < n {
			d2 = min(d2, abs(nums[j]-b))
		}
		if j > 0 {
			d2 = min(d2, abs(nums[j-1]-b))
		}
		mx = max(mx, d1-d2)
	}
	return (s - mx + mod) % mod
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1818: Minimum Absolute Sum Difference
class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        mod = 10**9 + 7
        nums = sorted(nums1)
        s = sum(abs(a - b) for a, b in zip(nums1, nums2)) % mod
        mx = 0
        for a, b in zip(nums1, nums2):
            d1, d2 = abs(a - b), inf
            i = bisect_left(nums, b)
            if i < len(nums):
                d2 = min(d2, abs(nums[i] - b))
            if i:
                d2 = min(d2, abs(nums[i - 1] - b))
            mx = max(mx, d1 - d2)
        return (s - mx + mod) % mod

// Accepted solution for LeetCode #1818: Minimum Absolute Sum Difference
struct Solution;

use std::collections::BTreeSet;
use std::iter::FromIterator;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn min_absolute_sum_diff(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let bts: BTreeSet<i32> = BTreeSet::from_iter(nums1.clone());
        let n = nums1.len();
        let mut max = 0;
        let mut sum: i64 = 0;
        for i in 0..n {
            let x = nums2[i];
            let y = nums1[i];
            let z = (x - y).abs();
            sum += z as i64;
            if let Some(left) = bts.range(..=x).next_back() {
                if (x - left) < z {
                    max = max.max(z - (x - left));
                }
            }
            if let Some(right) = bts.range(x..).next() {
                if (right - x) < z {
                    max = max.max(z - (right - x));
                }
            }
        }
        ((sum - max as i64) % MOD) as i32
    }
}

#[test]
fn test() {
    let nums1 = vec![1, 7, 5];
    let nums2 = vec![2, 3, 5];
    let res = 3;
    assert_eq!(Solution::min_absolute_sum_diff(nums1, nums2), res);
    let nums1 = vec![2, 4, 6, 8, 10];
    let nums2 = vec![2, 4, 6, 8, 10];
    let res = 0;
    assert_eq!(Solution::min_absolute_sum_diff(nums1, nums2), res);
    let nums1 = vec![1, 10, 4, 4, 2, 7];
    let nums2 = vec![9, 3, 5, 1, 7, 4];
    let res = 20;
    assert_eq!(Solution::min_absolute_sum_diff(nums1, nums2), res);
}

// Accepted solution for LeetCode #1818: Minimum Absolute Sum Difference
function minAbsoluteSumDiff(nums1: number[], nums2: number[]): number {
    const mod = 10 ** 9 + 7;
    const nums = [...nums1];
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let s = 0;
    for (let i = 0; i < n; ++i) {
        s = (s + Math.abs(nums1[i] - nums2[i])) % mod;
    }
    let mx = 0;
    for (let i = 0; i < n; ++i) {
        const d1 = Math.abs(nums1[i] - nums2[i]);
        let d2 = 1 << 30;
        let j = search(nums, nums2[i]);
        if (j < n) {
            d2 = Math.min(d2, Math.abs(nums[j] - nums2[i]));
        }
        if (j) {
            d2 = Math.min(d2, Math.abs(nums[j - 1] - nums2[i]));
        }
        mx = Math.max(mx, d1 - d2);
    }
    return (s - mx + mod) % mod;
}

function search(nums: number[], x: number): number {
    let left = 0;
    let right = nums.length;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (nums[mid] >= x) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.