LeetCode #1906 — MEDIUM

Minimum Absolute Difference Queries

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

The Problem

Problem Statement

The minimum absolute difference of an array a is defined as the minimum value of |a[i] - a[j]|, where 0 <= i < j < a.length and a[i] != a[j]. If all elements of a are the same, the minimum absolute difference is -1.

For example, the minimum absolute difference of the array [5,2,3,7,2] is |2 - 3| = 1. Note that it is not 0 because a[i] and a[j] must be different.

You are given an integer array nums and the array queries where queries[i] = [l_i, r_i]. For each query i, compute the minimum absolute difference of the subarray nums[l_i...r_i] containing the elements of nums between the 0-based indices l_i and r_i (inclusive).

Return an array ans where ans[i] is the answer to the i^th query.

A subarray is a contiguous sequence of elements in an array.

The value of |x| is defined as:

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

Example 1:

Input: nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]]
Output: [2,1,4,1]
Explanation: The queries are processed as follows:
- queries[0] = [0,1]: The subarray is [1,3] and the minimum absolute difference is |1-3| = 2.
- queries[1] = [1,2]: The subarray is [3,4] and the minimum absolute difference is |3-4| = 1.
- queries[2] = [2,3]: The subarray is [4,8] and the minimum absolute difference is |4-8| = 4.
- queries[3] = [0,3]: The subarray is [1,3,4,8] and the minimum absolute difference is |3-4| = 1.

Example 2:

Input: nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]]
Output: [-1,1,1,3]
Explanation: The queries are processed as follows:
- queries[0] = [2,3]: The subarray is [2,2] and the minimum absolute difference is -1 because all the
  elements are the same.
- queries[1] = [0,2]: The subarray is [4,5,2] and the minimum absolute difference is |4-5| = 1.
- queries[2] = [0,5]: The subarray is [4,5,2,2,7,10] and the minimum absolute difference is |4-5| = 1.
- queries[3] = [3,5]: The subarray is [2,7,10] and the minimum absolute difference is |7-10| = 3.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 100
1 <= queries.length <= 2 * 10⁴
0 <= l_i < r_i < nums.length

Step 01

Brute Force Baseline

Problem summary: The minimum absolute difference of an array a is defined as the minimum value of |a[i] - a[j]|, where 0 <= i < j < a.length and a[i] != a[j]. If all elements of a are the same, the minimum absolute difference is -1. For example, the minimum absolute difference of the array [5,2,3,7,2] is |2 - 3| = 1. Note that it is not 0 because a[i] and a[j] must be different. You are given an integer array nums and the array queries where queries[i] = [li, ri]. For each query i, compute the minimum absolute difference of the subarray nums[li...ri] containing the elements of nums between the 0-based indices li and ri (inclusive). Return an array ans where ans[i] is the answer to the ith query. A subarray is a contiguous sequence of elements in an array. The value of |x| is defined as: x if x >= 0. -x if x < 0.

Baseline thinking

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

Pattern signal: Array

Example 1

[1,3,4,8]
[[0,1],[1,2],[2,3],[0,3]]

Example 2

[4,5,2,2,7,10]
[[2,3],[0,2],[0,5],[3,5]]

Step 02

Core Insight

What unlocks the optimal approach

How does the maximum value being 100 help us?
How can we tell if a number exists in a given range?

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 #1906: Minimum Absolute Difference Queries
class Solution {
    public int[] minDifference(int[] nums, int[][] queries) {
        int m = nums.length, n = queries.length;
        int[][] preSum = new int[m + 1][101];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= 100; ++j) {
                int t = nums[i - 1] == j ? 1 : 0;
                preSum[i][j] = preSum[i - 1][j] + t;
            }
        }

        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int left = queries[i][0], right = queries[i][1] + 1;
            int t = Integer.MAX_VALUE;
            int last = -1;
            for (int j = 1; j <= 100; ++j) {
                if (preSum[right][j] > preSum[left][j]) {
                    if (last != -1) {
                        t = Math.min(t, j - last);
                    }
                    last = j;
                }
            }
            if (t == Integer.MAX_VALUE) {
                t = -1;
            }
            ans[i] = t;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1906: Minimum Absolute Difference Queries
func minDifference(nums []int, queries [][]int) []int {
	m, n := len(nums), len(queries)
	preSum := make([][101]int, m+1)
	for i := 1; i <= m; i++ {
		for j := 1; j <= 100; j++ {
			t := 0
			if nums[i-1] == j {
				t = 1
			}
			preSum[i][j] = preSum[i-1][j] + t
		}
	}

	ans := make([]int, n)
	for i := 0; i < n; i++ {
		left, right := queries[i][0], queries[i][1]+1
		t, last := 101, -1
		for j := 1; j <= 100; j++ {
			if preSum[right][j]-preSum[left][j] > 0 {
				if last != -1 {
					if t > j-last {
						t = j - last
					}
				}
				last = j
			}
		}
		if t == 101 {
			t = -1
		}
		ans[i] = t
	}
	return ans
}

# Accepted solution for LeetCode #1906: Minimum Absolute Difference Queries
class Solution:
    def minDifference(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        m, n = len(nums), len(queries)
        pre_sum = [[0] * 101 for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, 101):
                t = 1 if nums[i - 1] == j else 0
                pre_sum[i][j] = pre_sum[i - 1][j] + t

        ans = []
        for i in range(n):
            left, right = queries[i][0], queries[i][1] + 1
            t = inf
            last = -1
            for j in range(1, 101):
                if pre_sum[right][j] - pre_sum[left][j] > 0:
                    if last != -1:
                        t = min(t, j - last)
                    last = j
            if t == inf:
                t = -1
            ans.append(t)
        return ans

// Accepted solution for LeetCode #1906: Minimum Absolute Difference Queries
/**
 * [1906] Minimum Absolute Difference Queries
 *
 * The minimum absolute difference of an array a is defined as the minimum value of |a[i] - a[j]|, where 0 <= i < j < a.length and a[i] != a[j]. If all elements of a are the same, the minimum absolute difference is -1.
 *
 * 	For example, the minimum absolute difference of the array [5,<u>2</u>,<u>3</u>,7,2] is |2 - 3| = 1. Note that it is not 0 because a[i] and a[j] must be different.
 *
 * You are given an integer array nums and the array queries where queries[i] = [li, ri]. For each query i, compute the minimum absolute difference of the subarray nums[li...ri] containing the elements of nums between the 0-based indices li and ri (inclusive).
 * Return an array ans where ans[i] is the answer to the i^th query.
 * A subarray is a contiguous sequence of elements in an array.
 * The value of |x| is defined as:
 *
 * 	x if x >= 0.
 * 	-x if x < 0.
 *
 *  
 * Example 1:
 *
 * Input: nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]]
 * Output: [2,1,4,1]
 * Explanation: The queries are processed as follows:
 * - queries[0] = [0,1]: The subarray is [<u>1</u>,<u>3</u>] and the minimum absolute difference is |1-3| = 2.
 * - queries[1] = [1,2]: The subarray is [<u>3</u>,<u>4</u>] and the minimum absolute difference is |3-4| = 1.
 * - queries[2] = [2,3]: The subarray is [<u>4</u>,<u>8</u>] and the minimum absolute difference is |4-8| = 4.
 * - queries[3] = [0,3]: The subarray is [1,<u>3</u>,<u>4</u>,8] and the minimum absolute difference is |3-4| = 1.
 *
 * Example 2:
 *
 * Input: nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]]
 * Output: [-1,1,1,3]
 * Explanation: The queries are processed as follows:
 * - queries[0] = [2,3]: The subarray is [2,2] and the minimum absolute difference is -1 because all the
 *   elements are the same.
 * - queries[1] = [0,2]: The subarray is [<u>4</u>,<u>5</u>,2] and the minimum absolute difference is |4-5| = 1.
 * - queries[2] = [0,5]: The subarray is [<u>4</u>,<u>5</u>,2,2,7,10] and the minimum absolute difference is |4-5| = 1.
 * - queries[3] = [3,5]: The subarray is [2,<u>7</u>,<u>10</u>] and the minimum absolute difference is |7-10| = 3.
 *
 *  
 * Constraints:
 *
 * 	2 <= nums.length <= 10^5
 * 	1 <= nums[i] <= 100
 * 	1 <= queries.length <= 2 * 10^4
 * 	0 <= li < ri < nums.length
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-absolute-difference-queries/
// discuss: https://leetcode.com/problems/minimum-absolute-difference-queries/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn min_difference(nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1906_example_1() {
        let nums = vec![4, 5, 2, 2, 7, 10];
        let queries = vec![vec![2, 3], vec![0, 2], vec![0, 5], vec![3, 5]];

        let result = vec![2, 1, 4, 1];

        assert_eq!(Solution::min_difference(nums, queries), result);
    }

    #[test]
    #[ignore]
    fn test_1906_example_2() {
        let nums = vec![4, 5, 2, 2, 7, 10];
        let queries = vec![vec![2, 3], vec![0, 2], vec![0, 5], vec![3, 5]];

        let result = vec![-1, 1, 1, 3];

        assert_eq!(Solution::min_difference(nums, queries), result);
    }
}

// Accepted solution for LeetCode #1906: Minimum Absolute Difference Queries
function minDifference(nums: number[], queries: number[][]): number[] {
    let m = nums.length,
        n = queries.length;
    let max = 100;
    // let max = Math.max(...nums);
    let pre: number[][] = [];
    pre.push(new Array(max + 1).fill(0));
    for (let i = 0; i < m; ++i) {
        let num = nums[i];
        pre.push(pre[i].slice());
        pre[i + 1][num] += 1;
    }

    let ans = [];
    for (let [left, right] of queries) {
        let last = -1;
        let min = Infinity;
        for (let j = 1; j < max + 1; ++j) {
            if (pre[left][j] < pre[right + 1][j]) {
                if (last != -1) {
                    min = Math.min(min, j - last);
                }
                last = j;
            }
        }
        ans.push(min == Infinity ? -1 : min);
    }
    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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.