# 1906. Minimum Absolute Difference Queries

## Description

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.

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 <= li < ri < nums.length

## Solutions

• 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;
}
}

• class Solution {
public:
vector<int> minDifference(vector<int>& nums, vector<vector<int>>& queries) {
int m = nums.size(), n = queries.size();
int preSum[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;
}
}

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

• 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


• 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
}

• 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;
}