---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1906.Minimum%20Absolute%20Difference%20Queries/README.md
rating: 2146
source: 第 246 场周赛 Q4
tags:
- 数组
- 哈希表
---
<!-- problem:start -->
# [1906. 查询差绝对值的最小值](https://leetcode.cn/problems/minimum-absolute-difference-queries)
[English Version](/solution/1900-1999/1906.Minimum%20Absolute%20Difference%20Queries/README_EN.md)
## 题目描述
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<p>一个数组 <code>a</code> 的 <strong>差绝对值的最小值</strong> 定义为：<code>0 <= i < j < a.length</code> 且 <code>a[i] != a[j]</code> 的<strong> </strong><code><span style="">|a[i] - a[j]|</span></code> 的 <strong>最小值</strong>。如果 <code>a</code> 中所有元素都 <strong>相同</strong> ，那么差绝对值的最小值为 <code>-1</code> 。</p>
<ul>
<li>比方说，数组 <code>[5,<strong>2</strong>,<strong>3</strong>,7,2]</code> 差绝对值的最小值是 <code>|2 - 3| = 1</code> 。注意答案不为 <code>0</code> ，因为 <code>a[i]</code> 和 <code>a[j]</code> 必须不相等。</li>
</ul>
<p>给你一个整数数组 <code>nums</code> 和查询数组 <code>queries</code> ，其中 <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code> 。对于每个查询 <code>i</code> ，计算 <strong>子数组</strong> <code>nums[l<sub>i</sub>...r<sub>i</sub>]</code> 中 <strong>差绝对值的最小值</strong> ，子数组 <code>nums[l<sub>i</sub>...r<sub>i</sub>]</code> 包含 <code>nums</code> 数组（下标从 <strong>0</strong> 开始）中下标在 <code>l<sub>i</sub></code> 和 <code>r<sub>i</sub></code> 之间的所有元素（包含 <code>l<sub>i</sub></code> 和 <code>r<sub>i</sub></code> 在内）。</p>
<p>请你返回 <code>ans</code> <strong>数组</strong>，其中 <code>ans[i]</code> 是第 <code>i</code> 个查询的答案。</p>
<p><strong>子数组</strong> 是一个数组中连续的一段元素。</p>
<p><code>|x|</code> 的值定义为：</p>
<ul>
<li>如果 <code>x >= 0</code> ，那么值为 <code>x</code> 。</li>
<li>如果 <code>x < 0</code> ，那么值为 <code>-x</code> 。</li>
</ul>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<b>输入：</b>nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]]
<b>输出：</b>[2,1,4,1]
<b>解释：</b>查询结果如下：
- queries[0] = [0,1]：子数组是 [<strong>1</strong>,<strong>3</strong>] ，差绝对值的最小值为 |1-3| = 2 。
- queries[1] = [1,2]：子数组是 [<strong>3</strong>,<strong>4</strong>] ，差绝对值的最小值为 |3-4| = 1 。
- queries[2] = [2,3]：子数组是 [<strong>4</strong>,<strong>8</strong>] ，差绝对值的最小值为 |4-8| = 4 。
- queries[3] = [0,3]：子数组是 [1,<strong>3</strong>,<strong>4</strong>,8] ，差的绝对值的最小值为 |3-4| = 1 。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<b>输入：</b>nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]]
<b>输出：</b>[-1,1,1,3]
<strong>解释：</strong>查询结果如下：
- queries[0] = [2,3]：子数组是 [2,2] ，差绝对值的最小值为 -1 ，因为所有元素相等。
- queries[1] = [0,2]：子数组是 [<strong>4</strong>,<strong>5</strong>,2] ，差绝对值的最小值为 |4-5| = 1 。
- queries[2] = [0,5]：子数组是 [<strong>4</strong>,<strong>5</strong>,2,2,7,10] ，差绝对值的最小值为 |4-5| = 1 。
- queries[3] = [3,5]：子数组是 [2,<strong>7</strong>,<strong>10</strong>] ，差绝对值的最小值为 |7-10| = 3 。
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>2 <= nums.length <= 10<sup>5</sup></code></li>
<li><code>1 <= nums[i] <= 100</code></li>
<li><code>1 <= queries.length <= 2 * 10<sup>4</sup></code></li>
<li><code>0 <= l<sub>i</sub> < r<sub>i</sub> < nums.length</code></li>
</ul>
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## 解法
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### 方法一
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#### Python3
```python
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
```
#### Java
```java
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;
}
}
```
#### C++
```cpp
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;
}
};
```
#### Go
```go
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
}
```
#### TypeScript
```ts
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;
}
```
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