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] = [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

Solutions

Solution 1

Thinking

Sorting a query range to scan adjacent gaps costs \(O((r-l)\log(r-l))\). With \(n\le 10^5\) and \(q\le 2\times 10^4\) that is too slow.

Values lie in \([1,100]\), so the minimum distinct gap is the difference of two consecutive values that actually appear in \([l,r]\). We only need presence of each of the \(100\) numbers.

Prefix counts of every value answer presence in \(O(1)\) per value; a query then walks \(1\ldots 100\) and records the minimum gap between successive present numbers.

Python3JavaC++GoTypeScript

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

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;
    }
};

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;
}

1906. Minimum Absolute Difference Queries

Description

Solutions

Solution 1

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