Question

1906. Minimum Absolute Difference Queries

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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

Solution 1

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;
}