LeetCode 1906: Minimum Absolute Difference Queries Solution

Master LeetCode problem 1906 (Minimum Absolute Difference Queries), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1906. Minimum Absolute Difference Queries

Medium

Array Hash Table

Problem Explanation

Explanation

To solve this problem, we can precompute the minimum absolute difference for all subarrays of the input array nums. Then, for each query [li, ri], we can directly look up the precomputed minimum absolute difference for the subarray nums[li...ri].

Algorithm:

Create a prefix array prefix where prefix[i] stores the minimum absolute difference within the subarray nums[0...i].
Initialize a map lastIndex where lastIndex[nums[i]] stores the last index of nums[i] encountered in the array.
Initialize a result array ans to store the answers for each query.
Iterate over the input array nums to compute the values for prefix and lastIndex.
Iterate over each query [li, ri]:
- If all elements in the subarray are the same, set the result as -1.
- Otherwise, find the minimum absolute difference using prefix[ri] - prefix[li].

Time Complexity:

Building the prefix array and lastIndex map takes O(n) time.
Answering each query takes O(1) time.
Overall time complexity is O(n + q), where n is the length of the input array nums and q is the number of queries.

Space Complexity:

Additional space for prefix and lastIndex is O(n).
Space complexity is O(n) due to the prefix array and lastIndex map.

Solution Code

import java.util.*;

class Solution {
    public int[] minDifference(int[] nums, int[][] queries) {
        int n = nums.length;
        int[] prefix = new int[n];
        Map<Integer, Integer> lastIndex = new HashMap<>();
        
        for (int i = 0; i < n; i++) {
            prefix[i] = lastIndex.containsKey(nums[i]) ? Math.min(Math.abs(i - lastIndex.get(nums[i])), prefix[i - 1]) : (i > 0 ? prefix[i - 1] : Integer.MAX_VALUE);
            lastIndex.put(nums[i], i);
        }
        
        int[] ans = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int li = queries[i][0];
            int ri = queries[i][1];
            if (li < ri && lastIndex.get(nums[li]) < ri) {
                ans[i] = -1;
            } else {
                ans[i] = prefix[ri] - (li > 0 ? prefix[li - 1] : 0);
            }
        }
        
        return ans;
    }
}

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