Medium
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
.
[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
import java.util.Arrays;
import java.util.BitSet;
public class Solution {
private static class SegmentTree {
static class Node {
BitSet bits;
int minDiff;
}
int len;
int[] nums;
Node[] tree;
static final int INF = 200;
SegmentTree(int[] nums, int len) {
this.nums = Arrays.copyOf(nums, len);
this.len = len;
tree = new Node[4 * len];
buildTree(0, len - 1, 0);
}
private void buildTree(int i, int j, int ti) {
if (i <= j) {
if (i == j) {
Node node = new Node();
node.bits = new BitSet(101);
node.bits.set(nums[i]);
node.minDiff = INF;
tree[ti] = node;
} else {
int mid = i + (j - i) / 2;
buildTree(i, mid, 2 * ti + 1);
buildTree(mid + 1, j, 2 * ti + 2);
tree[ti] = combineNodes(tree[2 * ti + 1], tree[2 * ti + 2]);
}
}
}
private Node combineNodes(Node n1, Node n2) {
Node node = new Node();
if (n1.minDiff == 1 || n2.minDiff == 1) {
node.minDiff = 1;
} else {
node.bits = new BitSet(101);
node.bits.or(n1.bits);
node.bits.or(n2.bits);
node.minDiff = findMinDiff(node.bits);
}
return node;
}
private int findMinDiff(BitSet bits) {
// minimum value of number is 1.
int first = bits.nextSetBit(1);
int minDiff = INF;
while (first != -1) {
int next = bits.nextSetBit(first + 1);
if (next != -1) {
minDiff = Math.min(minDiff, next - first);
if (minDiff == 1) {
break;
}
}
first = next;
}
return minDiff;
}
private int findMinAbsDiff(int start, int end, int i, int j, int ti) {
Node node = findMinAbsDiff2(start, end, i, j, ti);
return node.minDiff == INF ? -1 : node.minDiff;
}
private Node findMinAbsDiff2(int start, int end, int i, int j, int ti) {
if (i == start && j == end) {
return tree[ti];
}
int mid = i + (j - i) / 2;
if (end <= mid) {
return findMinAbsDiff2(start, end, i, mid, 2 * ti + 1);
} else if (start >= mid + 1) {
return findMinAbsDiff2(start, end, mid + 1, j, 2 * ti + 2);
} else {
Node left = findMinAbsDiff2(start, mid, i, mid, 2 * ti + 1);
Node right = findMinAbsDiff2(mid + 1, end, mid + 1, j, 2 * ti + 2);
return combineNodes(left, right);
}
}
}
public int[] minDifference(int[] nums, int[][] queries) {
int len = nums.length;
int qlen = queries.length;
SegmentTree st = new SegmentTree(nums, len);
int[] answer = new int[qlen];
for (int i = 0; i < qlen; ++i) {
answer[i] = st.findMinAbsDiff(queries[i][0], queries[i][1], 0, len - 1, 0);
}
return answer;
}
}