Lintcode: Interval Minimum Number
Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers [start, end]. For each query, calculate the minimum number between index start and end in the given array, return the result list. Have you met this question in a real interview? Yes Example For array [1,2,7,8,5], and queries [(1,2),(0,4),(2,4)], return [2,1,5] Note We suggest you finish problem Segment Tree Build, Segment Tree Query and Segment Tree Modify first. Challenge O(logN) time for each query
SegmentTree time complexity: build O(N), query O(logN), update O(logN), space cost O(NlogN)
1 /** 2 * Definition of Interval: 3 * public classs Interval { 4 * int start, end; 5 * Interval(int start, int end) { 6 * this.start = start; 7 * this.end = end; 8 * } 9 */ 10 public class Solution { 11 /** 12 *@param A, queries: Given an integer array and an query list 13 *@return: The result list 14 */ 15 public class SegmentTreeNode { 16 int min; 17 int start; 18 int end; 19 SegmentTreeNode left; 20 SegmentTreeNode right; 21 public SegmentTreeNode(int start, int end) { 22 this.start = start; 23 this.end = end; 24 this.min = Integer.MAX_VALUE; 25 this.left = null; 26 this.right = null; 27 } 28 } 29 30 SegmentTreeNode root; 31 32 public ArrayList<Integer> intervalMinNumber(int[] A, 33 ArrayList<Interval> queries) { 34 // write your code here 35 ArrayList<Integer> res = new ArrayList<Integer>(); 36 root = buildTree(A, 0, A.length-1); 37 query(res, queries); 38 return res; 39 } 40 41 public SegmentTreeNode buildTree(int[] A, int start, int end) { 42 SegmentTreeNode cur = new SegmentTreeNode(start, end); 43 if (start == end) { 44 cur.min = A[start]; 45 } 46 else { 47 int mid = (start+end)/2; 48 cur.left = buildTree(A, start, mid); 49 cur.right = buildTree(A, mid+1, end); 50 cur.min = Math.min(cur.left.min, cur.right.min); 51 } 52 return cur; 53 } 54 55 public void query(ArrayList<Integer> res, ArrayList<Interval> queries) { 56 for (Interval interval : queries) { 57 int result = queryTree(root, interval.start, interval.end); 58 res.add(result); 59 } 60 } 61 62 public int queryTree(SegmentTreeNode cur, int start, int end) { 63 if (start==cur.start && end==cur.end) { 64 return cur.min; 65 } 66 int mid = (cur.start + cur.end)/2; 67 if (end <= mid) return queryTree(cur.left, start, end); 68 else if (start > mid) return queryTree(cur.right, start, end); 69 else return Math.min(queryTree(cur.left, start, mid), queryTree(cur.right, mid+1, end)); 70 } 71 }
