# Interval Minimum Number

## Problem

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.

Notice

We suggest you finish problem Segment Tree Build, Segment Tree Query and Segment Tree Modify first.

## Example

For array [1,2,7,8,5], and queries [(1,2),(0,4),(2,4)], return [2,1,5]

## Code - Java

/**
* Definition of Interval:
* public class Interval {
*   int start, end;
*   Interval(int start, int end) {
*     this.start = start;
*     this.end = end;
*   }
* }
*/
public class Solution {
/**
* @param A, queries: Given an integer array and an query list
* @return: The result list
*/
public ArrayList<Integer> intervalMinNumber(int[] A,
ArrayList<Interval> queries) {
Node root = build(A, 0, A.length - 1);

ArrayList<Integer> result = new ArrayList<>();
for (Interval q : queries) {
}
return result;
}

private int query(Node root, Interval query) {
if (root.start > query.end || root.end < query.start) {
return Integer.MAX_VALUE;
}
if (root.start >= query.start && root.end <= query.end) {
return root.val;
}
return Math.min(query(root.left, query), query(root.right, query));
}

private Node build(int[] A, int start, int end) {
Node node = new Node(start, end, A[start]);
if (start == end) {
return node;
}
int mid = (start + end) / 2;
node.left = build(A, start, mid);
node.right = build(A, mid + 1, end);
node.val = Math.min(node.left.val, node.right.val);
return node;
}

class Node {
int start, end, val;
Node left, right;

public Node(int start, int end, int val) {
this.start = start;
this.end = end;
this.val = val;
}
}
}