You are given a 2D integer array items where items[i] = [pricei, beautyi] denotes the price and beauty of an item respectively.

You are also given a 0-indexed integer array queries. For each queries[j], you want to determine the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item exists, then the answer to this query is 0.

Return an array answer of the same length as queries where answer[j] is the answer to the jth query.

&nbsp;
Example 1:

Input: items = [[1,2],[3,2],[2,4],[5,6],[3,5]], queries = [1,2,3,4,5,6]
Output: [2,4,5,5,6,6]
Explanation:
- For queries[0]=1, [1,2] is the only item which has price &lt;= 1. Hence, the answer for this query is 2.
- For queries[1]=2, the items which can be considered are [1,2] and [2,4]. 
  The maximum beauty among them is 4.
- For queries[2]=3 and queries[3]=4, the items which can be considered are [1,2], [3,2], [2,4], and [3,5].
  The maximum beauty among them is 5.
- For queries[4]=5 and queries[5]=6, all items can be considered.
  Hence, the answer for them is the maximum beauty of all items, i.e., 6.


Example 2:

Input: items = [[1,2],[1,2],[1,3],[1,4]], queries = [1]
Output: [4]
Explanation: 
The price of every item is equal to 1, so we choose the item with the maximum beauty 4. 
Note that multiple items can have the same price and/or beauty.  


Example 3:

Input: items = [[10,1000]], queries = [5]
Output: [0]
Explanation:
No item has a price less than or equal to 5, so no item can be chosen.
Hence, the answer to the query is 0.


&nbsp;
Constraints:


	1 &lt;= items.length, queries.length &lt;= 105
	items[i].length == 2
	1 &lt;= pricei, beautyi, queries[j] &lt;= 109


class Solution:
    def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:

        items.sort()
        dic = dict()
        res = []
        gmax = 0
        for p,b in items:
            gmax = max(b,gmax)
            dic[p] = gmax

        keys = sorted(dic.keys())
        for q in queries:
            ind = bisect.bisect_left(keys,q)
            if ind<len(keys) and keys[ind]==q:
                res.append(dic[q])
            elif ind==0:
                res.append(0)
            else:
                res.append(dic[keys[ind-1]])

        return res
class Solution {
    public int[] maximumBeauty(int[][] items, int[] queries) {
        int[] ans = new int[queries.length];
        Arrays.sort(items, (a, b) -> (a[0] - b[0]));
        int maxBeautySoFar = Integer.MIN_VALUE;
        int[] maxBeauty = new int[items.length];
        
        for(int i = 0; i < items.length; i++) {
            if(maxBeautySoFar < items[i][1]) maxBeautySoFar = items[i][1];
            maxBeauty[i] = maxBeautySoFar;
        }
        
        for(int i = 0; i < queries.length; i++) {
            int idx = findLargestIdxWithPriceLessThan(items, queries[i]);
            if(idx != Integer.MIN_VALUE) ans[i] = maxBeauty[idx];
        }
        return ans;
    }
    
    public int findLargestIdxWithPriceLessThan(int[][] items, int price) {
        int l = 0;
        int r = items.length - 1;
        int maxIdxLessThanEqualToPrice = Integer.MIN_VALUE; 
        while(l <= r) {
            int mid = (l + r)/2;
            if(items[mid][0] > price) {
                r = mid - 1;
            } else {
                maxIdxLessThanEqualToPrice = Math.max(maxIdxLessThanEqualToPrice, mid);
                l = mid + 1;
            }
        }
        return maxIdxLessThanEqualToPrice;
    }
}
class Solution {
public:
    vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
       vector<vector<int>> v;
        int n = queries.size();
        for(int i = 0; i < n; i++){
            v.push_back({queries[i],i});
        }
        sort(v.begin(),v.end());
        sort(items.begin(),items.end());
        vector<int> ans(n);
        int j=0;
        n = items.size();
        int mx = 0;
        for(auto &i: v){
            while(j<n && items[j][0]<=i[0]){
                mx = max(mx,items[j][1]);
                j++;
            }
            ans[i[1]] = mx;
        }
        return ans;
    }
};
var maximumBeauty = function(items, queries) {
    items.sort((a,b) => a[0]-b[0]);
    const n = items.length;
    
    
    let mx = items[0][1];
    
    for (let i = 0; i<n; i++) {
        mx = Math.max(mx, items[i][1]);
        items[i][1] = mx;
    }
    
    
    const ans = [];
    
    for (const q of queries) {
        let l = 0, r = n-1, a = 0;
        while (l<=r) {
            let mid = Math.floor(l+(r-l)/2);
            if (items[mid][0]<=q) {
                a = items[mid][1]
                l = mid+1;
            } else r = mid-1;
        }
        ans.push(a)
    }
    
    return ans;
};
Most Beautiful Item for Each Query