Permute to Make List Larger - Facebook Top Interview Questions


Problem Statement :


You are given two lists of integers a and b, both of the same length.

Given that you can first permute a in any order, return the maximum number of indices where a[i] > b[i].

Constraints

n ≤ 100,000 where n is the length of a and b

Example 1


Input

a = [3, 5, 8, 1]

b = [4, 7, 2, 1]

Output

3

Explanation

If we permute a to [5, 8, 3, 1].



Solution :



title-img




                        Solution in C++ :

int solve(vector<int>& a, vector<int>& b) {
    sort(a.begin(), a.end(), greater<int>());
    sort(b.begin(), b.end(), greater<int>());

    int res = 0;
    auto ai = a.begin(), bi = b.begin();
    while (bi != b.end())
        if (*ai > *bi)
            res++, ai++, bi++;
        else
            bi++;

    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] a, int[] b) {
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for (int v : a) {
            map.put(v, map.getOrDefault(v, 0) + 1);
        }
        int ret = 0;
        for (int v : b) {
            Integer find = map.higherKey(v);
            if (find == null) {
                continue;
            }
            map.put(find, map.get(find) - 1);
            if (map.get(find) == 0) {
                map.remove(find);
            }
            ret++;
        }
        return ret;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, a, b):
        a = sorted(a)
        b = sorted(b)
        c = 0
        while a and b:
            while b and b[-1] >= a[-1]:
                b.pop()
            if b:
                a.pop()
                b.pop()
                c += 1
        return c
                    


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