# Max Multiplied Pairings - Google Top Interview Questions

### Problem Statement :

```You are given two lists of integers nums and multipliers.
Consider an operation where we remove any number in nums and remove any number in multipliers then multiply them together.
Given that you must repeat this operation until one of the lists is empty, return the maximum sum of the multiplied numbers.

Note that the integers may be negative, zero, and/or positive.

Constraints

0 ≤ n ≤ 100,000 where n is the length of nums

0 ≤ m ≤ 100,000 where m is the length of multipliers

Example 1

Input

nums = [-5, 3, 2]

multipliers = [-3, 1]

Output

18

Explanation

We can match -5 with -3 and 3 with 1 to get -5 * -3 + 3 * 1.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<int>& a, vector<int>& b) {
int ans = 0;
sort(a.begin(), a.end());
sort(b.begin(), b.end());
int lhsa = 0;
int rhsa = (int)a.size() - 1;
int lhsb = 0;
int rhsb = (int)b.size() - 1;
while (lhsa <= rhsa && lhsb <= rhsb) {
if (a[lhsa] * b[lhsb] >= a[rhsa] * b[rhsb]) {
ans += a[lhsa++] * b[lhsb++];
} else {
ans += a[rhsa--] * b[rhsb--];
}
}
return ans;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] nums, int[] muls) {
Arrays.sort(nums);
Arrays.sort(muls);
int nums_l = 0;
int nums_r = nums.length - 1;

int muls_l = 0;
int muls_r = muls.length - 1;
int res = 0;
while (nums_l <= nums_r && muls_l <= muls_r) {
int cand1 = nums[nums_l] * muls[muls_l];
int cand2 = nums[nums_r] * muls[muls_r];

if (cand1 > cand2) {
res += cand1;
nums_l++;
muls_l++;
continue;
}
res += cand2;
nums_r--;
muls_r--;
}
return res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, A, B):
A = deque(sorted(A))
B = deque(sorted(B))

ans = 0
while A and B:
p1 = A[0] * B[0]
p2 = A[-1] * B[-1]
if p1 > p2:
ans += p1
A.popleft()
B.popleft()
else:
ans += p2
A.pop()
B.pop()

return ans```
```

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