数组的最小相等和

You are given two arrays nums1 and nums2 consisting of positive integers.

You have to replace all the 0's in both arrays with strictly positive integers such that the sum of elements of both arrays becomes equal.

Return the minimum equal sum you can obtain, or -1 if it is impossible.

 

Example 1:

Input: nums1 = [3,2,0,1,0], nums2 = [6,5,0]
Output: 12
Explanation: We can replace 0's in the following way:
- Replace the two 0's in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4].
- Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1].
Both arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain.

Example 2:

Input: nums1 = [2,0,2,0], nums2 = [1,4]
Output: -1
Explanation: It is impossible to make the sum of both arrays equal.

 

Constraints:

• 1 <= nums1.length, nums2.length <= 105
• 0 <= nums1[i], nums2[i] <= 106

/*
 * 思路：
 * 1. 使用流计算数组的非零元素和与零的数量。
 * 2. 计算两个数组和的差值与零的数量。
 * 3. 返回最小的相等和或-1。
 */

import java.util.Arrays;

public int minEqualSumStream(int[] nums1, int[] nums2) {
    int sum1 = Arrays.stream(nums1).filter(num -> num != 0).sum();
    int sum2 = Arrays.stream(nums2).filter(num -> num != 0).sum();
    long zeros1 = Arrays.stream(nums1).filter(num -> num == 0).count();
    long zeros2 = Arrays.stream(nums2).filter(num -> num == 0).count();
    long totalZeros = zeros1 + zeros2;
    int diff = Math.abs(sum1 - sum2);
    if (totalZeros < diff) return -1;
    return Math.max(sum1, sum2) + diff;
}

这个题解的核心思路是首先计算每个数组中除零外所有元素的和，并加上零的个数，因为零至少需要被替换成1来使得和有效。这样，对于每个数组，计算的结果是如果所有的零被替换成1后数组的最小可能和。然后检查两个计算结果a和b。如果其中一个数组的和（假设为a）小于另一个数组的和（假设为b），同时第一个数组中不包含0（即无法通过增加零的值来调整总和），则返回-1，表示无法通过任何正整数替换0使两数组和相等。如果不是这种情况，返回a和b中的较大值，因为这就是将所有零替换为1后，能使得两数组和相等的最小值。

O(n + m)

O(1)