You are given an array nums consisting of positive integers where all integers have the same number of digits.
The digit difference between two integers is the count of different digits that are in the same position in the two integers.
\n\nReturn the sum of the digit differences between all pairs of integers in nums.
\n
Example 1:
\n\nInput: nums = [13,23,12]
\n\nOutput: 4
\n\nExplanation:
\nWe have the following:
\n- The digit difference between 13 and 23 is 1.
\n- The digit difference between 13 and 12 is 1.
\n- The digit difference between 23 and 12 is 2.
\nSo the total sum of digit differences between all pairs of integers is 1 + 1 + 2 = 4.
Example 2:
\n\nInput: nums = [10,10,10,10]
\n\nOutput: 0
\n\nExplanation:
\nAll the integers in the array are the same. So the total sum of digit differences between all pairs of integers will be 0.
\n
Constraints:
\n\n2 <= nums.length <= 1051 <= nums[i] < 109nums have the same number of digits.车尔尼有一个数组 nums ，它只包含 正 整数，所有正整数的数位长度都 相同 。
两个整数的 数位不同 指的是两个整数 相同 位置上不同数字的数目。
\n\n请车尔尼返回 nums 中 所有 整数对里，数位不同之和。
\n\n
示例 1：
\n\n输入：nums = [13,23,12]
\n\n输出：4
\n\n解释：
\n计算过程如下：
\n- 13 和 23 的数位不同为 1 。
\n- 13 和 12 的数位不同为 1 。
\n- 23 和 12 的数位不同为 2 。
\n所以所有整数数对的数位不同之和为 1 + 1 + 2 = 4 。
示例 2：
\n\n输入：nums = [10,10,10,10]
\n\n输出：0
\n\n解释：
\n数组中所有整数都相同，所以所有整数数对的数位不同之和为 0 。
\n\n
提示：
\n\n2 <= nums.length <= 1051 <= nums[i] < 109nums 中的整数都有相同的数位长度。c be the number of occurences of a digit on a position, that will contribute with c * (n - c) to the final answer, where n is the number of integers in nums."
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