Simple Brute force solution that works !!!!
Intuition\n Describe your first thoughts on how to solve this problem. \n taking all possibilities of nums1 by excluding different combinations of two elements 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n* taking all possibilities of nums1 by excluding different combinations of two elements \n* for each pair we are find the diff and validating the diff for remaining elements of nums2 in the k loop \n* if the diff is valid it is used for calculating the min difference \n* else discarded\n* O(n^2) solution\n* upvote if you can understand the solution \n* will be coding out an optimized approach soon and posting it in the solution for the community \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\no(n*2)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\no(n)\n\n# Code\n```\n/**\n * @param {number[]} nums1\n * @param {number[]} nums2\n * @return {number}\n */\nvar minimumAddedInteger = function(nums1, nums2) {\n   let n=nums1.length;\n   nums1.sort((a,b)=>a-b);\n   nums2.sort((a,b)=>a-b);\n   let min=Infinity;\n    for(let i=0;i<n-1;i++){\n        for(let j=i+1;j<n;j++){\n        let new_ar=(nums1.slice(0,i).concat(nums1.slice(i+1,j))).concat(nums1.slice(j+1));\n        let v=nums2[0]-new_ar[0];\n        let cnt=0;\n        for(let k=1;k<nums2.length;k++){\n            let diff=nums2[k]-new_ar[k];\n            if(diff!==v) cnt++;\n        }\n        if(!cnt) min=Math.min(min,nums2[0]-new_ar[0]);\n\n\n        }\n    }\n    return min;\n    \n};\n```
['Iterator', 'JavaScript']