[JS] O(N^3) - bruteforce to delete 2 nums
Intuition\nBruteForce to try to delete any 2 numbers.\n\n#### Code\n\n/**\n * @param {number[]} nums1\n * @param {number[]} nums2\n * @return {number}\n */\nvar
#### Intuition\nBruteForce to try to delete any 2 numbers.\n\n#### Code\n```\n/**\n * @param {number[]} nums1\n * @param {number[]} nums2\n * @return {number}\n */\nvar minimumAddedInteger = function(nums1, nums2) {\n    const m = nums1.length;\n    const n = nums2.length;\n    nums2.sort((a,b)=>a-b);\n    nums1.sort((a,b)=>a-b);\n    if(n===1) {\n        return nums2[0] - nums1[m-1];\n    }\n    console.log(nums1, nums2);\n    for(let i = 0; i<m; i++){\n        for(let j = i+1; j<m; j++) {\n            // delete i & j\n            const first1 = i === 0 ? (j === 1 ? 2 : 1) : 0;\n            const diff = nums2[0] - nums1[first1];\n            // check\n            let flag = true;\n            let q = 0;\n            for(let p = 0; p<m; p++) {\n                if(p===j || p===i){\n                    continue;\n                }\n                if(nums2[q] !== nums1[p]+diff) {\n                    flag = false;\n                    break;\n                }\n                q++;\n            }\n            if(flag) {\n                return diff;\n            }\n        }\n    }\n};\n```
['JavaScript']