JavaScript solution, 88ms
With comments\n\nvar maxSubsequence = function(nums, k) {\n\t// instantiate the result array, also the set the length of nums to a variable\n    let result = []
**With comments**\n```\nvar maxSubsequence = function(nums, k) {\n\t// instantiate the result array, also the set the length of nums to a variable\n    let result = [], len = nums.length\n\t\n\t// clone the nums array into a new array. Use this to retain the original index of each number\n    numsIndex = nums.map(i => i)\n\t\n\t// sort the nums array in ascending order\n    nums.sort((a, b) => { return a - b })\n\t\n\t// create a variable to store the largest \'k\' amount of numbers\n\t// slice the nums array with the value of \'nums.length\' minus\'k\'. \n\t// This will return an array with the \'k\' amount of largest numbers, but they may be in the wrong order\n    let largeNums = nums.slice(len-k)\n\t\n\t// loop through the cloned nums array\n    for(let i = 0; i < len; i++){\n\t\n\t\t// if the large numbers array contains the number in the cloned nums array at index \'i\'...\n        if(largeNums.includes(numsIndex[i])){\n\t\t\n\t\t\t// push the number to the result array\n            result.push(numsIndex[i])\n\t\t\t\n\t\t\t// this next step is important, grab the index of the number inside of the large nums array...\n            let largeNumsIndex = largeNums.indexOf(numsIndex[i])\n\t\t\t\n\t\t\t// and remove it. This is done so that if there are duplicates of the number we just pushed to our results array\n\t\t\t// present in the cloned array they will not be found **again** in the large nums array\n            largeNums.splice(largeNumsIndex, 1)\n        }\n    } \n\t// return the result array\n    return result\n};\n```\n**Just code**\n```\nvar maxSubsequence = function(nums, k) {\n    let result = [], len = nums.length\n    numsIndex = nums.map(i => i)\n    nums.sort((a, b) => { return a - b })\n    let largeNums = nums.slice(len-k)\n    for(let i = 0; i < len; i++){\n        if(largeNums.includes(numsIndex[i])){\n            result.push(numsIndex[i])\n            let largeNumsIndex = largeNums.indexOf(numsIndex[i])\n            largeNums.splice(largeNumsIndex, 1)\n        }\n    }            \n    return result\n};\n```
['JavaScript']