Run results for: Array duplicate removal for duplicates exceeding `N`-number

Run details:

User agent: Mozilla/5.0 (Windows NT 10.0; WOW64; rv:54.0) Gecko/20100101 Firefox/54.0

Browser: Firefox 54

Operating system: Windows

Device Platform: Other

Date tested: 6 years ago

Test name	Executions per second
original duplicate removal	7212.3 Ops/sec
2-pass approach	7775.4 Ops/sec
2-pass with filtering	7766.6 Ops/sec

Tests:

original duplicate removal
// Remove duplicates that occur 3 or more times in an array // keeping unique values and those with less than 3 function removeMany(arr) { const newArr = Array.from(arr).sort() let count = 0; let result = [] newArr.forEach((value, index, ar) => { count += 1; // refactored afterwards from (ar[index + 1] !== value) if (ar.lastIndexOf(value) <= index && count <= 2) { for (var i = 0; i < arr.length; i++) { if (arr[i] === value) { result.push(arr[i]) } } count = 0 } else if (ar[index + 1] !== value) { count = 0; } }); // +1 is there anyway to return a result that mimicks the original order of `numbers`? return result; // [1, 2, 2, 3, 4, 4] } const numbers = [1, 2, 3, 2, 4, 4, 5, 5, 5, 5]; console.log(removeMany(numbers));
2-pass approach
// Remove duplicates that occur 3 or more times in an array // keeping unique values and those with less than 3 function removeMany(arr) { let countMappings = arr.reduce(function(carry, item) { if (carry[item]!== undefined) { carry[item]++; } else { carry[item] = 1; } return carry; }, {}); return arr.reduce(function(final, item) { if (countMappings[item] <3) { final.push(item); } return final; }, []); } const numbers = [1, 2, 3, 2, 4, 4, 5, 5, 5, 5]; console.log(removeMany(numbers));
2-pass with filtering
function removeMany(numbers, max) { const numberMap = numbers.reduce((map, num) => { map[num] = map[num] ? map[num] + 1 : 1; return map; }, []); return numbers.filter(num => numberMap[num] < max); } const numbers = [1, 2, 3, 2, 4, 4, 5, 5, 5, 5]; console.log(removeMany(numbers, 3));

Run results for: Array duplicate removal for duplicates exceeding `N`-number

https://codereview.stackexchange.com/q/175067/120114

original duplicate removal

2-pass approach

2-pass with filtering