Make 2 arrays ($B and $C) using $A elements that [ (sum of $B elements) - (sum of $C elements) ] equals minimum number (
Tag : php , By : sgmichelsen
Date : March 29 2020, 07:55 AM
hope this fix your issue Sort the numbers in $A, then start moving them into $B and $C from the front or end of $A as needed, keeping them sorted as well. When done, you can do binary search and shift items back and forth as needed to balance. I'll add a code example if you'd like
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How to remove duplicate part-elements from elements without being biased towards elements that occur later
Date : March 29 2020, 07:55 AM
I hope this helps . As you state, since theList is sorted, you have a bias of removing lower runs than higher runs. If you would like symmetry, then one way would be to randomly shuffle theList before starting the removal phase. Randomly shuffling a (Javascript) array is a well-known problem. See this community Wiki question, for ready-to-use answers.
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Selecting 5 elements out of an array of n elements and applying Bubble sort on those 5 elements
Date : March 29 2020, 07:55 AM
this one helps. Any algorithm that doesn't depend on the input size, is constant. So in your case, it doesn't depend on the array size, so yes, it is constant.
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Compare two lists to find the new elements, the removed elements and the existing elements
Date : March 29 2020, 07:55 AM
To fix this issue I would build your array operations atop set ones. Since there are not supposed to be duplicates, this is the correct structure for internal calculations. I don't know why Set.prototype does not have on it at least union, intersection, and difference, but it's quite trivial to write them yourself. // Set operations
const intersection = (set1, set2) => new Set([...set1].filter(x => set2.has(x)))
const difference = (set1, set2) => new Set([...set1].filter(x => !set2.has(x)))
// Array operations, builtt using the set ones
// NB: Arrays are NOT Sets, and there is some information lost in the conversion.
// But Sets are the proper data structure for unordered collections of unique values.
const intersectionA = (arr1, arr2) => Array.from(intersection(new Set(arr1), new Set(arr2)))
const differenceA = (arr1, arr2) => Array.from(difference(new Set(arr1), new Set(arr2)))
// Main code
const breakdown = (prev, curr) => ({
delete_id: differenceA(prev, curr),
new_id: differenceA(curr, prev),
existing_id: intersectionA(prev, curr)
})
let previous_id = ["5b7b5498ab3f510e307e7d04", "5b7ae97cc6d75e1331e28d9c", "5b7a0c207fab2722a2081caf"];
let current_id = ["5b7b5498ab3f510e307e7d04", "5b83e3b4412f370bd7b9a05d"];
console.log(breakdown(previous_id, current_id))
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Find the minimum sum of 'P' elements in an array of N elements such that no more than 'k' consecutive elements are selec
Tag : cpp , By : Ivan Kitanovski
Date : March 29 2020, 07:55 AM
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