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

How to remove duplicate partelements 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 wellknown problem. See this community Wiki question, for readytouse answers.

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.

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))

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

