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Python: algorithm to find four numbers in an unsorted array which satisfy desired sum

By : user2947986
Date : November 14 2020, 04:48 PM
Any of those help Given an array of unsorted positive integers and a desired sum, what is the best way to print in ascending order 4 distinct 0-based indexes of elements in the array that add up to an input sum, if such indexes exist. If there are multiple solutions I just need to print the lowest one according to the lexicographical order
code :
``````In [36]: A
Out[36]: [3, 2, 1, 4, 5, 7, 6, 9, 7, 8]

In [37]: S
Out[37]: 30

In [38]: N
Out[38]: 4

In [39]: for tup in itertools.combinations(A,N):
....:     if sum(tup) == S:
....:          for perm in itertools.product(*[[i for i,num in enumerate(A) if num==n] for n in tup]):
....:              print(*sorted(perm))
....:
5 6 7 9
6 7 8 9
5 6 7 9
6 7 8 9
``````

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Unsorted n array- algorithm to find duplicates using 2 nested loops

By : Jayson Aquino
Date : March 29 2020, 07:55 AM
should help you out The following question was:
code :
``````bool IsHaveDup(int[] myArray, int arraySize)
{
int i, j;
bool isHaveDup = false;

for (i = 0; i < arraySize - 1; i++)
{
for (j = i + 1; j < arraySize; j++)
{
if (myArray[i] == myArray[j])
{
isHaveDup = true;
break;
}
}

if (isHaveDup)
{
break;
}
}

return isHaveDup;
}
``````

Find Median in unsorted array using Quickselect algorithm

By : Fermata
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I believe you need to get the kth position of the pivot and cross check if its at the same position as the median, by performing this "int k=j-low+1;" with respect from the variable low in the array. For example a value at the 2nd index will be the 3rd smallest element in the array.
Also for the second recursive call, since we know the median is on the right hand side of the pivot(which is at the kth position) we expect the result to be at the (n-k)th position of the right hand sided subarray
code :
`````` public static int progress(int[] a, int low, int high,int n){
if (low==high)
return a[low];
//partition array and return index of pivot
int j= partition(a, low, high, n);

//find the kth position of the pivot
int k=j-low+1;
//if the kth position of the pivot is the same as the required ith smallest int return pivot.
if (k==n){
return a[j];
}
else
if (n<k)
return progress(a, low, j-1, n, i);
else if (n>k)
return progress(a, j+1, high, n-k, i);
}
``````
``````int n= (a.length-1)/2;
``````

Algorithm - Find maximum sum of two numbers in unsorted array that have a minimum distance

By : Pedro Braconnot
Date : March 29 2020, 07:55 AM
this one helps. For each position in the array, find and store the maximum of the elements up to that position. This can be done in O(n) for all positions by updating the maximum in each step.
By scanning right-to-left in the same manner, find and store for each position the maximum of all elements from that position to the end.

Complexity of the following algorithm to find minimum value in an unsorted array

By : chko.luevano
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , By using divide & conquer approach, if we divide an array into two halves repeatedly, until they reduce into size of two -after which we can in O(1) time return the minimum of the two. Extending the approach, in order to merge two subarrays A & B with their minimum 'a' & 'b' respectively, we can directly return their minimum in O(1) time -making merging step a constant time operation. ,
after which we can in O(1) time return the minimum of the two.

How to find 2 numbers and their sum in an unsorted array

By : Eskimu
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , This is very similar to the standard problem 3SUM, which many of the related questions along the right are about.
Your solution is O(n^2 lg n); there are O(n^2) algorithms based on sorting the array, which work with slight modification for this variant. The best known lower bound is O(n lg n) (because you can use it to perform a comparison sort, if you're clever about it). If you can find a subquadratic algorithm or a tighter lower bound, you'll get some publications out of it. :)
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