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How Can I Find the MIN, MAX, SUM and AVG of A Sorted List


How Can I Find the MIN, MAX, SUM and AVG of A Sorted List

By : balgood
Date : November 22 2020, 09:00 AM
around this issue The program is intended to read elements from a file then sort those elements using shell sort, save the sorted elements into an array and then find the minimum, maximum, sum and average. , Well if the list is sorted then:
min is first element
code :


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How to find an index at which a new item can be inserted into sorted list and keep it sorted?

How to find an index at which a new item can be inserted into sorted list and keep it sorted?


By : Szocs Barni
Date : March 29 2020, 07:55 AM
Hope that helps Use bisect. It's not the most beautiful API, but it's exactly what you need.
You'll want to use bisect.bisect, which returns exactly what you want.
How can I insert a sorted linked list into another sorted one while maintaining it sorted?

How can I insert a sorted linked list into another sorted one while maintaining it sorted?


By : Nikolaos N
Date : March 29 2020, 07:55 AM
I hope this helps you . You are looking for the merge phase of merge sort.
it can be done in LinkedList simply by iterating the two together, and pushing to a result list the smallest element every time.
code :
iter1 = list1.iterator()
iter2 = list2.iterator()
//before these commands make sure the lists are not empty:
curr1 = iter1.next(); 
curr2 = iter2.next();
List result = new LinkedList();
while curr1 != null && curr2 != null {
   if curr1 < curr2 { 
          result.add(curr1);
          curr1 = list1.next();
   } else { 
          result.add(curr2);
          curr2 = list2.next();
   }
}
while (curr1 != null) {
   result.add(curr1);
   curr1 = curr1.next();
}
while (curr2 != null) {
   result.add(curr2);
   curr2 = curr1.next();
}
In a python list which is sorted, find the closest value to target value and its index in the list

In a python list which is sorted, find the closest value to target value and its index in the list


By : icanc
Date : March 29 2020, 07:55 AM
should help you out bisect wasn't used in the linked question because the list was not sorted. Here, we don't have the same problem, and we can use bisect for the speed it provides:
code :
import bisect

def find_closest_index(a, x):
    i = bisect.bisect_left(a, x)
    if i >= len(a):
        i = len(a) - 1
    elif i and a[i] - x > x - a[i - 1]:
        i = i - 1
    return (i, a[i])

find_closest_index([1, 2, 3, 7, 10, 11], 0)   # => 0, 1
find_closest_index([1, 2, 3, 7, 10, 11], 7)   # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 8)   # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 9)   # => 4, 10
find_closest_index([1, 2, 3, 7, 10, 11], 12)  # => 5, 11
def bisect_left_rev(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] > x: lo = mid+1
        else: hi = mid
    return lo

def find_closest_index_rev(a, x):
    i = bisect_left_rev(a, x)
    if i >= len(a):
        i = len(a) - 1
    elif i and a[i] - x < x - a[i - 1]:
        i = i - 1
    return (i, a[i])

find_closest_index_rev([11, 10, 7, 3, 2, 1], 0)   # => 5, 1
find_closest_index_rev([11, 10, 7, 3, 2, 1], 7)   # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 8)   # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 9)   # => 1, 10
find_closest_index_rev([11, 10, 7, 3, 2, 1], 12)  # => 0, 11
Binary search of sorted list that find the closest value in list to targeted value provided by the user Python 3

Binary search of sorted list that find the closest value in list to targeted value provided by the user Python 3


By : ThisIsErick
Date : March 29 2020, 07:55 AM
Any of those help can someone help me how to access all the float point values and do a binary search so that if one match it will output the title which is the second on in the list. For example, looking at the output below if 0.369 was matched it will output selfish. Thank you. , If I understand you properly here is what you are looking for:
code :
def binary_search(storage, target):
    first = 0
    last = len(storage) - 1
    while first <= last:
        mid = (first + last) // 2
        value = float(storage[mid][0])
        if target == value:
            return storage[mid][1]
        elif value < target:
            first = mid + 1
        else:
            last = mid - 1
    return -1


storage = [["0.369", "selfish", "Future"], ["0.412", "family", "Future"]]

target = float(input("Please enter the desired float "))

result = binary_search(storage, target)
print(result)
Insert a node into a sorted linked list of integers such that the list still remains sorted with final members for next

Insert a node into a sorted linked list of integers such that the list still remains sorted with final members for next


By : Rodrive
Date : September 07 2020, 12:00 AM
it fixes the issue When the next link is final:
The only theoretical way would be to add a new node in front of the list and shift the data:
code :
void insertSorted(MyList list, int data) {
    list.head = new Node(0, list.head); // Insert in front;
    Node prior = list.head;
    // Invariant condition: prior points to a node (not null) and soon data >= prior.data
    Node current = prior.next;
    while (current != null) {
        if (data < current.data) {
            break;
        }
        prior.data = current.data; // Shift smaller
        prior = current;
        current = current.next;
    }
    prior.data = data;
}

insert:      d
list.head:   a ; b ; c ; e ; f ; g
--------------------------------------
list.head:   X ; a ; b ; c ; e ; f ; g
             a <-´   |   |
                 b <-´   |
                     c <-´
                         d
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