Find index of less or equal value in list recursively (python)

Question

Got task to find indexes of value and double that value in input list.
In input first line we get list range, second – list of values, third – value to find.
The output is 2 numbers – indexes of equal or higher value and double the value. If there is none, return -1

Example input:

6
1 2 4 4 6 8
3

Example output:

3 5

What i got so far is standart binary search func, but i dont get how to make it search not only for exact number but nearest higher.

def binarySearch(arr, x, left, right):
    if right <= left:
        return -1
    mid = (left + right) // 2
    if arr[mid] >= x: 
        return mid
    elif x < arr[mid]:          
        return binarySearch(arr, x, left, mid)
    else: 
        return binarySearch(arr, x, mid + 1, right)

def main():
    n = int(input())
    k = input().split()
    q = []
    for i in k:
        q.append(int(i))
    s = int(input())
    res1 = binarySearch(q, s, q[0], (n-1))
    res2 = binarySearch(q, (s*2), q[0], (n-1))
    print(res1, res2)

if __name__ == "__main__":
    main()

The input is:

6
1 2 4 4 6 8
3

And output:

3 4

Asked By: r3tr0

||

Source

Answer 1

Here’s a modified binary search which will return the base zero index of a value if found or the index of the next highest value in the list.

def bsearch(lst, x):
    L = 0
    R = len(lst) - 1
    while L <= R:
        m = (L + R) // 2
        if (v := lst[m]) == x:
            return m
        if v < x:
            L = m + 1
        else:
            R = m - 1
    return L if L < len(lst) else -1


data = list(map(int, '1 2 4 4 6 8'.split()))

for x in range(10):
    print(x, bsearch(data, x))

Output:

Answered By: OldBill

Find index of less or equal value in list recursively (python)

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