Python Code Optimization Problem (Lintcode Problem 1886 Moving Targets)

Question

I have been working on a problem on https://www.lintcode.com/ and I have ran into a problem while doing one of the questions. The problem requires me to write a function with two parameters. A list of nums and a target num. You have to take all instances of the target from the list and move them to the front of the original list and the function cannot have a return value. The length of the list is between 1 and 1000000. You also have to do it within a time limit, which is around 400 milliseconds. I can solve the problem, I can't pass the last test case where the length of the list is 1000000. Does anyone know how I can make my code faster?

Original Problem Description for anyone who still isn't clear:

Current Code:

def MoveTarget(nums, target):
    if len(set(nums)) == 1:
        return nums
    index = [i for i in range(len(nums)) if nums[i] == target]
    for i in index:
        nums.insert(0, nums.pop(i))

It works if you do:

def MoveTarget(nums, target): 
    count = 0
    left, right = len(nums) - 1, len(nums) - 1
    
    while left >= 0:
        if nums[left] != target:
            nums[right] = nums[left]
            right -= 1
        else:
            count += 1
        left -= 1
        
    for i in range(count):
        nums[i] = target

but I was wondering if there was another, less complicated way.

You may try this - A[:] = sorted(A, key=lambda x: x!= target) A is your list. — Daniel Hao, Aug 14 '21 at 23:00
@DanielHao Or just `A.sort(key=target.__ne__)`. Less memory, too. — Kelly Bundy, Aug 14 '21 at 23:31

score 2 · Accepted Answer · edited Aug 15 '21 at 19:36

2

Here is a simple and relatively efficient implementation:

def MoveTarget(nums, target):
    n = nums.count(target)
    nums[:] = [target] * n + [e for e in nums if e != target]

It creates a new list with the n target values in the front and append all the other values that are not target. The input list nums is mutated thanks to the expression nums[:] = ....

The solution run in linear time as opposed to the previously proposed implementations (running in quadratic time). Indeed, insert runs in linear time in CPython.

edited Aug 15 '21 at 19:36

ignassew

83
5

answered Aug 14 '21 at 23:19

Jérôme Richard

41,678
6
29
59

Wow! I didn't even think of doing this! Thanks! – CoderTang Aug 15 '21 at 18:33

score -1 · Answer 2 · edited Aug 14 '21 at 23:08

-1

Your code uses 2 loops. One in:

index = [i for i in range(len(nums)) if nums[i] == target]

And one in:

for i in index:
    nums.insert(0, nums.pop(i))

Instead, you can combine finding the target and moving it to the front of array with only one loop, which will greatly reduce the execution time:

def MoveTarget(nums, target):
    if len(set(nums)) == 1:
        return nums
    for num in nums:
        if num == target:
            nums.insert(0, nums.pop(num))

edited Aug 14 '21 at 23:08

Jérôme Richard

41,678
6
29
59

answered Aug 14 '21 at 23:01

ignassew

83
5

2

Run some more test on long list, and it failed. – Daniel Hao Aug 14 '21 at 23:09

Python Code Optimization Problem (Lintcode Problem 1886 Moving Targets)

2 Answers2