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  1. Practice
  2. Data Structures
  3. Arrays
  4. Array Manipulation
  5. Discussions

Array Manipulation

  • m_zayed + 2 comments

    Python 3 more clear code

    def arrayManipulation(n, queries):
    
    my_array = [0] * (n+1)
    count = 0
    temp = 0
    for first,last,value in queries:
        
        my_array[first-1] += value
        my_array[last] -= value
    
    for item in my_array:
        count += item
        if count > temp:
            temp = count
       
    return temp
    • perfectcodder + 2 comments

      can you explain your logic ??

      • brayoni_ian + 0 comments
        [deleted]
      • Kanahaiya + 0 comments

        Hi,

        Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)

        I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.

        Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

        https://youtu.be/hDhf04AJIRs

        Would really appreciate your feedback like, dislike , comment etc. on my video.

    • brayoni_ian + 0 comments

      Thank you. This is clear.

      You could also find the max using: max(accumulate(my_array))

      See below:

      def arrayManipulation(n, queries):
    # initialise list
    arr = [0] * (n + 1)

    # update
    for a, b, k in queries:
        arr[a - 1] += k
        arr[b] -= k
    
    # accumulate values
    return max(accumulate(arr))

      However, I have not understood the theory behind the difference array algorithm, why it works the way it works and how to derive it.