Range sum query - immutable

Time: ctor:O(N),lookup:O(1); Space: O(N); easy

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example 1:

Input/Output: nums = [-2, 0, 3, -5, 2, -1]

  • sumRange(0, 2) -> 1

  • sumRange(2, 5) -> -1

  • sumRange(0, 5) -> -3

Explanation:

  • sumRange(0, 2) -> (-2) + 0 + 3 = 1

  • sumRange(2, 5) -> 3 + (-5) + 2 + (-1) = -1

  • sumRange(0, 5) -> (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Example 2:

Input/Output: nums = [-4, -5]

  • sumRange(0, 0) -> -4

  • sumRange(1, 1) -> -5

  • sumRange(0, 1) -> -9

  • sumRange(1, 1) -> -5

  • sumRange(0, 0) -> -4

Explanation:

  • sumRange(0, 0) -> -4

  • sumRange(1, 1) -> -5

  • sumRange(0, 1) -> (-4) + (-5) = -9

  • sumRange(1, 1) -> -5

  • sumRange(0, 0) -> -4

Notes:

  • You may assume that the array does not change.

  • There are many calls to sumRange function.

[1]:

class NumArray(object):
    """
    Time: ctor:O(N), lookup:O(1)
    Space: O(N)
    """
    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.accu = [0]
        for num in nums:
            self.accu.append(self.accu[-1] + num),

    def sumRange(self, i, j):
        """
        sum of elements nums[i..j], inclusive.
        :type i: int
        :type j: int
        :rtype: int
        """
        return self.accu[j + 1] - self.accu[i]

[2]:

nums = [-2, 0, 3, -5, 2, -1]
n = NumArray(nums)
assert n.sumRange(0, 2) == 1
assert n.sumRange(2, 5) == -1
assert n.sumRange(0, 5) == -3

nums = [-4, -5]
n = NumArray(nums)
assert n.sumRange(0, 0) == -4
assert n.sumRange(1, 1) == -5
assert n.sumRange(0, 1) == -9
assert n.sumRange(1, 1) == -5
assert n.sumRange(0, 0) == -4