0303 Range Sum Query

0303 Range Sum Query#

Problem#

Given an integer array nums, handle multiple queries of the following type:

Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

NumArray(int[] nums) Initializes the object with the integer array nums.

int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + … + nums[right]).

Examples#

Example 1:

Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]

Explanation
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraint#

1 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= left <= right < nums.length
At most 104 calls will be made to sumRange.

Analysis#

  • we can definitely use for loop to do the summation, but it will be inefficient if the sumRange method will be called thousands of times as each call is \(O(n)\).

  • Here we precalculate the accumalative sum in \(O(n)\), and then can call sumRange in \(O(1)\).

Solution#

# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(left,right)
class NumArray:
    def __init__(self, nums: list):
        self.nums = nums
        self.pre_sums = self.presum()
    
    def presum(self):
        pre_sums = [0]
        for num in self.nums:
            pre_sums.append(pre_sums[-1]+num)
        
        return pre_sums 

    def sumRange(self, left: int, right: int) -> int:
        return self.pre_sums[right+1] - self.pre_sums[left]

# tests
nums = [-2, 0, 3, -5, 2, -1]
array = NumArray(nums)
print(array.sumRange(0, 2))
print(array.sumRange(2, 5))
print(array.sumRange(0, 5))

1
-1
-3
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