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Abhishek Chaudhary
Abhishek Chaudhary

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Range Sum Query - Immutable

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

  1. 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]).

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

Constraints:

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

SOLUTION:

class NumArray:

    def __init__(self, nums: List[int]):
        currSum = 0
        self.sumTill = [currSum]
        for num in nums:
            currSum += num
            self.sumTill.append(currSum)

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


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(left,right)
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