Given an integer array nums
, handle multiple queries of the following type:
- Calculate the sum of the elements of
nums
between indicesleft
andright
inclusive whereleft <= right
.
Implement the NumArray
class:
-
NumArray(int[] nums)
Initializes the object with the integer arraynums
. -
int sumRange(int left, int right)
Returns the sum of the elements ofnums
between indicesleft
andright
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 tosumRange
.
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)
Top comments (0)