# 303. Range Sum Query - Immutable
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
Example:
Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
Note:
- You may assume that the array does not change.
- There are many calls to sumRange function.
# Solution
Approach 1: Caching.
# Code (Python)
Approach 1:
class NumArray:
def __init__(self, nums: List[int]):
self.sums = [0]
for num in nums:
self.sums.append(self.sums[-1] + num)
def sumRange(self, i: int, j: int) -> int:
return self.sums[j+1] - self.sums[i]
# Code (C++)
Approach 1:
class NumArray {
private:
vector<int> sum;
public:
NumArray(vector<int> nums) {
int tmp = 0;
for (int i = 0; i < nums.size(); ++i)
{
tmp += nums[i];
sum.push_back(tmp);
}
}
int sumRange(int i, int j) {
if (i > 0)
return sum[j] - sum[i-1];
else
return sum[j];
}
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/