README_EN.md

303. Range Sum Query - Immutable

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Description

Given an integer array nums, find the sum of the elements 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 the nums array in the range [left, right] inclusive (i.e., sum(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 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) 
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

 

Constraints:

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

Solutions

Python3

class NumArray:

    def __init__(self, nums: List[int]):
        n = len(nums)
        self.sums = [0] * (n + 1)
        for i in range(n):
            self.sums[i + 1] = nums[i] + self.sums[i]


    def sumRange(self, i: int, j: int) -> int:
        return self.sums[j + 1] - self.sums[i]


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

Java

class NumArray {

    private int[] sums;

    public NumArray(int[] nums) {
        int n = nums.length;
        sums = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            sums[i + 1] = nums[i] + sums[i];
        }
    }

    public int sumRange(int i, int j) {
        return sums[j + 1] - sums[i];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

...