303

303. Range Sum Query - Immutable (Easy)

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.

Companies: Facebook, Amazon, Bloomberg, Snowflake, Palantir Technologies

Related Topics:
Array, Design, Prefix Sum

Similar Questions:

• Range Sum Query 2D - Immutable (Medium)
• Range Sum Query - Mutable (Medium)
• Maximum Size Subarray Sum Equals k (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/range-sum-query-immutable
// Author: github.com/lzl124631x
// Time:
//      NumArray: O(N)
//      sumRange: O(1)
// Space: O(N)
class NumArray {
    vector<int> sum;
public:
    NumArray(vector<int>& A) : sum(A.size() + 1) {
        for (int i = 0; i < A.size(); ++i) sum[i + 1] = sum[i] + A[i];
    }
    int sumRange(int left, int right) {
        return sum[right + 1] - sum[left];
    }
};