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    LeetCode link: 303. Range Sum Query - Immutable, difficulty: 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 <= 10^4
    • -10^5 <= nums[i] <= 10^5
    • 0 <= left <= right < nums.length
    • At most 10^4 calls will be made to sumRange.

    Intuition
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    • Use a new array prefix_sums to save the sum of the previous elements.
    • The first element of prefix_sums is 0 because the prefix sum does not include the current element.
    • To find the sum of the elements from index left to right (inclusive), just use prefix_sums[right + 1] - prefix_sums[left].

    Complexity

    Time complexity

    O(N)

    Space complexity

    O(N)

    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Java # 

    class NumArray {
    private int[] prefixSums;

    public NumArray(int[] nums) {
        prefixSums = new int[nums.length + 1];
        var sum = 0;

        for (var i = 0; i < nums.length; i++) {
            sum += nums[i];
            prefixSums[i + 1] = sum;
        }
    }

    public int sumRange(int left, int right) {
        return prefixSums[right + 1] - prefixSums[left];
    }
}

    Python # 

    class NumArray:
    def __init__(self, nums: List[int]):
        self.prefix_sums = [0]
        sum_ = 0

        for num in nums:
            sum_ += num
            self.prefix_sums.append(sum_)

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

    C++ # 

    class NumArray {
private:
    vector<int> prefixSums;

public:
    NumArray(vector<int>& nums) {
        prefixSums.push_back(0);
        auto sum = 0;

        for (auto num : nums) {
            sum += num;
            prefixSums.push_back(sum);
        }
    }

    int sumRange(int left, int right) {
        return prefixSums[right + 1] - prefixSums[left];
    }
};

    JavaScript # 

    let prefixSums

var NumArray = function (nums) {
  prefixSums = [0]
  let sum = 0

  nums.forEach((num) => {
    sum += num
    prefixSums.push(sum)
  })
};

NumArray.prototype.sumRange = function (left, right) {
  return prefixSums[right + 1] - prefixSums[left]
};

    C# # 

    public class NumArray
{
    int[] prefixSums;

    public NumArray(int[] nums)
    {
        prefixSums = new int[nums.Length + 1];
        int sum = 0;

        for (int i = 0; i < nums.Length; i++)
        {
            sum += nums[i];
            prefixSums[i + 1] = sum;
        }
    }

    public int SumRange(int left, int right)
    {
        return prefixSums[right + 1] - prefixSums[left];
    }
}

    Go # 

    type NumArray struct {
    prefixSums []int
}

func Constructor(nums []int) NumArray {
    prefixSums := make([]int, len(nums) + 1)
    sum := 0

    for i, num := range nums {
        sum += num
        prefixSums[i + 1] = sum
    }

    return NumArray{prefixSums}
}

func (this *NumArray) SumRange(left int, right int) int {
    return this.prefixSums[right + 1] - this.prefixSums[left]
}

    Ruby # 

    class NumArray
  def initialize(nums)
    @prefix_sums = [0]
    sum = 0

    nums.each do |num|
      sum += num
      @prefix_sums << sum
    end
  end

  def sum_range(left, right)
    @prefix_sums[right + 1] - @prefix_sums[left]
  end
end
    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 303. Range Sum Query - Immutable.

    Intuition of solution 2: Directly Returning the Sum of the Array Range (Slow)

    Directly returning the sum of the values ​​in the array range can pass the test, but it will fail if the test case is stricter.
    So we still need to learn a more efficient solution: Prefix Sum solution.

    Complexity

    Time complexity

    O(M * N)

    Space complexity

    O(M * N)

    Python # 

    class NumArray:
    def __init__(self, nums: List[int]):
        self.nums = nums

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

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 303. Range Sum Query - Immutable.
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