# 303. Range Sum Query - Immutable

## Description

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.

## Solutions

Solution 1: Prefix Sum

We create a prefix sum array $s$ of length $n + 1$, where $s[i]$ represents the prefix sum of the first $i$ elements, that is, $s[i] = \sum_{j=0}^{i-1} nums[j]$. Therefore, the sum of the elements between the indices $[left, right]$ can be expressed as $s[right + 1] - s[left]$.

The time complexity for initializing the prefix sum array $s$ is $O(n)$, and the time complexity for querying is $O(1)$. The space complexity is $O(n)$.

• class NumArray {
private int[] s;

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

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

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

• class NumArray {
public:
NumArray(vector<int>& nums) {
int n = nums.size();
s.resize(n + 1);
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
}

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

private:
vector<int> s;
};

/**
* Your NumArray object will be instantiated and called as such:
* NumArray* obj = new NumArray(nums);
* int param_1 = obj->sumRange(left,right);
*/

• '''
>>> from itertools import accumulate
>>> accumulate([1,2,3])
<itertools.accumulate object at 0x108f38340>
>>> list(accumulate([1,2,3]))
[1, 3, 6]
>>> list(accumulate([1,2,3], initial=0))
[0, 1, 3, 6]
>>> list(accumulate([1,2,3], initial=10))
[10, 11, 13, 16]
'''
# note: when using python2, I always got error when importing it, via itertools.accumulate()
#       switching to python3, then all good for itertools.accumulate()
class NumArray:
def __init__(self, nums: List[int]):
self.s = list(accumulate(nums, initial=0))

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

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

############

class NumArray(object):
def __init__(self, nums):
"""
:type nums: List[int]
"""
self.dp = [0] * (len(nums) + 1)
for i in range(0, len(nums)):
self.dp[i + 1] = self.dp[i] + nums[i]

def sumRange(self, i, j):
"""
sum of elements nums[i..j], inclusive.
:type i: int
:type j: int
:rtype: int
"""
return self.dp[j + 1] - self.dp[i]

# Your NumArray object will be instantiated and called as such:
# numArray = NumArray(nums)
# numArray.sumRange(0, 1)
# numArray.sumRange(1, 2)


• type NumArray struct {
s []int
}

func Constructor(nums []int) NumArray {
n := len(nums)
s := make([]int, n+1)
for i, v := range nums {
s[i+1] = s[i] + v
}
return NumArray{s}
}

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

/**
* Your NumArray object will be instantiated and called as such:
* obj := Constructor(nums);
* param_1 := obj.SumRange(left,right);
*/

• class NumArray {
private s: number[];

constructor(nums: number[]) {
const n = nums.length;
this.s = Array(n + 1).fill(0);
for (let i = 0; i < n; ++i) {
this.s[i + 1] = this.s[i] + nums[i];
}
}

sumRange(left: number, right: number): number {
return this.s[right + 1] - this.s[left];
}
}

/**
* Your NumArray object will be instantiated and called as such:
* var obj = new NumArray(nums)
* var param_1 = obj.sumRange(left,right)
*/


• /**
* @param {number[]} nums
*/
var NumArray = function (nums) {
const n = nums.length;
this.s = Array(n + 1).fill(0);
for (let i = 0; i < n; ++i) {
this.s[i + 1] = this.s[i] + nums[i];
}
};

/**
* @param {number} left
* @param {number} right
* @return {number}
*/
NumArray.prototype.sumRange = function (left, right) {
return this.s[right + 1] - this.s[left];
};

/**
* Your NumArray object will be instantiated and called as such:
* var obj = new NumArray(nums)
* var param_1 = obj.sumRange(left,right)
*/


• class NumArray {
/**
* @param Integer[] $nums */ function __construct($nums) {
$this->s = [0]; foreach ($nums as $x) {$this->s[] = $this->s[count($this->s) - 1] + $x; } } /** * @param Integer$left
* @param Integer $right * @return Integer */ function sumRange($left, $right) { return$this->s[$right + 1] -$this->s[$left]; } } /** * Your NumArray object will be instantiated and called as such: *$obj = NumArray($nums); *$ret_1 = $obj->sumRange($left, \$right);
*/

• struct NumArray {
s: Vec<i32>,
}

/**
* &self means the method takes an immutable reference.
* If you need a mutable reference, change it to &mut self instead.
*/
impl NumArray {
fn new(mut nums: Vec<i32>) -> Self {
let n = nums.len();
let mut s = vec![0; n + 1];
for i in 0..n {
s[i + 1] = s[i] + nums[i];
}
Self { s }
}

fn sum_range(&self, left: i32, right: i32) -> i32 {
self.s[(right + 1) as usize] - self.s[left as usize]
}
}/**
* Your NumArray object will be instantiated and called as such:
* let obj = NumArray::new(nums);
* let ret_1: i32 = obj.sum_range(left, right);
*/