LeetCode #303 — EASY

Range Sum Query - Immutable

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

Given an integer array nums, handle multiple queries of the following type:

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.

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, handle multiple queries of the following type: 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]).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Design

Example 1

["NumArray","sumRange","sumRange","sumRange"]
[[[-2,0,3,-5,2,-1]],[0,2],[2,5],[0,5]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #303: Range Sum Query - Immutable
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);
 */

// Accepted solution for LeetCode #303: Range Sum Query - Immutable
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);
 */

# Accepted solution for LeetCode #303: Range Sum Query - Immutable
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)

// Accepted solution for LeetCode #303: Range Sum Query - Immutable
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]
    }
}

// Accepted solution for LeetCode #303: Range Sum Query - Immutable
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)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.