What is the brute force approach and complexity for Range Sum Query – Immutable?

Keep the array; each query adds nums[left..right] one by one with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Range Sum Query – Immutable?

Build prefix of size n+1 once; each query is prefix[r+1] − prefix[l] with optimal time complexity O(1) and space complexity O(N).

How do you identify the pattern for Range Sum Query – Immutable?

Look for key signals in the prompt: "array never changes" indicates Totals computed once stay correct forever → precompute.; "many queries" indicates Spend O(N) once so each query is O(1).; "sum from left to right" indicates A range sum = difference of two totals from the start..

Range Sum Query – Immutable

Easy·LeetCode #303 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given an integer array nums that never changes. After that, you will be asked many questions of the form "what is the sum of nums[left] through nums[right], both ends included?"

Write a class NumArray:

NumArray(vector<int>& nums) receives the array once.
int sumRange(int left, int right) returns nums[left] + nums[left + 1] + … + nums[right].

Input

The array nums (given once, to the constructor), followed by many calls sumRange(left, right).

Output

For each sumRange call, the sum of the elements from index left to index right, inclusive.

Example 1

Input: nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2), sumRange(2, 5), sumRange(0, 5)

Output: 1, -1, -3

(-2) + 0 + 3 = 1. 3 + (-5) + 2 + (-1) = -1. The whole array adds up to -3.

Constraints

• 1 ≤ nums.length ≤ 10^4
• −10^5 ≤ nums[i] ≤ 10^5
• 0 ≤ left ≤ right < nums.length
• At most 10^4 calls to sumRange
• Every total is at most 10^4 · 10^5 = 10^9 in size, so `int` is enough.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“array never changes”

→ Totals computed once stay correct forever → precompute.

“many queries”

→ Spend O(N) once so each query is O(1).

“sum from left to right”

→ A range sum = difference of two totals from the start.

Approach Complexity Ladder

Brute Force — loop over the range per queryO(N)O(N)

Keep the array; each query adds nums[left..right] one by one

Optimal — prefix array with a leading 0O(1)O(N)

Build prefix of size n+1 once; each query is prefix[r+1] − prefix[l]

Brute Force Intuition

Store the array as it is. For each query, loop from left to right and add up the elements.

Recognition rule for this sub-pattern:

Many range-sum questions on data that never changes. Build prefix once, answer each query in O(1).

Remember this · Prefix Sum

Trigger

Range sums, "subarray sum equals K" (especially with negatives), equal counts, divisible by K.

Think

If I knew the total up to every index, would the answer become a difference of two totals?

Key question

Which earlier prefix value am I looking for when I stand at index i?

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