LeetCode #307 — MEDIUM

Range Sum Query - Mutable

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Update the value of an element in nums.
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.
void update(int index, int val) Updates the value of nums[index] to be val.
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", "update", "sumRange"]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
Output
[null, 9, null, 8]

Explanation
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // return 1 + 3 + 5 = 9
numArray.update(1, 2);   // nums = [1, 2, 5]
numArray.sumRange(0, 2); // return 1 + 2 + 5 = 8

Constraints:

1 <= nums.length <= 3 * 10⁴
-100 <= nums[i] <= 100
0 <= index < nums.length
-100 <= val <= 100
0 <= left <= right < nums.length
At most 3 * 10⁴ calls will be made to update and sumRange.

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, handle multiple queries of the following types: Update the value of an element in nums. 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. void update(int index, int val) Updates the value of nums[index] to be val. 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 · Segment Tree

Example 1

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

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 #307: Range Sum Query - Mutable
class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += x & -x;
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class NumArray {
    private BinaryIndexedTree tree;

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

    public void update(int index, int val) {
        int prev = sumRange(index, index);
        tree.update(index + 1, val - prev);
    }

    public int sumRange(int left, int right) {
        return tree.query(right + 1) - tree.query(left);
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * obj.update(index,val);
 * int param_2 = obj.sumRange(left,right);
 */

// Accepted solution for LeetCode #307: Range Sum Query - Mutable
type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (t *BinaryIndexedTree) update(x, delta int) {
	for ; x <= t.n; x += x & -x {
		t.c[x] += delta
	}
}

func (t *BinaryIndexedTree) query(x int) (s int) {
	for ; x > 0; x -= x & -x {
		s += t.c[x]
	}
	return s
}

type NumArray struct {
	tree *BinaryIndexedTree
}

func Constructor(nums []int) NumArray {
	tree := newBinaryIndexedTree(len(nums))
	for i, v := range nums {
		tree.update(i+1, v)
	}
	return NumArray{tree}
}

func (t *NumArray) Update(index int, val int) {
	prev := t.SumRange(index, index)
	t.tree.update(index+1, val-prev)
}

func (t *NumArray) SumRange(left int, right int) int {
	return t.tree.query(right+1) - t.tree.query(left)
}

/**
 * Your NumArray object will be instantiated and called as such:
 * obj := Constructor(nums);
 * obj.Update(index,val);
 * param_2 := obj.SumRange(left,right);
 */

# Accepted solution for LeetCode #307: Range Sum Query - Mutable
class BinaryIndexedTree:
    __slots__ = ["n", "c"]

    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, delta: int):
        while x <= self.n:
            self.c[x] += delta
            x += x & -x

    def query(self, x: int) -> int:
        s = 0
        while x > 0:
            s += self.c[x]
            x -= x & -x
        return s


class NumArray:
    __slots__ = ["tree"]

    def __init__(self, nums: List[int]):
        self.tree = BinaryIndexedTree(len(nums))
        for i, v in enumerate(nums, 1):
            self.tree.update(i, v)

    def update(self, index: int, val: int) -> None:
        prev = self.sumRange(index, index)
        self.tree.update(index + 1, val - prev)

    def sumRange(self, left: int, right: int) -> int:
        return self.tree.query(right + 1) - self.tree.query(left)


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# obj.update(index,val)
# param_2 = obj.sumRange(left,right)

// Accepted solution for LeetCode #307: Range Sum Query - Mutable
struct BitTree {
    tree: Vec<i32>,
    data: Vec<i32>,
    n: usize,
}

impl BitTree {
    fn new(n: usize) -> Self {
        let tree = vec![0; n];
        let data = vec![0; n];
        let n = n;
        BitTree { tree, data, n }
    }

    fn get(&self, i: usize) -> i32 {
        self.data[i]
    }

    fn sum(&self, i: usize) -> i32 {
        let mut res = 0;
        let down_iter = std::iter::successors(Some(i), |&i| {
            let j = i & (i + 1);
            if j > 0 {
                Some(j - 1)
            } else {
                None
            }
        });
        for j in down_iter {
            res += self.tree[j];
        }
        res
    }

    fn add(&mut self, i: usize, v: i32) {
        self.data[i] += v;
        let n = self.n;
        let up_iter = std::iter::successors(Some(i), |&i| {
            let j = i | (i + 1);
            if j < n {
                Some(j)
            } else {
                None
            }
        });
        for j in up_iter {
            self.tree[j] += v;
        }
    }
}

struct NumArray {
    bit_tree: BitTree,
}

impl NumArray {
    fn new(nums: Vec<i32>) -> Self {
        let n = nums.len();
        let mut bit_tree = BitTree::new(n);
        for i in 0..n {
            bit_tree.add(i, nums[i]);
        }
        NumArray { bit_tree }
    }

    fn update(&mut self, i: i32, val: i32) {
        let i = i as usize;
        self.bit_tree.add(i as usize, val - self.bit_tree.get(i))
    }

    fn sum_range(&self, i: i32, j: i32) -> i32 {
        let i = i as usize;
        let j = j as usize;
        if i > 0 {
            self.bit_tree.sum(j) - self.bit_tree.sum(i - 1)
        } else {
            self.bit_tree.sum(j)
        }
    }
}

#[test]
fn test() {
    let nums = vec![1, 3, 5];
    let mut obj = NumArray::new(nums);
    assert_eq!(obj.sum_range(0, 2), 9);
    obj.update(1, 2);
    assert_eq!(obj.sum_range(0, 2), 8);
}

// Accepted solution for LeetCode #307: Range Sum Query - Mutable
class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, delta: number): void {
        while (x <= this.n) {
            this.c[x] += delta;
            x += x & -x;
        }
    }

    query(x: number): number {
        let s = 0;
        while (x > 0) {
            s += this.c[x];
            x -= x & -x;
        }
        return s;
    }
}

class NumArray {
    private tree: BinaryIndexedTree;

    constructor(nums: number[]) {
        const n = nums.length;
        this.tree = new BinaryIndexedTree(n);
        for (let i = 0; i < n; ++i) {
            this.tree.update(i + 1, nums[i]);
        }
    }

    update(index: number, val: number): void {
        const prev = this.sumRange(index, index);
        this.tree.update(index + 1, val - prev);
    }

    sumRange(left: number, right: number): number {
        return this.tree.query(right + 1) - this.tree.query(left);
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * var obj = new NumArray(nums)
 * obj.update(index,val)
 * var param_2 = 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.