LeetCode #2080 — MEDIUM

Range Frequency Queries

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

Solve on LeetCode

The Problem

Problem Statement

Design a data structure to find the frequency of a given value in a given subarray.

The frequency of a value in a subarray is the number of occurrences of that value in the subarray.

Implement the RangeFreqQuery class:

RangeFreqQuery(int[] arr) Constructs an instance of the class with the given 0-indexed integer array arr.
int query(int left, int right, int value) Returns the frequency of value in the subarray arr[left...right].

A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes the subarray that contains the elements of nums between indices left and right (inclusive).

Example 1:

Input
["RangeFreqQuery", "query", "query"]
[[[12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]], [1, 2, 4], [0, 11, 33]]
Output
[null, 1, 2]

Explanation
RangeFreqQuery rangeFreqQuery = new RangeFreqQuery([12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]);
rangeFreqQuery.query(1, 2, 4); // return 1. The value 4 occurs 1 time in the subarray [33, 4]
rangeFreqQuery.query(0, 11, 33); // return 2. The value 33 occurs 2 times in the whole array.

Constraints:

1 <= arr.length <= 10⁵
1 <= arr[i], value <= 10⁴
0 <= left <= right < arr.length
At most 10⁵ calls will be made to query

Step 01

Brute Force Baseline

Problem summary: Design a data structure to find the frequency of a given value in a given subarray. The frequency of a value in a subarray is the number of occurrences of that value in the subarray. Implement the RangeFreqQuery class: RangeFreqQuery(int[] arr) Constructs an instance of the class with the given 0-indexed integer array arr. int query(int left, int right, int value) Returns the frequency of value in the subarray arr[left...right]. A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes the subarray that contains the elements of nums between indices left and right (inclusive).

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Design · Segment Tree

Example 1

["RangeFreqQuery","query","query"]
[[[12,33,4,56,22,2,34,33,22,12,34,56]],[1,2,4],[0,11,33]]

Step 02

Core Insight

What unlocks the optimal approach

The queries must be answered efficiently to avoid time limit exceeded verdict.
Store the elements of the array in a data structure that helps answering the queries efficiently.
Use a hash table that stored for each value, the indices where that value appeared.
Use binary search over the indices of a value to find its range frequency.

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 #2080: Range Frequency Queries
class RangeFreqQuery {
    private Map<Integer, List<Integer>> g = new HashMap<>();

    public RangeFreqQuery(int[] arr) {
        for (int i = 0; i < arr.length; ++i) {
            g.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
        }
    }

    public int query(int left, int right, int value) {
        if (!g.containsKey(value)) {
            return 0;
        }
        var idx = g.get(value);
        int l = Collections.binarySearch(idx, left);
        l = l < 0 ? -l - 1 : l;
        int r = Collections.binarySearch(idx, right + 1);
        r = r < 0 ? -r - 1 : r;
        return r - l;
    }
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * RangeFreqQuery obj = new RangeFreqQuery(arr);
 * int param_1 = obj.query(left,right,value);
 */

// Accepted solution for LeetCode #2080: Range Frequency Queries
type RangeFreqQuery struct {
	g map[int][]int
}

func Constructor(arr []int) RangeFreqQuery {
	g := make(map[int][]int)
	for i, v := range arr {
		g[v] = append(g[v], i)
	}
	return RangeFreqQuery{g}
}

func (this *RangeFreqQuery) Query(left int, right int, value int) int {
	if idx, ok := this.g[value]; ok {
		l := sort.SearchInts(idx, left)
		r := sort.SearchInts(idx, right+1)
		return r - l
	}
	return 0
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * obj := Constructor(arr);
 * param_1 := obj.Query(left,right,value);
 */

# Accepted solution for LeetCode #2080: Range Frequency Queries
class RangeFreqQuery:

    def __init__(self, arr: List[int]):
        self.g = defaultdict(list)
        for i, x in enumerate(arr):
            self.g[x].append(i)

    def query(self, left: int, right: int, value: int) -> int:
        idx = self.g[value]
        l = bisect_left(idx, left)
        r = bisect_left(idx, right + 1)
        return r - l


# Your RangeFreqQuery object will be instantiated and called as such:
# obj = RangeFreqQuery(arr)
# param_1 = obj.query(left,right,value)

// Accepted solution for LeetCode #2080: Range Frequency Queries
use std::collections::HashMap;

struct RangeFreqQuery {
    g: HashMap<i32, Vec<usize>>,
}

impl RangeFreqQuery {
    fn new(arr: Vec<i32>) -> Self {
        let mut g = HashMap::new();
        for (i, &value) in arr.iter().enumerate() {
            g.entry(value).or_insert_with(Vec::new).push(i);
        }
        RangeFreqQuery { g }
    }

    fn query(&self, left: i32, right: i32, value: i32) -> i32 {
        if let Some(idx) = self.g.get(&value) {
            let l = idx.partition_point(|&x| x < left as usize);
            let r = idx.partition_point(|&x| x <= right as usize);
            return (r - l) as i32;
        }
        0
    }
}

// Accepted solution for LeetCode #2080: Range Frequency Queries
class RangeFreqQuery {
    private g: Map<number, number[]> = new Map();

    constructor(arr: number[]) {
        for (let i = 0; i < arr.length; ++i) {
            if (!this.g.has(arr[i])) {
                this.g.set(arr[i], []);
            }
            this.g.get(arr[i])!.push(i);
        }
    }

    query(left: number, right: number, value: number): number {
        const idx = this.g.get(value);
        if (!idx) {
            return 0;
        }
        const l = _.sortedIndex(idx, left);
        const r = _.sortedIndex(idx, right + 1);
        return r - l;
    }
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * var obj = new RangeFreqQuery(arr)
 * var param_1 = obj.query(left,right,value)
 */

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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.