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.
Move from brute-force thinking to an efficient approach using array strategy.
A spreadsheet is a grid with 26 columns (labeled from 'A' to 'Z') and a given number of rows. Each cell in the spreadsheet can hold an integer value between 0 and 105.
Implement the Spreadsheet class:
Spreadsheet(int rows) Initializes a spreadsheet with 26 columns (labeled 'A' to 'Z') and the specified number of rows. All cells are initially set to 0.void setCell(String cell, int value) Sets the value of the specified cell. The cell reference is provided in the format "AX" (e.g., "A1", "B10"), where the letter represents the column (from 'A' to 'Z') and the number represents a 1-indexed row.void resetCell(String cell) Resets the specified cell to 0.int getValue(String formula) Evaluates a formula of the form "=X+Y", where X and Y are either cell references or non-negative integers, and returns the computed sum.Note: If getValue references a cell that has not been explicitly set using setCell, its value is considered 0.
Example 1:
Input:
["Spreadsheet", "getValue", "setCell", "getValue", "setCell", "getValue", "resetCell", "getValue"]
[[3], ["=5+7"], ["A1", 10], ["=A1+6"], ["B2", 15], ["=A1+B2"], ["A1"], ["=A1+B2"]]
Output:
[null, 12, null, 16, null, 25, null, 15]
Explanation
Spreadsheet spreadsheet = new Spreadsheet(3); // Initializes a spreadsheet with 3 rows and 26 columnsConstraints:
1 <= rows <= 1030 <= value <= 105"=X+Y", where X and Y are either valid cell references or non-negative integers with values less than or equal to 105.'A' to 'Z' followed by a row number between 1 and rows.104 calls will be made in total to setCell, resetCell, and getValue.Problem summary: A spreadsheet is a grid with 26 columns (labeled from 'A' to 'Z') and a given number of rows. Each cell in the spreadsheet can hold an integer value between 0 and 105. Implement the Spreadsheet class: Spreadsheet(int rows) Initializes a spreadsheet with 26 columns (labeled 'A' to 'Z') and the specified number of rows. All cells are initially set to 0. void setCell(String cell, int value) Sets the value of the specified cell. The cell reference is provided in the format "AX" (e.g., "A1", "B10"), where the letter represents the column (from 'A' to 'Z') and the number represents a 1-indexed row. void resetCell(String cell) Resets the specified cell to 0. int getValue(String formula) Evaluates a formula of the form "=X+Y", where X and Y are either cell references or non-negative integers, and returns the computed sum. Note: If getValue references a cell that has not been explicitly set
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Design
["Spreadsheet","getValue","setCell","getValue","setCell","getValue","resetCell","getValue"] [[3],["=5+7"],["A1",10],["=A1+6"],["B2",15],["=A1+B2"],["A1"],["=A1+B2"]]
excel-sheet-column-title)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3484: Design Spreadsheet
class Spreadsheet {
private Map<String, Integer> d = new HashMap<>();
public Spreadsheet(int rows) {
}
public void setCell(String cell, int value) {
d.put(cell, value);
}
public void resetCell(String cell) {
d.remove(cell);
}
public int getValue(String formula) {
int ans = 0;
for (String cell : formula.substring(1).split("\\+")) {
ans += Character.isDigit(cell.charAt(0)) ? Integer.parseInt(cell)
: d.getOrDefault(cell, 0);
}
return ans;
}
}
/**
* Your Spreadsheet object will be instantiated and called as such:
* Spreadsheet obj = new Spreadsheet(rows);
* obj.setCell(cell,value);
* obj.resetCell(cell);
* int param_3 = obj.getValue(formula);
*/
// Accepted solution for LeetCode #3484: Design Spreadsheet
type Spreadsheet struct {
d map[string]int
}
func Constructor(rows int) Spreadsheet {
return Spreadsheet{d: make(map[string]int)}
}
func (this *Spreadsheet) SetCell(cell string, value int) {
this.d[cell] = value
}
func (this *Spreadsheet) ResetCell(cell string) {
delete(this.d, cell)
}
func (this *Spreadsheet) GetValue(formula string) int {
ans := 0
cells := strings.Split(formula[1:], "+")
for _, cell := range cells {
if val, err := strconv.Atoi(cell); err == nil {
ans += val
} else {
ans += this.d[cell]
}
}
return ans
}
/**
* Your Spreadsheet object will be instantiated and called as such:
* obj := Constructor(rows);
* obj.SetCell(cell,value);
* obj.ResetCell(cell);
* param_3 := obj.GetValue(formula);
*/
# Accepted solution for LeetCode #3484: Design Spreadsheet
class Spreadsheet:
def __init__(self, rows: int):
self.d = {}
def setCell(self, cell: str, value: int) -> None:
self.d[cell] = value
def resetCell(self, cell: str) -> None:
self.d.pop(cell, None)
def getValue(self, formula: str) -> int:
ans = 0
for cell in formula[1:].split("+"):
ans += int(cell) if cell[0].isdigit() else self.d.get(cell, 0)
return ans
# Your Spreadsheet object will be instantiated and called as such:
# obj = Spreadsheet(rows)
# obj.setCell(cell,value)
# obj.resetCell(cell)
# param_3 = obj.getValue(formula)
// Accepted solution for LeetCode #3484: Design Spreadsheet
use std::collections::HashMap;
struct Spreadsheet {
d: HashMap<String, i32>,
}
impl Spreadsheet {
fn new(_rows: i32) -> Self {
Spreadsheet { d: HashMap::new() }
}
fn set_cell(&mut self, cell: String, value: i32) {
self.d.insert(cell, value);
}
fn reset_cell(&mut self, cell: String) {
self.d.remove(&cell);
}
fn get_value(&self, formula: String) -> i32 {
let mut ans = 0;
for cell in formula[1..].split('+') {
if cell.chars().next().unwrap().is_ascii_digit() {
ans += cell.parse::<i32>().unwrap();
} else {
ans += *self.d.get(cell).unwrap_or(&0);
}
}
ans
}
}
// Accepted solution for LeetCode #3484: Design Spreadsheet
class Spreadsheet {
private d: Map<string, number>;
constructor(rows: number) {
this.d = new Map<string, number>();
}
setCell(cell: string, value: number): void {
this.d.set(cell, value);
}
resetCell(cell: string): void {
this.d.delete(cell);
}
getValue(formula: string): number {
let ans = 0;
const cells = formula.slice(1).split('+');
for (const cell of cells) {
ans += isNaN(Number(cell)) ? this.d.get(cell) || 0 : Number(cell);
}
return ans;
}
}
/**
* Your Spreadsheet object will be instantiated and called as such:
* var obj = new Spreadsheet(rows)
* obj.setCell(cell,value)
* obj.resetCell(cell)
* var param_3 = obj.getValue(formula)
*/
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.
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.