Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Move from brute-force thinking to an efficient approach using math strategy.
There is only one character 'A' on the screen of a notepad. You can perform one of two operations on this notepad for each step:
Given an integer n, return the minimum number of operations to get the character 'A' exactly n times on the screen.
Example 1:
Input: n = 3 Output: 3 Explanation: Initially, we have one character 'A'. In step 1, we use Copy All operation. In step 2, we use Paste operation to get 'AA'. In step 3, we use Paste operation to get 'AAA'.
Example 2:
Input: n = 1 Output: 0
Constraints:
1 <= n <= 1000Problem summary: There is only one character 'A' on the screen of a notepad. You can perform one of two operations on this notepad for each step: Copy All: You can copy all the characters present on the screen (a partial copy is not allowed). Paste: You can paste the characters which are copied last time. Given an integer n, return the minimum number of operations to get the character 'A' exactly n times on the screen.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming
3
1
4-keys-keyboard)broken-calculator)smallest-value-after-replacing-with-sum-of-prime-factors)distinct-prime-factors-of-product-of-array)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #650: 2 Keys Keyboard
class Solution {
private int[] f;
public int minSteps(int n) {
f = new int[n + 1];
Arrays.fill(f, -1);
return dfs(n);
}
private int dfs(int n) {
if (n == 1) {
return 0;
}
if (f[n] != -1) {
return f[n];
}
int ans = n;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
ans = Math.min(ans, dfs(n / i) + i);
}
}
f[n] = ans;
return ans;
}
}
// Accepted solution for LeetCode #650: 2 Keys Keyboard
func minSteps(n int) int {
f := make([]int, n+1)
for i := range f {
f[i] = -1
}
var dfs func(int) int
dfs = func(n int) int {
if n == 1 {
return 0
}
if f[n] != -1 {
return f[n]
}
ans := n
for i := 2; i*i <= n; i++ {
if n%i == 0 {
ans = min(ans, dfs(n/i)+i)
}
}
return ans
}
return dfs(n)
}
# Accepted solution for LeetCode #650: 2 Keys Keyboard
class Solution:
def minSteps(self, n: int) -> int:
@cache
def dfs(n):
if n == 1:
return 0
i, ans = 2, n
while i * i <= n:
if n % i == 0:
ans = min(ans, dfs(n // i) + i)
i += 1
return ans
return dfs(n)
// Accepted solution for LeetCode #650: 2 Keys Keyboard
struct Solution;
impl Solution {
fn min_steps(n: i32) -> i32 {
let n = n as usize;
let mut dp = vec![0; n + 1];
for i in 2..=n {
dp[i] = i;
for j in (2..i).rev() {
if i % j == 0 {
dp[i] = dp[j] + i / j;
break;
}
}
}
dp[n] as i32
}
}
#[test]
fn test() {
let n = 3;
let res = 3;
assert_eq!(Solution::min_steps(n), res);
}
// Accepted solution for LeetCode #650: 2 Keys Keyboard
function minSteps(n: number): number {
const dp = Array(n + 1).fill(1000);
dp[1] = 0;
for (let i = 2; i <= n; i++) {
for (let j = 1, half = i / 2; j <= half; j++) {
if (i % j === 0) {
dp[i] = Math.min(dp[i], dp[j] + i / j);
}
}
}
return dp[n];
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.