Description
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.
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:
Solutions
Solution 1
PythonJavaC++Go
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)(code-box)
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;
}
}(code-box)
class Solution {
public:
vector<int> f;
int minSteps(int n) {
f.assign(n + 1, -1);
return dfs(n);
}
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 = min(ans, dfs(n / i) + i);
}
}
f[n] = ans;
return ans;
}
};(code-box)
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)
}(code-box)
Solution 2
PythonJavaC++GoTypeScriptJavaScript
class Solution:
def minSteps(self, n: int) -> int:
dp = list(range(n + 1))
dp[1] = 0
for i in range(2, n + 1):
j = 2
while j * j <= i:
if i % j == 0:
dp[i] = min(dp[i], dp[i // j] + j)
j += 1
return dp[-1](code-box)
class Solution {
public int minSteps(int n) {
int[] dp = new int[n + 1];
for (int i = 0; i < n + 1; ++i) {
dp[i] = i;
}
dp[1] = 0;
for (int i = 2; i < n + 1; ++i) {
for (int j = 2; j * j <= i; ++j) {
if (i % j == 0) {
dp[i] = Math.min(dp[i], dp[i / j] + j);
}
}
}
return dp[n];
}
}(code-box)
class Solution {
public:
int minSteps(int n) {
vector<int> dp(n + 1);
iota(dp.begin(), dp.end(), 0);
dp[1] = 0;
for (int i = 2; i < n + 1; ++i) {
for (int j = 2; j * j <= i; ++j) {
if (i % j == 0) {
dp[i] = min(dp[i], dp[i / j] + j);
}
}
}
return dp[n];
}
};(code-box)
func minSteps(n int) int {
dp := make([]int, n+1)
for i := range dp {
dp[i] = i
}
dp[1] = 0
for i := 2; i < n+1; i++ {
for j := 2; j*j <= i; j++ {
if i%j == 0 {
dp[i] = min(dp[i], dp[i/j]+j)
}
}
}
return dp[n]
}(code-box)
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];
}(code-box)
/**
* @param {number} n
* @return {number}
*/
var minSteps = function (n) {
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];
};(code-box)
Solution 3
Java
class Solution {
public int minSteps(int n) {
int res = 0;
for (int i = 2; n > 1; ++i) {
while (n % i == 0) {
res += i;
n /= i;
}
}
return res;
}
}(code-box)
