Description#
Given a positive integer n, return the punishment number of n.
The punishment number of n is defined as the sum of the squares of all integers i such that:
1 <= i <= n- The decimal representation of
i * i can be partitioned into contiguous substrings such that the sum of the integer values of these substrings equals i.
Example 1:
Input: n = 10
Output: 182
Explanation: There are exactly 3 integers i that satisfy the conditions in the statement:
- 1 since 1 * 1 = 1
- 9 since 9 * 9 = 81 and 81 can be partitioned into 8 + 1.
- 10 since 10 * 10 = 100 and 100 can be partitioned into 10 + 0.
Hence, the punishment number of 10 is 1 + 81 + 100 = 182
Example 2:
Input: n = 37
Output: 1478
Explanation: There are exactly 4 integers i that satisfy the conditions in the statement:
- 1 since 1 * 1 = 1.
- 9 since 9 * 9 = 81 and 81 can be partitioned into 8 + 1.
- 10 since 10 * 10 = 100 and 100 can be partitioned into 10 + 0.
- 36 since 36 * 36 = 1296 and 1296 can be partitioned into 1 + 29 + 6.
Hence, the punishment number of 37 is 1 + 81 + 100 + 1296 = 1478
Constraints:
Solutions#
Solution 1: Enumeration + DFS#
We enumerate $i$, where $1 \leq i \leq n$. For each $i$, we split the decimal representation string of $x = i^2$, and then check whether it meets the requirements of the problem. If it does, we add $x$ to the answer.
After the enumeration ends, we return the answer.
The time complexity is $O(n^{1 + 2 \log_{10}^2})$, and the space complexity is $O(\log n)$, where $n$ is the given positive integer.
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| class Solution:
def punishmentNumber(self, n: int) -> int:
def check(s: str, i: int, x: int) -> bool:
m = len(s)
if i >= m:
return x == 0
y = 0
for j in range(i, m):
y = y * 10 + int(s[j])
if y > x:
break
if check(s, j + 1, x - y):
return True
return False
ans = 0
for i in range(1, n + 1):
x = i * i
if check(str(x), 0, i):
ans += x
return ans
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| class Solution {
public int punishmentNumber(int n) {
int ans = 0;
for (int i = 1; i <= n; ++i) {
int x = i * i;
if (check(x + "", 0, i)) {
ans += x;
}
}
return ans;
}
private boolean check(String s, int i, int x) {
int m = s.length();
if (i >= m) {
return x == 0;
}
int y = 0;
for (int j = i; j < m; ++j) {
y = y * 10 + (s.charAt(j) - '0');
if (y > x) {
break;
}
if (check(s, j + 1, x - y)) {
return true;
}
}
return false;
}
}
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| class Solution {
public:
int punishmentNumber(int n) {
int ans = 0;
for (int i = 1; i <= n; ++i) {
int x = i * i;
string s = to_string(x);
if (check(s, 0, i)) {
ans += x;
}
}
return ans;
}
bool check(const string& s, int i, int x) {
int m = s.size();
if (i >= m) {
return x == 0;
}
int y = 0;
for (int j = i; j < m; ++j) {
y = y * 10 + s[j] - '0';
if (y > x) {
break;
}
if (check(s, j + 1, x - y)) {
return true;
}
}
return false;
}
};
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| func punishmentNumber(n int) (ans int) {
var check func(string, int, int) bool
check = func(s string, i, x int) bool {
m := len(s)
if i >= m {
return x == 0
}
y := 0
for j := i; j < m; j++ {
y = y*10 + int(s[j]-'0')
if y > x {
break
}
if check(s, j+1, x-y) {
return true
}
}
return false
}
for i := 1; i <= n; i++ {
x := i * i
s := strconv.Itoa(x)
if check(s, 0, i) {
ans += x
}
}
return
}
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| function punishmentNumber(n: number): number {
const check = (s: string, i: number, x: number): boolean => {
const m = s.length;
if (i >= m) {
return x === 0;
}
let y = 0;
for (let j = i; j < m; ++j) {
y = y * 10 + Number(s[j]);
if (y > x) {
break;
}
if (check(s, j + 1, x - y)) {
return true;
}
}
return false;
};
let ans = 0;
for (let i = 1; i <= n; ++i) {
const x = i * i;
const s = x.toString();
if (check(s, 0, i)) {
ans += x;
}
}
return ans;
}
|
