# 2522. Partition String Into Substrings With Values at Most K

## Description

You are given a string s consisting of digits from 1 to 9 and an integer k.

A partition of a string s is called good if:

• Each digit of s is part of exactly one substring.
• The value of each substring is less than or equal to k.

Return the minimum number of substrings in a good partition of s. If no good partition of s exists, return -1.

Note that:

• The value of a string is its result when interpreted as an integer. For example, the value of "123" is 123 and the value of "1" is 1.
• A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "165462", k = 60
Output: 4
Explanation: We can partition the string into substrings "16", "54", "6", and "2". Each substring has a value less than or equal to k = 60.
It can be shown that we cannot partition the string into less than 4 substrings.


Example 2:

Input: s = "238182", k = 5
Output: -1
Explanation: There is no good partition for this string.


Constraints:

• 1 <= s.length <= 105
• s[i] is a digit from '1' to '9'.
• 1 <= k <= 109

## Solutions

• class Solution {
private Integer[] f;
private int n;
private String s;
private int k;
private int inf = 1 << 30;

public int minimumPartition(String s, int k) {
n = s.length();
f = new Integer[n];
this.s = s;
this.k = k;
int ans = dfs(0);
return ans < inf ? ans : -1;
}

private int dfs(int i) {
if (i >= n) {
return 0;
}
if (f[i] != null) {
return f[i];
}
int res = inf;
long v = 0;
for (int j = i; j < n; ++j) {
v = v * 10 + (s.charAt(j) - '0');
if (v > k) {
break;
}
res = Math.min(res, dfs(j + 1));
}
return f[i] = res + 1;
}
}

• class Solution {
public:
int minimumPartition(string s, int k) {
int n = s.size();
int f[n];
memset(f, 0, sizeof f);
const int inf = 1 << 30;
function<int(int)> dfs = [&](int i) -> int {
if (i >= n) return 0;
if (f[i]) return f[i];
int res = inf;
long v = 0;
for (int j = i; j < n; ++j) {
v = v * 10 + (s[j] - '0');
if (v > k) break;
res = min(res, dfs(j + 1));
}
return f[i] = res + 1;
};
int ans = dfs(0);
return ans < inf ? ans : -1;
}
};

• class Solution:
def minimumPartition(self, s: str, k: int) -> int:
@cache
def dfs(i):
if i >= n:
return 0
res, v = inf, 0
for j in range(i, n):
v = v * 10 + int(s[j])
if v > k:
break
res = min(res, dfs(j + 1))
return res + 1

n = len(s)
ans = dfs(0)
return ans if ans < inf else -1


• func minimumPartition(s string, k int) int {
n := len(s)
f := make([]int, n)
const inf int = 1 << 30
var dfs func(int) int
dfs = func(i int) int {
if i >= n {
return 0
}
if f[i] > 0 {
return f[i]
}
res, v := inf, 0
for j := i; j < n; j++ {
v = v*10 + int(s[j]-'0')
if v > k {
break
}
res = min(res, dfs(j+1))
}
f[i] = res + 1
return f[i]
}
ans := dfs(0)
if ans < inf {
return ans
}
return -1
}