Description#
You are given a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing.
A 0-indexed string num of length n + 1 is created using the following conditions:
num consists of the digits '1' to '9', where each digit is used at most once.- If
pattern[i] == 'I', then num[i] < num[i + 1]. - If
pattern[i] == 'D', then num[i] > num[i + 1].
Return the lexicographically smallest possible string num that meets the conditions.
Example 1:
Input: pattern = "IIIDIDDD"
Output: "123549876"
Explanation:
At indices 0, 1, 2, and 4 we must have that num[i] < num[i+1].
At indices 3, 5, 6, and 7 we must have that num[i] > num[i+1].
Some possible values of num are "245639871", "135749862", and "123849765".
It can be proven that "123549876" is the smallest possible num that meets the conditions.
Note that "123414321" is not possible because the digit '1' is used more than once.
Example 2:
Input: pattern = "DDD"
Output: "4321"
Explanation:
Some possible values of num are "9876", "7321", and "8742".
It can be proven that "4321" is the smallest possible num that meets the conditions.
Constraints:
1 <= pattern.length <= 8pattern consists of only the letters 'I' and 'D'.
Solutions#
Solution 1#
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| class Solution:
def smallestNumber(self, pattern: str) -> str:
def dfs(u):
nonlocal ans
if ans:
return
if u == len(pattern) + 1:
ans = ''.join(t)
return
for i in range(1, 10):
if not vis[i]:
if u and pattern[u - 1] == 'I' and int(t[-1]) >= i:
continue
if u and pattern[u - 1] == 'D' and int(t[-1]) <= i:
continue
vis[i] = True
t.append(str(i))
dfs(u + 1)
vis[i] = False
t.pop()
vis = [False] * 10
t = []
ans = None
dfs(0)
return ans
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| class Solution {
private boolean[] vis = new boolean[10];
private StringBuilder t = new StringBuilder();
private String p;
private String ans;
public String smallestNumber(String pattern) {
p = pattern;
dfs(0);
return ans;
}
private void dfs(int u) {
if (ans != null) {
return;
}
if (u == p.length() + 1) {
ans = t.toString();
return;
}
for (int i = 1; i < 10; ++i) {
if (!vis[i]) {
if (u > 0 && p.charAt(u - 1) == 'I' && t.charAt(u - 1) - '0' >= i) {
continue;
}
if (u > 0 && p.charAt(u - 1) == 'D' && t.charAt(u - 1) - '0' <= i) {
continue;
}
vis[i] = true;
t.append(i);
dfs(u + 1);
t.deleteCharAt(t.length() - 1);
vis[i] = false;
}
}
}
}
|
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| class Solution {
public:
string ans = "";
string pattern;
vector<bool> vis;
string t = "";
string smallestNumber(string pattern) {
this->pattern = pattern;
vis.assign(10, false);
dfs(0);
return ans;
}
void dfs(int u) {
if (ans != "") return;
if (u == pattern.size() + 1) {
ans = t;
return;
}
for (int i = 1; i < 10; ++i) {
if (!vis[i]) {
if (u && pattern[u - 1] == 'I' && t.back() - '0' >= i) continue;
if (u && pattern[u - 1] == 'D' && t.back() - '0' <= i) continue;
vis[i] = true;
t += to_string(i);
dfs(u + 1);
t.pop_back();
vis[i] = false;
}
}
}
};
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| func smallestNumber(pattern string) string {
vis := make([]bool, 10)
t := []byte{}
ans := ""
var dfs func(u int)
dfs = func(u int) {
if ans != "" {
return
}
if u == len(pattern)+1 {
ans = string(t)
return
}
for i := 1; i < 10; i++ {
if !vis[i] {
if u > 0 && pattern[u-1] == 'I' && int(t[len(t)-1]-'0') >= i {
continue
}
if u > 0 && pattern[u-1] == 'D' && int(t[len(t)-1]-'0') <= i {
continue
}
vis[i] = true
t = append(t, byte('0'+i))
dfs(u + 1)
vis[i] = false
t = t[:len(t)-1]
}
}
}
dfs(0)
return ans
}
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| function smallestNumber(pattern: string): string {
const n = pattern.length;
const res = new Array(n + 1).fill('');
const vis = new Array(n + 1).fill(false);
const dfs = (i: number, num: number) => {
if (i === n) {
return;
}
if (vis[num]) {
vis[num] = false;
if (pattern[i] === 'I') {
dfs(i - 1, num - 1);
} else {
dfs(i - 1, num + 1);
}
return;
}
vis[num] = true;
res[i] = num;
if (pattern[i] === 'I') {
for (let j = res[i] + 1; j <= n + 1; j++) {
if (!vis[j]) {
dfs(i + 1, j);
return;
}
}
vis[num] = false;
dfs(i, num - 1);
} else {
for (let j = res[i] - 1; j > 0; j--) {
if (!vis[j]) {
dfs(i + 1, j);
return;
}
}
vis[num] = false;
dfs(i, num + 1);
}
};
dfs(0, 1);
for (let i = 1; i <= n + 1; i++) {
if (!vis[i]) {
res[n] = i;
break;
}
}
return res.join('');
}
|
