# 943. Find the Shortest Superstring

## Description

Given an array of strings words, return the smallest string that contains each string in words as a substring. If there are multiple valid strings of the smallest length, return any of them.

You may assume that no string in words is a substring of another string in words.

Example 1:

Input: words = ["alex","loves","leetcode"]
Output: "alexlovesleetcode"
Explanation: All permutations of "alex","loves","leetcode" would also be accepted.


Example 2:

Input: words = ["catg","ctaagt","gcta","ttca","atgcatc"]
Output: "gctaagttcatgcatc"


Constraints:

• 1 <= words.length <= 12
• 1 <= words[i].length <= 20
• words[i] consists of lowercase English letters.
• All the strings of words are unique.

## Solutions

• class Solution {
public String shortestSuperstring(String[] words) {
int n = words.length;
int[][] g = new int[n][n];
for (int i = 0; i < n; ++i) {
String a = words[i];
for (int j = 0; j < n; ++j) {
String b = words[j];
if (i != j) {
for (int k = Math.min(a.length(), b.length()); k > 0; --k) {
if (a.substring(a.length() - k).equals(b.substring(0, k))) {
g[i][j] = k;
break;
}
}
}
}
}
int[][] dp = new int[1 << n][n];
int[][] p = new int[1 << n][n];
for (int i = 0; i < 1 << n; ++i) {
Arrays.fill(p[i], -1);
for (int j = 0; j < n; ++j) {
if (((i >> j) & 1) == 1) {
int pi = i ^ (1 << j);
for (int k = 0; k < n; ++k) {
if (((pi >> k) & 1) == 1) {
int v = dp[pi][k] + g[k][j];
if (v > dp[i][j]) {
dp[i][j] = v;
p[i][j] = k;
}
}
}
}
}
}
int j = 0;
for (int i = 0; i < n; ++i) {
if (dp[(1 << n) - 1][i] > dp[(1 << n) - 1][j]) {
j = i;
}
}
List<Integer> arr = new ArrayList<>();
for (int i = (1 << n) - 1; p[i][j] != -1;) {
int k = i;
i ^= (1 << j);
j = p[k][j];
}
Set<Integer> vis = new HashSet<>(arr);
for (int i = 0; i < n; ++i) {
if (!vis.contains(i)) {
}
}
Collections.reverse(arr);
StringBuilder ans = new StringBuilder(words[arr.get(0)]);
for (int i = 1; i < n; ++i) {
int k = g[arr.get(i - 1)][arr.get(i)];
ans.append(words[arr.get(i)].substring(k));
}
return ans.toString();
}
}

• class Solution {
public:
string shortestSuperstring(vector<string>& words) {
int n = words.size();
vector<vector<int>> g(n, vector<int>(n));
for (int i = 0; i < n; ++i) {
auto a = words[i];
for (int j = 0; j < n; ++j) {
auto b = words[j];
if (i != j) {
for (int k = min(a.size(), b.size()); k > 0; --k) {
if (a.substr(a.size() - k) == b.substr(0, k)) {
g[i][j] = k;
break;
}
}
}
}
}
vector<vector<int>> dp(1 << n, vector<int>(n));
vector<vector<int>> p(1 << n, vector<int>(n, -1));
for (int i = 0; i < 1 << n; ++i) {
for (int j = 0; j < n; ++j) {
if ((i >> j) & 1) {
int pi = i ^ (1 << j);
for (int k = 0; k < n; ++k) {
if ((pi >> k) & 1) {
int v = dp[pi][k] + g[k][j];
if (v > dp[i][j]) {
dp[i][j] = v;
p[i][j] = k;
}
}
}
}
}
}
int j = 0;
for (int i = 0; i < n; ++i) {
if (dp[(1 << n) - 1][i] > dp[(1 << n) - 1][j]) {
j = i;
}
}
vector<int> arr = {j};
for (int i = (1 << n) - 1; p[i][j] != -1;) {
int k = i;
i ^= (1 << j);
j = p[k][j];
arr.push_back(j);
}
unordered_set<int> vis(arr.begin(), arr.end());
for (int i = 0; i < n; ++i) {
if (!vis.count(i)) {
arr.push_back(i);
}
}
reverse(arr.begin(), arr.end());
string ans = words[arr[0]];
for (int i = 1; i < n; ++i) {
int k = g[arr[i - 1]][arr[i]];
ans += words[arr[i]].substr(k);
}
return ans;
}
};

• class Solution:
def shortestSuperstring(self, words: List[str]) -> str:
n = len(words)
g = [[0] * n for _ in range(n)]
for i, a in enumerate(words):
for j, b in enumerate(words):
if i != j:
for k in range(min(len(a), len(b)), 0, -1):
if a[-k:] == b[:k]:
g[i][j] = k
break
dp = [[0] * n for _ in range(1 << n)]
p = [[-1] * n for _ in range(1 << n)]
for i in range(1 << n):
for j in range(n):
if (i >> j) & 1:
pi = i ^ (1 << j)
for k in range(n):
if (pi >> k) & 1:
v = dp[pi][k] + g[k][j]
if v > dp[i][j]:
dp[i][j] = v
p[i][j] = k
j = 0
for i in range(n):
if dp[-1][i] > dp[-1][j]:
j = i
arr = [j]
i = (1 << n) - 1
while p[i][j] != -1:
i, j = i ^ (1 << j), p[i][j]
arr.append(j)
arr = arr[::-1]
vis = set(arr)
arr.extend([j for j in range(n) if j not in vis])
ans = [words[arr[0]]] + [words[j][g[i][j] :] for i, j in pairwise(arr)]
return ''.join(ans)


• func shortestSuperstring(words []string) string {
n := len(words)
g := make([][]int, n)
for i, a := range words {
g[i] = make([]int, n)
for j, b := range words {
if i != j {
for k := min(len(a), len(b)); k > 0; k-- {
if a[len(a)-k:] == b[:k] {
g[i][j] = k
break
}
}
}
}
}
dp := make([][]int, 1<<n)
p := make([][]int, 1<<n)
for i := 0; i < 1<<n; i++ {
dp[i] = make([]int, n)
p[i] = make([]int, n)
for j := 0; j < n; j++ {
p[i][j] = -1
if ((i >> j) & 1) == 1 {
pi := i ^ (1 << j)
for k := 0; k < n; k++ {
if ((pi >> k) & 1) == 1 {
v := dp[pi][k] + g[k][j]
if v > dp[i][j] {
dp[i][j] = v
p[i][j] = k
}
}
}
}
}
}
j := 0
for i := 0; i < n; i++ {
if dp[(1<<n)-1][i] > dp[(1<<n)-1][j] {
j = i
}
}
arr := []int{j}
vis := make([]bool, n)
vis[j] = true
for i := (1 << n) - 1; p[i][j] != -1; {
k := i
i ^= (1 << j)
j = p[k][j]
arr = append(arr, j)
vis[j] = true
}
for i := 0; i < n; i++ {
if !vis[i] {
arr = append(arr, i)
}
}
ans := &strings.Builder{}
ans.WriteString(words[arr[n-1]])
for i := n - 2; i >= 0; i-- {
k := g[arr[i+1]][arr[i]]
ans.WriteString(words[arr[i]][k:])
}
return ans.String()
}