Weekly Contest 111-------->943. Find the Shortest Superstring(can't understand)

Given an array A of strings, find any smallest string that contains each string in A as a substring.

We may assume that no string in A is substring of another string in A.

 

Example 1:

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

Example 2:

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

Note:

1. 1 <= A.length <= 12
2. 1 <= A[i].length <= 20

Approach #1:

class Solution {
    public String shortestSuperstring(String[] A) {
        int N = A.length;

        int [][] overlaps = new int[N][N];
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                int m = Math.min(A[i].length(), A[j].length());
                for (int k = m; k >= 0; --k) {
                    if (A[i].endsWith(A[j].substring(0, k))) {
                        overlaps[i][j] = k;
                        break;
                    }
                }
            }
        }

        int[][] dp = new int[1<<N][N];
        int[][] parent = new int[1<<N][N];
        for (int mask = 0; mask < (1<<N); ++mask) {
            Arrays.fill(parent[mask], -1);

            for (int bit = 0; bit < N; ++bit) if (((mask >> bit) & 1) > 0) {
                int pmask = mask ^ (1 << bit);
                if (pmask == 0) continue;
                for (int i = 0; i < N; ++i) if (((pmask >> i) & 1) > 0){
                    int val = dp[pmask][i] + overlaps[i][bit];
                    if (val > dp[mask][bit]) {
                        dp[mask][bit] = val;
                        parent[mask][bit] = i;
                    }
                }
            }
        }

        int[] perm = new int[N];
        boolean[] seen = new boolean[N];
        int t = 0;
        int mask = (1 << N) - 1;

        int p = 0;
        for (int j = 0; j < N; ++j)
            if (dp[(1 << N) - 1][j] > dp[(1 << N) - 1][p])
                p = j;

        while (p != -1) {
            perm[t++] = p;
            seen[p] = true;
            int p2 = parent[mask][p];
            mask ^= 1 << p;
            p = p2;
        }

        for (int i = 0; i < t/2; ++i) {
            int v = perm[i];
            perm[i] = perm[t-1-i];
            perm[t-1-i] = v;
        }

        for (int i = 0; i < N; ++i) if (!seen[i])
            perm[t++] = i;

        StringBuilder ans = new StringBuilder(A[perm[0]]);
        for (int i = 1; i < N; ++i) {
            int overlap = overlaps[perm[i-1]][perm[i]];
            ans.append(A[perm[i]].substring(overlap));
        }

        return ans.toString();
    }
}