# Construct an Array of Strings having Longest Common Prefix specified by the given Array

Given an integer array arr[] of size N, the task is to construct an array consisting of N+1 strings of length N such that arr[i] is equal to the Longest Common Prefix of ith String and (i+1)th String.

Examples:

Input: arr[] = {1, 2, 3}
Output: {“abb”, “aab”, “aaa”, “aaa”}
Explanation:
Strings “abb” and “aab” have a single character “a” as Longest Common Prefix.
Strings “aab” and “aaa” have “aa” as Longest Common Prefix.
Strings “aaa” and “aaa” have “aa” as Longest Common Prefix.

Input : arr[]={2, 0, 3}
Output: {“bab”, “baa”, “aaa”, “aaa”}
Explanation:
Strings “bab” and “baa” have “ba” as Longest Common Prefix.
Strings “baa” and “aaa” have no common prefix.
Strings “aaa” and “aaa” have “aaa” as Longest Common Prefix.

Approach:

Follow the steps below to solve the problem:

• The idea is to observe that if ith string is known then (i-1)th string can be formed from ith string by changing N – arr[i-1] characters from ith string.
• Start constructing strings from right to left and generate the N + 1 strings.

Illustration:
N = 3, arr[] = {2, 0, 3}
Let the (N + 1)th string is “aaa”
Therefore, the remaining strings from right to left are {“aaa”, “baa”, “bab”}

Below is the implementation of the above approach:

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the array of strings` `vector solve(``int` `n, ``int` `arr[])` `{` `    ``// Marks the (N+1)th string` `    ``string s = string(n, ``'a'``);`   `    ``vector ans;` `    ``ans.push_back(s);`   `    ``// To generate remaining N strings` `    ``for` `(``int` `i = n - 1; i >= 0; i--) {`   `        ``// Find i-th string using` `        ``// (i+1)-th string` `        ``char` `ch = s[arr[i]];`   `        ``// Check if current character` `        ``// is b` `        ``if` `(ch == ``'b'``)` `            ``ch = ``'a'``;`   `        ``// Otherwise` `        ``else` `            ``ch = ``'b'``;` `        ``s[arr[i]] = ch;`   `        ``// Insert the string` `        ``ans.push_back(s);` `    ``}`   `    ``// Return the answer` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `arr[] = { 2, 0, 3 };` `    ``int` `n = ``sizeof` `arr / ``sizeof` `arr[0];` `    ``vector ans = solve(n, arr);`   `    ``// Print the strings` `    ``for` `(``int` `i = ans.size() - 1; i >= 0; i--) {` `        ``cout << ans[i] << endl;` `    ``}`   `    ``return` `0;` `}`

 `# Python3 Program to implement` `# the above approach`   `# Function to find the array ` `# of strings` `def` `solve(n, arr):`   `    ``# Marks the (N+1)th ` `    ``# string` `    ``s ``=` `'a'` `*` `(n)` `    ``ans ``=` `[]` `    ``ans.append(s)`   `    ``# To generate remaining ` `    ``# N strings` `    ``for` `i ``in` `range``(n ``-` `1``, ` `                   ``-``1``, ``-``1``):`   `        ``# Find i-th string using` `        ``# (i+1)-th string    ` `        ``if` `len``(s) ``-` `1` `>``=` `arr[i]:` `           ``ch ``=` `s[arr[i]]`   `           ``# Check if current ` `           ``# character` `           ``# is b` `           ``if` `(ch ``=``=` `'b'``):` `               ``ch ``=` `'a'`   `           ``# Otherwise` `           ``else``:` `               ``ch ``=` `'b'` `           `  `           ``p ``=` `list``(s)` `           ``p[arr[i]] ``=` `ch` `           ``s ``=` `''.join(p)`   `        ``# Insert the string` `        ``ans.append(s)`   `    ``# Return the answer` `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``arr ``=` `[``2``, ``0``, ``3``]` `    ``n ``=` `len``(arr)` `    ``ans ``=` `solve(n, arr)`   `    ``# Print the strings` `    ``for` `i ``in` `range``(``len``(ans) ``-` `1``,` `                   ``-``1``, ``-``1``):` `        ``print``(ans[i])`   `# This code is contributed by Chitranayal`

Output:
```bab
baa
aaa
aaa

```

Time Complexity: O(N)
Auxiliary Space: O(N)

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