# Maximize product of lengths of strings having no common characters

• Difficulty Level : Hard
• Last Updated : 17 Jun, 2022

Given an array arr[] consisting of N strings, the task is to find the maximum product of the length of the strings arr[i] and arr[j] for all unique pairs (i, j), where the strings arr[i] and arr[j] contain no common characters.

Examples:

Input: arr[] = {“abcw”, “baz”, “foo”, “bar”, “xtfn”, “abcdef”}
Output: 16
Explanation: The strings “abcw” and “xtfn” have no common characters in it. Therefore, the product of the length of both the strings = 4 * 4 = 16, which is maximum among all possible pairs.

Input: arr[] = {“a”, “aa”, “aaa”, “aaaa”}
Output: 0

Naive Approach: The idea is to generate all the pairs of strings and use a map to find the common character between them, if there are no common character between them, then use the product of their length to maximize the result.

Follow the steps below to implement the above idea:

• Generate all pairs of strings say, s1 and s2.
• Use a map to find the common character between the strings s1 and s2.
• Check if there is any common character between them:
• If there is no common character between them then use the product of their length and maximize the result.
• Finally, return the result.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the find the maximum product of the``// length of the strings arr[i] and arr[j] for all unique``// pairs (i, j), where the strings arr[i] and arr[j] contain``// no common characters.``#include ``using` `namespace` `std;` `// Function to find the maximum product of the length of the``// strings``int` `maximizeProduct(vector& arr)``{``    ``int` `n = arr.size();` `    ``// This store the maximum product of string's length``    ``int` `result = 0;` `    ``// Iterate to find the 1st string``    ``for` `(``int` `i = 0; i < n; i++) {``        ``string s1 = arr[i];``        ``int` `len1 = arr[i].size();` `        ``// Map to store the unique character``        ``unordered_map<``char``, ``int``> unmap;``        ``for` `(``auto` `c : s1)``            ``unmap++;` `        ``// Iterate to find the 2nd string``        ``for` `(``int` `j = i + 1; j < n; j++) {``            ``string s2 = arr[j];``            ``int` `len2 = arr[j].size();` `            ``bool` `flag = ``false``;` `            ``for` `(``int` `k = 0; k < s2.size(); k++) {` `                ``// Check if the characters of s2 are common``                ``// with s1 characters or not``                ``if` `(unmap.count(s2[k])) {``                    ``flag = ``true``;``                    ``break``;``                ``}``            ``}` `            ``// This verify that there is no common character``            ``// between s1 and s2.``            ``if` `(flag == ``false``) {``                ``result = max(result, len1 * len2);``            ``}``        ``}``    ``}` `    ``return` `result;``}` `// Driver code``int` `main()``{``    ``vector arr = { ``"abcw"``, ``"baz"``,   ``"foo"``,``                           ``"bar"``,  ``"xtfn"``, ``"abcdef"` `};` `    ``int` `result = maximizeProduct(arr);` `    ``// Print the output``    ``cout << result;` `    ``return` `0;``}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {``// Java program to find the find the maximum product of the``// length of the strings arr[i] and arr[j] for all unique``// pairs (i, j), where the strings arr[i] and arr[j] contain``// no common characters.` `// Function to find the maximum product of the length of the``// strings``static` `int` `maximizeProduct(String[] arr)``{``    ``int` `n = arr.length;` `    ``// This store the maximum product of string's length``    ``int` `result = ``0``;` `    ``// Iterate to find the 1st string``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``String s1 = arr[i];``        ``int` `len1 = arr[i].length();` `        ``// Map to store the unique character``        ``HashMap unmap = ``new` `HashMap<>();``        ``for` `(``char` `c : s1.toCharArray()){``            ``if``(unmap.containsKey(c)){``                ``unmap.put(c, unmap.get(c)+``1``);``            ``}``            ``else` `unmap.put(c,``1``);``        ``}``        ``// Iterate to find the 2nd string``        ``for` `(``int` `j = i + ``1``; j < n; j++) {``            ``String s2 = arr[j];``            ``int` `len2 = arr[j].length();` `            ``boolean` `flag = ``false``;` `            ``for` `(``int` `k = ``0``; k < s2.length(); k++) {` `                ``// Check if the characters of s2 are common``                ``// with s1 characters or not``                ``if` `(unmap.containsKey(s2.charAt(k))) {``                    ``flag = ``true``;``                    ``break``;``                ``}``            ``}` `            ``// This verify that there is no common character``            ``// between s1 and s2.``            ``if` `(flag == ``false``) {``                ``result = Math.max(result, len1 * len2);``            ``}``        ``}``    ``}` `    ``return` `result;``}``    ` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``String[] arr = { ``"abcw"``, ``"baz"``,   ``"foo"``,``                           ``"bar"``,  ``"xtfn"``, ``"abcdef"` `};` `    ``int` `result = maximizeProduct(arr);` `    ``// Print the output``    ``System.out.println(result);``}``}`

## Python3

 `# Python3 program to find the find the maximum product of the``# length of the strings arr[i] and arr[j] for all unique``# pairs (i, j), where the strings arr[i] and arr[j] contain``# no common characters.` `# Function to find the maximum product of the length of the``# strings``def` `maximizeProduct(arr):` `    ``n ``=` `len``(arr)` `    ``# This store the maximum product of string's length``    ``result ``=` `0` `    ``# Iterate to find the 1st string``    ``for` `i ``in` `range``(n):``        ``s1 ``=` `arr[i]``        ``len1 ``=` `len``(arr[i])` `        ``# Map to store the unique character``        ``unmap ``=` `{}``        ``for` `c ``in` `s1:``            ``if``(c ``in` `unmap):``                ``unmap ``+``=` `1``            ` `            ``unmap ``=` `1` `        ``# Iterate to find the 2nd string``        ``for` `j ``in` `range``(i``+``1``,n):``            ``s2 ``=` `arr[j]``            ``len2 ``=` `len``(arr[j])` `            ``flag ``=` `False` `            ``for` `k ``in` `range``(``len``(s2)):` `                ``# Check if the characters of s2 are common``                ``# with s1 characters or not``                ``if` `(s2[k] ``in`  `unmap):``                    ``flag ``=` `True``                    ``break``                `  `            ``# This verify that there is no common character``            ``# between s1 and s2.``            ``if` `(flag ``=``=` `False``):``                ``result ``=` `max``(result, len1 ``*` `len2)``            `  `    ``return` `result` `# Driver code``arr ``=` `[ ``"abcw"``, ``"baz"``,  ``"foo"``,  ``"bar"``,  ``"xtfn"``, ``"abcdef"` `]``result ``=` `maximizeProduct(arr)` `# Print the output``print``(result)` `# This code is contributed by shinjanpatra`

## Javascript

 ``

Output

`16`

Time Complexity: O(N2 * M), where M is the maximum length of the string.
Auxiliary Space: O(M)

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