Given an array arr[] consisting of N strings, the task is to find the maximum sum of length of the strings arr[i] and arr[j] for all unique pairs (i, j), where the strings arr[i] and arr[j] contains no common characters.
Examples:
Input: arr[] = [“abcd”, “cat”, “lto”, “car”, “wxyz”, “abcdef”]
Output: 8
Explanation:
The strings “abcd” and “wxyz” have no common characters in it. Therefore, the sum of the length of both the strings = 4 + 4 = 8, which is maximum among all possible pairs.Input: arr[] = [“abcd”, “def”, “fghi”, “ijklm”]
Output: 8
Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs of the array of strings and print the maximum value of the sum of the length of strings of pairs having no common characters between them.
Time Complexity: O(N2 * M), where M is the maximum length of the string.
Auxiliary Space: O(M)
Efficient Approach: The above approach can also be optimized by using the idea of Bit Manipulation. the idea is to convert each string into its bitmask integer equivalent and then find the pair of strings having no common characters having the maximum sum of their lengths. Follow the steps below to solve the problem:
- Initialize a vector mask of size N to store the bitwise OR of a string in the array of strings words[].
- Initialize the variable maxLength as 0 to store the answer.
-
Iterate over the range [0, N] using the variable i and performing the following tasks:
- Iterate over the range [0, M] where M is the length of the string using the variable j and set the value of mask[i] as mask[i]|1<<(words[i][j] – ‘a’).
- Iterate over the range [0, i] using the variable j and if the value bitwise AND of mask[i] and mask[j] is not 0, then set the value of maxLength as the maximum of maxLength or words[i].length() + words[j].length().
- After completing the above steps, print the value of the maxLength as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;
// Function to find the maximum sum of// length of pair of strings having// no common charactersint maxSum(vector<string>& words)
{ // Stores the bitmask of each strings
vector<int> mask(words.size());
// Initialize the result as zero
int result = 0;
// Iterate the given vector
for (int i = 0; i < words.size(); ++i) {
// For each of character
for (char c : words[i]) {
// If the ith value of
// mask |= 1 left shift
// that character - a
mask[i] |= 1 << (c - 'a');
}
// Check for each ith character,
// if the ith and jth value of
// mask are not same, then add
// and maximize them
for (int j = 0; j < i; ++j) {
if (!(mask[i] & mask[j])) {
result
= max(result, int(words[i].size()
+ words[j].size()));
}
}
}
// Return maximum sum of lengths
// of strings
return result;
}// Driver Codeint main()
{ vector<string> words = { "abcd", "def",
"fghi", "ijklm" };
cout << maxSum(words);
return 0;
} |
Java
// Java program for the above approachclass GFG {
// Function to find the maximum sum of
// length of pair of strings having
// no common characters
public static int maxSum(String[] words) {
// Stores the bitmask of each strings
int[] mask = new int[words.length];
// Initialize the result as zero
int result = 0;
// Iterate the given vector
for (int i = 0; i < words.length; ++i) {
// For each of character
for (char c : words[i].toCharArray()) {
// If the ith value of
// mask |= 1 left shift
// that character - a
mask[i] |= 1 << (c - 'a');
}
// Check for each ith character,
// if the ith and jth value of
// mask are not same, then add
// and maximize them
for (int j = 0; j < i; ++j) {
if ((mask[i] & mask[j]) < 1) {
result = Math.max(result, (int) words[i].length() + words[j].length());
}
}
}
// Return maximum sum of lengths
// of strings
return result;
}
// Driver Code
public static void main(String args[]) {
String[] words = { "abcd", "def", "fghi", "ijklm" };
System.out.println(maxSum(words));
}
}// This code is contributed by saurabh_jaiswal. |
Python3
# Function to find the maximum sum of# length of pair of strings having# no common charactersdef maxSum(words):
# Stores the bitmask of each strings
mask = [0]*len(words)
# Initialize the result as zero
result = 0
# Iterate the given vector
for i in range(len(words)):
for c in words[i]:
# If the ith value of
# mask |= 1 left shift
# that character - a
mask[i] |= 1 << (ord(c)-97)
# Check for each ith character,
# if the ith and jth value of
# mask are not same, then add
# and maximize them
for j in range(i):
if not(mask[i] & mask[j]):
result = max(result, len(words[i])+len(words[j]))
# Return maximum sum of lengths
# of strings
return result
# Driver codewords = ["abcd", "def", "fghi", "ijklm"]
print(maxSum(words))
# This code is contributed by Parth Manchanda |
C#
using System;
public class GFG {
public static int maxSum(String[] words)
{
// Stores the bitmask of each strings
int[] mask = new int[words.Length];
// Initialize the result as zero
int result = 0;
// Iterate the given vector
for (int i = 0; i < words.Length; ++i) {
// For each of character
foreach (char c in words[i]) {
// If the ith value of
// mask |= 1 left shift
// that character - a
mask[i] |= 1 << (c - 'a');
}
// Check for each ith character,
// if the ith and jth value of
// mask are not same, then add
// and maximize them
for (int j = 0; j < i; ++j) {
if ((mask[i] & mask[j]) < 1) {
result = Math.Max(
result, (int)words[i].Length
+ words[j].Length);
}
}
}
// Return maximum sum of lengths
// of strings
return result;
}
static public void Main()
{
String[] words = { "abcd", "def", "fghi", "ijklm" };
Console.WriteLine(maxSum(words));
}
}// This code is contributed by maddler. |
Javascript
<script>// JavaScript program for the above approach// Function to find the maximum sum of// length of pair of strings having// no common charactersfunction maxSum(words)
{ // Stores the bitmask of each strings
let mask = new Array(words.length);
// Initialize the result as zero
let result = 0;
// Iterate the given vector
for(let i = 0; i < words.length; ++i)
{
// For each of character
for(let c = words[i]; c < words[i].length; c++)
{
// If the ith value of
// mask |= 1 left shift
// that character - a
mask[i] |= 1 << (words[i].charCodeAt(0) -
'a'.charCodeAt(0));
}
// Check for each ith character,
// if the ith and jth value of
// mask are not same, then add
// and maximize them
for(let j = 0; j < i; ++j)
{
if (!(mask[i] & mask[j]))
{
result = Math.max(result, words[i].length +
words[j].length);
}
}
}
// Return maximum sum of lengths
// of strings
return result;
}// Driver Codelet words = [ "abcd", "def",
"fghi", "ijklm" ];
document.write(maxSum(words));// This code is contributed by Potta Lokesh</script> |
Output:
9
Time Complexity: O(max(N*M, N2)), where M is the maximum length of the string.
Auxiliary Space: O(N)
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