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Count of distinct groups of strings formed after performing equivalent operation

  • Difficulty Level : Medium
  • Last Updated : 07 Oct, 2021

Given an array arr[] of N strings consisting of lowercase alphabets, the task is to find the number of distinct groups of strings formed after performing the equivalent operation.

Two strings are said to be equivalent if there exists the same character in both the strings and if there exists another string that is equivalent to one of the strings in the group of equivalent string then that string is also equivalent to that group.

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Examples:



Input: arr[] = {“a”, “b”, “ab”, “d”}
Output: 2
Explanation:
As strings “b” and “ab” have ‘b’ as the same character, they are equivalent also “ab” and  the string”a” have ‘a’ as the same character, so the strings “a”, “b”, “ab” are equivalent and “d” is another string.

Therefore, the count of distinct group of strings formed is 2.

Input: arr[] = {“ab”, “bc”, “abc”}
Output: 1

Approach: The given problem can be solved using the Disjoint Set Union, the idea is to traverse the string and mark all the characters of the current string as true and perform the union operation on the first character of the current string with the character ‘a’, and count the different number of parents in the parent vector and store it. Follow the below steps to solve the problem:

  • Initialize the vectors parent(27), rank(27, 0), total(26, false), and current(26, false).
  • Initialize a variable, say distCount as 0 that stores the count of distinct strings.
  • Iterate over the range [0, 27) using the variable i and set the value of parent[i] as i.
  • Iterate over the range [0, N) using the variable i and perform the following steps:
    • Iterate over the range [0, 26) using the variable j and set current[j] to false.
    • Iterate over the characters of the string arr[i] using the variable ch and set current[ch – ‘a’] to true.
    • Iterate over the range [0, 26) using the variable j and if current[j] is true then set total[j] to true and call for the function Union(parent, rank, arr[i][0] – ‘a’, j).
  • Iterate over the range [0, 26) using the variable i and check if total[i] is true and Find(parent, i) is I if it is true then increment the value of distCount by 1.
  • Finally, print the value of distCount.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to perform the find operation
// to find the parent of a disjoint set
int Find(vector<int>& parent, int a)
{
    return parent[a]
           = (parent[a] == a ? a : Find(parent, parent[a]));
}
 
// Function to perform union operation
// of disjoint set union
void Union(vector<int>& parent,
           vector<int>& rank, int a,
           int b)
{
 
    // Find the parent of node a and b
    a = Find(parent, a);
    b = Find(parent, b);
 
    // Update the rank
    if (rank[a] == rank[b])
        rank[a]++;
    if (rank[a] > rank[b])
        parent[b] = a;
    else
        parent[a] = b;
}
 
// Function to find the number of distinct
// strings after performing the
// given operations
void numOfDistinctStrings(string arr[],
                          int N)
{
    // Stores the parent elements
    // of the sets
    vector<int> parent(27);
 
    // Stores the rank of the sets
    vector<int> rank(27, 0);
 
    for (int j = 0; j < 27; j++) {
        // Update parent[i] to i
        parent[j] = j;
    }
 
    // Stores the total characters
    // traversed through the strings
    vector<bool> total(26, false);
 
    // Stores the current characters
    // traversed through a string
    vector<bool> current(26, false);
 
    for (int i = 0; i < N; i++) {
 
        for (int j = 0; j < 26; j++) {
 
            // Update current[i] to false
            current[j] = false;
        }
 
        for (char ch : arr[i]) {
 
            // Update current[ch - 'a'] to true
            current[ch - 'a'] = true;
        }
 
        for (int j = 0; j < 26; j++) {
 
            // Check if current[j] is true
            if (current[j]) {
 
                // Update total[j] to true
                total[j] = true;
 
                // Add arr[i][0] - 'a' and
                // j elements to same set
                Union(parent, rank,
                      arr[i][0] - 'a', j);
            }
        }
    }
 
    // Stores the count of distinct strings
    int distCount = 0;
    for (int i = 0; i < 26; i++) {
 
        // Check total[i] is true and
        // parent of i is i only
        if (total[i] && Find(parent, i) == i) {
 
            // Increment the value of
            // distCount by 1
            distCount++;
        }
    }
 
    // Print the value of distCount
    cout << distCount << endl;
}
 
// Driver Code
int main()
{
    string arr[] = { "a", "ab", "b", "d" };
    int N = sizeof(arr) / sizeof(arr[0]);
    numOfDistinctStrings(arr, N);
 
    return 0;
}

Python3




# python program for the above approach
 
# Function to perform the find operation
# to find the parent of a disjoint set
def Find(parent, a):
    if parent[a] == a:
        parent[a] = a
        return parent[a]
    else:
        parent[a] = Find(parent, parent[a])
        return parent[a]
 
# Function to perform union operation
# of disjoint set union
def Union(parent, rank, a, b):
 
        # Find the parent of node a and b
    a = Find(parent, a)
    b = Find(parent, b)
 
    # Update the rank
    if (rank[a] == rank[b]):
        rank[a] += 1
    if (rank[a] > rank[b]):
        parent[b] = a
    else:
        parent[a] = b
 
# Function to find the number of distinct
# strings after performing the
# given operations
def numOfDistinctStrings(arr, N):
 
    # Stores the parent elements
    # of the sets
    parent = [0 for _ in range(27)]
 
    # Stores the rank of the sets
    rank = [0 for _ in range(27)]
 
    for j in range(0, 27):
        # Update parent[i] to i
        parent[j] = j
 
    # Stores the total characters
    # traversed through the strings
    total = [False for _ in range(26)]
 
    # Stores the current characters
    # traversed through a string
    current = [False for _ in range(26)]
 
    for i in range(0, N):
 
        for j in range(0, 26):
 
            # Update current[i] to false
            current[j] = False
 
        for ch in arr[i]:
 
            # Update current[ch - 'a'] to true
            current[ord(ch) - ord('a')] = True
 
        for j in range(0, 26):
 
            # Check if current[j] is true
            if (current[j]):
 
                # Update total[j] to true
                total[j] = True
 
                # Add arr[i][0] - 'a' and
                # j elements to same set
                Union(parent, rank, ord(arr[i][0]) - ord('a'), j)
 
    # Stores the count of distinct strings
    distCount = 0
    for i in range(0, 26):
 
        # Check total[i] is true and
        # parent of i is i only
        if (total[i] and Find(parent, i) == i):
 
            # Increment the value of
            # distCount by 1
            distCount += 1
 
    # Print the value of distCount
    print(distCount)
 
# Driver Code
if __name__ == "__main__":
 
    arr = ["a", "ab", "b", "d"]
    N = len(arr)
    numOfDistinctStrings(arr, N)
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
class GFG {
 
    // Function to perform the find operation
    // to find the parent of a disjoint set
    static int Find(int[] parent, int a)
    {
        return parent[a]
            = (parent[a] == a ? a
                              : Find(parent, parent[a]));
    }
 
    // Function to perform union operation
    // of disjoint set union
    static void Union(int[] parent, int[] rank, int a,
                      int b)
    {
 
        // Find the parent of node a and b
        a = Find(parent, a);
        b = Find(parent, b);
 
        // Update the rank
        if (rank[a] == rank[b])
            rank[a]++;
        if (rank[a] > rank[b])
            parent[b] = a;
        else
            parent[a] = b;
    }
 
    // Function to find the number of distinct
    // strings after performing the
    // given operations
    static void numOfDistinctStrings(string[] arr, int N)
    {
        // Stores the parent elements
        // of the sets
        int[] parent = new int[(27)];
 
        // Stores the rank of the sets
        int[] rank = new int[(27)];
 
        for (int j = 0; j < 27; j++) {
            // Update parent[i] to i
            parent[j] = j;
        }
 
        // Stores the total characters
        // traversed through the strings
        bool[] total = new bool[26];
 
        // Stores the current characters
        // traversed through a string
        bool[] current = new bool[26];
 
        for (int i = 0; i < N; i++) {
 
            for (int j = 0; j < 26; j++) {
 
                // Update current[i] to false
                current[j] = false;
            }
 
            foreach(char ch in arr[i])
            {
 
                // Update current[ch - 'a'] to true
                current[ch - 'a'] = true;
            }
 
            for (int j = 0; j < 26; j++) {
 
                // Check if current[j] is true
                if (current[j]) {
 
                    // Update total[j] to true
                    total[j] = true;
 
                    // Add arr[i][0] - 'a' and
                    // j elements to same set
                    Union(parent, rank, arr[i][0] - 'a', j);
                }
            }
        }
 
        // Stores the count of distinct strings
        int distCount = 0;
        for (int i = 0; i < 26; i++) {
 
            // Check total[i] is true and
            // parent of i is i only
            if (total[i] && Find(parent, i) == i) {
 
                // Increment the value of
                // distCount by 1
                distCount++;
            }
        }
 
        // Print the value of distCount
        Console.WriteLine(distCount);
    }
 
    // Driver Code
    public static void Main()
    {
        string[] arr = { "a", "ab", "b", "d" };
        int N = arr.Length;
        numOfDistinctStrings(arr, N);
    }
}
 
// This code is contributed by ukasp.

 
 

Output: 
2

 

 

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

 




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