Count of subsequence of an Array having all unique digits

Last Updated : 30 Nov, 2021

Given an array A containing N positive integers, the task is to find the number of subsequences of this array such that in each subsequence , no digit is repeated twice, i.e. all the digits of the subsequences must be unique.
Examples:

Input: A = [1, 12, 23, 34]
Output: 7
The subsequences are: {1}, {12}, {23}, {34}, {1, 23}, {1, 34}, {12, 34}
Therefore the count of such subsequences = 7

Input: A = [5, 12, 2, 1, 165, 2323, 7]
Output: 33

Naive approach: Generate all subsequences of the array and traverse through them to check whether the given condition is satisfied or not. Print the count of such subsequences at the end.

Below is the implementation of the above approach:

C++

// C++ program to find the count
// of subsequences  of an Array
// having all unique digits
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether
// the subsequences  has all unique digits
bool check(vector<int>& v)
{
 
    // Storing all digits occurred
    set<int> digits;
 
    // Traversing all the numbers of v
    for (int i = 0; i < v.size(); i++) {
        // Storing all digits of v[i]
        set<int> d;
 
        while (v[i]) {
            d.insert(v[i] % 10);
            v[i] /= 10;
        }
 
        // Checking whether digits of v[i]
        // have already occurred
        for (auto it : d) {
            if (digits.count(it))
                return false;
        }
 
        // Inserting digits of v[i] in the set
        for (auto it : d)
            digits.insert(it);
    }
 
    return true;
}
 
// Function to count the number
// subarray with all digits unique
int numberOfSubarrays(int a[], int n)
{
 
    int answer = 0;
 
    // Traverse through all the subarrays
    for (int i = 1; i < (1 << n); i++) {
        // To store elements of this subarray
        vector<int> temp;
 
        // Generate all subarray
        // and store it in vector
        for (int j = 0; j < n; j++) {
            if (i & (1 << j))
                temp.push_back(a[j]);
        }
 
        // Check whether this subarray
        // has all digits unique
        if (check(temp))
 
            // Increase the count
            answer++;
    }
 
    // Return the count
    return answer;
}
 
// Driver code
int main()
{
    int N = 4;
    int A[] = { 1, 12, 23, 34 };
 
    cout << numberOfSubarrays(A, N);
    return 0;
}

Java

// Java program to find the count
// of subarrays of an Array
// having all unique digits
  
 
import java.util.*;
 
class GFG{
  
// Function to check whether
// the subarray has all unique digits
static boolean check(Vector<Integer> v)
{
  
    // Storing all digits occurred
    HashSet<Integer> digits = new HashSet<Integer>();
  
    // Traversing all the numbers of v
    for (int i = 0; i < v.size(); i++) {
        // Storing all digits of v[i]
        HashSet<Integer> d = new HashSet<Integer>();
  
        while (v.get(i)>0) {
            d.add(v.get(i) % 10);
            v.set(i, v.get(i)/10);
        }
  
        // Checking whether digits of v[i]
        // have already occurred
        for (int it : d) {
            if (digits.contains(it))
                return false;
        }
  
        // Inserting digits of v[i] in the set
        for (int it : d)
            digits.add(it);
    }
  
    return true;
}
  
// Function to count the number
// subarray with all digits unique
static int numberOfSubarrays(int a[], int n)
{
  
    int answer = 0;
  
    // Traverse through all the subarrays
    for (int i = 1; i < (1 << n); i++) {
        // To store elements of this subarray
        Vector<Integer> temp = new Vector<Integer>();
  
        // Generate all subarray
        // and store it in vector
        for (int j = 0; j < n; j++) {
            if ((i & (1 << j))>0)
                temp.add(a[j]);
        }
  
        // Check whether this subarray
        // has all digits unique
        if (check(temp))
  
            // Increase the count
            answer++;
    }
  
    // Return the count
    return answer;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 4;
    int A[] = { 1, 12, 23, 34 };
  
    System.out.print(numberOfSubarrays(A, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the count
# of subarrays of an Array
# having all unique digits
 
# Function to check whether
# the subarray has all unique digits
def check(v):
  
    # Storing all digits occurred
    digits = set()
  
    # Traversing all the numbers of v
    for i in range(len(v)):
         
        # Storing all digits of v[i]
        d = set()
  
        while (v[i] != 0):
            d.add(v[i] % 10)
            v[i] //= 10
  
        # Checking whether digits of v[i]
        # have already occurred
        for it in d:
            if it in digits:
                return False
         
        # Inserting digits of v[i] in the set
        for it in d:
            digits.add(it)
     
    return True
 
# Function to count the number
# subarray with all digits unique
def numberOfSubarrays(a, n):
  
    answer = 0
  
    # Traverse through all the subarrays
    for i in range(1, 1 << n):
     
        # To store elements of this subarray
        temp = []
  
        # Generate all subarray
        # and store it in vector
        for j in range(n):
            if (i & (1 << j)):
                temp.append(a[j])
         
        # Check whether this subarray
        # has all digits unique
        if (check(temp)):
  
            # Increase the count
            answer += 1
     
    # Return the count
    return answer
 
# Driver code
if __name__=="__main__":
     
    N = 4
    A = [ 1, 12, 23, 34 ]
  
    print(numberOfSubarrays(A, N))
 
# This code is contributed by rutvik_56

C#

// C# program to find the count
// of subarrays of an Array
// having all unique digits
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check whether
// the subarray has all unique digits
static bool check(List<int> v)
{
 
    // Storing all digits occurred
    HashSet<int> digits = new HashSet<int>();
 
    // Traversing all the numbers of v
    for(int i = 0; i < v.Count; i++) 
    {
         
        // Storing all digits of v[i]
        HashSet<int> d = new HashSet<int>();
 
        while (v[i] > 0)
        {
            d.Add(v[i] % 10);
            v[i] = v[i] / 10;
        }
 
        // Checking whether digits of v[i]
        // have already occurred
        foreach(int it in d)
        {
            if (digits.Contains(it))
                return false;
        }
 
        // Inserting digits of v[i] in the set
        foreach(int it in d)
            digits.Add(it);
    }
    return true;
}
 
// Function to count the number
// subarray with all digits unique
static int numberOfSubarrays(int []a, int n)
{
    int answer = 0;
 
    // Traverse through all the subarrays
    for(int i = 1; i < (1 << n); i++)
    {
         
        // To store elements of this subarray
        List<int> temp = new List<int>();
 
        // Generate all subarray
        // and store it in vector
        for(int j = 0; j < n; j++) 
        {
            if ((i & (1 << j)) > 0)
                temp.Add(a[j]);
        }
 
        // Check whether this subarray
        // has all digits unique
        if (check(temp))
 
            // Increase the count
            answer++;
    }
 
    // Return the count
    return answer;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 4;
    int []A = { 1, 12, 23, 34 };
 
    Console.Write(numberOfSubarrays(A, N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
// Javascript program to find the count
// of subarrays of an Array
// having all unique digits
 
// Function to check whether
// the subarray has all unique digits
function check(v) {
 
    // Storing all digits occurred
    let digits = new Set();
 
    // Traversing all the numbers of v
    for (let i = 0; i < v.length; i++) {
        // Storing all digits of v[i]
        let d = new Set();
 
        while (v[i]) {
            d.add(v[i] % 10);
            v[i] = Math.floor(v[i] / 10);
        }
 
        // Checking whether digits of v[i]
        // have already occurred
        for (let it of d) {
            if (digits.has(it))
                return false;
        }
 
        // Inserting digits of v[i] in the set
        for (let it of d)
            digits.add(it);
    }
 
    return true;
}
 
// Function to count the number
// subarray with all digits unique
function numberOfSubarrays(a, n) {
 
    let answer = 0;
 
    // Traverse through all the subarrays
    for (let i = 1; i < (1 << n); i++) {
        // To store elements of this subarray
        let temp = new Array();
 
        // Generate all subarray
        // and store it in vector
        for (let j = 0; j < n; j++) {
            if (i & (1 << j))
                temp.push(a[j]);
        }
 
        // Check whether this subarray
        // has all digits unique
        if (check(temp))
 
            // Increase the count
            answer++;
    }
 
    // Return the count
    return answer;
}
 
// Driver code
 
let N = 4;
let A = [1, 12, 23, 34];
 
document.write(numberOfSubarrays(A, N));
 
// This code is contributed by gfgking
</script>

Output:

Time Complexity: O(N * 2^N)

Efficient Approach: This approach depends upon the fact that there exist only 10 unique digits in the Decimal number system. Therefore the longest subsequence will have only 10 digits in it, to meet the required condition.

We will use Bitmasking and Dynamic Programming to solve the problem.
Since there are only 10 digits, consider a 10-bit representation of every number where each bit is 1 if digit corresponding to that bit is present in that number.
Let, i be the current array element (elements from 1 to i-1 are already processed). An integer variable ‘mask‘ indicates the digits which have already occurred in the subsequence. If i’th bit is set in the mask, then i’th digit has occurred, else not.
At each step of recurrence relation, the element can either be included in the subsequence or not. If the element is not included in the subarray, then simply move to the next index. If it is included, change the mask by setting all the bits corresponding to the current element’s digit, ON in the mask.
Note: The current element can only be included if all of its digits have not occurred previously.
This condition will be satisfied only if the bits corresponding to the current element’s digits in the mask are OFF.
If we draw the complete recursion tree, we can observe that many subproblems are being solved again and again. So we use Dynamic Programming. A table dp[][] is used such that for every index dp[i][j], i is the position of the element in the array and j is the mask.

Below is the implementation of the above approach:

C++

// C++ program to find the count
// of subsequences  of an Array
// having all unique digits
 
#include <bits/stdc++.h>
using namespace std;
 
// Dynamic programming table
int dp[5000][(1 << 10) + 5];
 
// Function to obtain
// the mask for any integer
int getmask(int val)
{
    int mask = 0;
 
    if (val == 0)
        return 1;
 
    while (val) {
        int d = val % 10;
        mask |= (1 << d);
        val /= 10;
    }
    return mask;
}
 
// Function to count the number of ways
int countWays(int pos, int mask,
              int a[], int n)
{
    // Subarray must not be empty
    if (pos == n)
        return (mask > 0 ? 1 : 0);
 
    // If subproblem has been solved
    if (dp[pos][mask] != -1)
        return dp[pos][mask];
 
    int count = 0;
 
    // Excluding this element in the subarray
    count = count
            + countWays(pos + 1, mask, a, n);
 
    // If there are no common digits
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0) {
 
        // Calculate the new mask
        // if this element is included
        int new_mask
            = (mask | (getmask(a[pos])));
 
        count = count
                + countWays(pos + 1,
                            new_mask,
                            a, n);
    }
 
    // Store and return the answer
    return dp[pos][mask] = count;
}
 
// Function to find the count of
// subarray with all digits unique
int numberOfSubarrays(int a[], int n)
{
    // initializing dp
    memset(dp, -1, sizeof(dp));
 
    return countWays(0, 0, a, n);
}
 
// Driver code
int main()
{
    int N = 4;
    int A[] = { 1, 12, 23, 34 };
 
    cout << numberOfSubarrays(A, N);
    return 0;
}

Java

// Java program to find the count
// of subarrays of an Array
// having all unique digits
import java.util.*;
 
class GFG{
  
// Dynamic programming table
static int [][]dp = new int[5000][(1 << 10) + 5];
  
// Function to obtain
// the mask for any integer
static int getmask(int val)
{
    int mask = 0;
  
    if (val == 0)
        return 1;
  
    while (val > 0) {
        int d = val % 10;
        mask |= (1 << d);
        val /= 10;
    }
    return mask;
}
  
// Function to count the number of ways
static int countWays(int pos, int mask,
              int a[], int n)
{
    // Subarray must not be empty
    if (pos == n)
        return (mask > 0 ? 1 : 0);
  
    // If subproblem has been solved
    if (dp[pos][mask] != -1)
        return dp[pos][mask];
  
    int count = 0;
  
    // Excluding this element in the subarray
    count = count
            + countWays(pos + 1, mask, a, n);
  
    // If there are no common digits
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0) {
  
        // Calculate the new mask
        // if this element is included
        int new_mask
            = (mask | (getmask(a[pos])));
  
        count = count
                + countWays(pos + 1,
                            new_mask,
                            a, n);
    }
  
    // Store and return the answer
    return dp[pos][mask] = count;
}
  
// Function to find the count of
// subarray with all digits unique
static int numberOfSubarrays(int a[], int n)
{
    // initializing dp
    for(int i = 0;i<5000;i++)
    {
        for (int j = 0; j < (1 << 10) + 5; j++) {
            dp[i][j] = -1;
        }
    }
  
    return countWays(0, 0, a, n);
}
  
// Driver code
public static void main(String[] args)
{
    int N = 4;
    int A[] = { 1, 12, 23, 34 };
  
    System.out.print(numberOfSubarrays(A, N));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to find the count 
# of subarrays of an Array having all
# unique digits 
 
# Function to obtain 
# the mask for any integer 
def getmask(val):
     
    mask = 0
     
    if val == 0:
        return 1
 
    while (val): 
        d = val % 10; 
        mask |= (1 << d) 
        val = val // 10
 
    return mask
 
# Function to count the number of ways 
def countWays(pos, mask, a, n):
 
    # Subarray must not be empty 
    if pos == n : 
        if mask > 0:
            return 1
        else:
            return 0
 
    # If subproblem has been solved 
    if dp[pos][mask] != -1:
        return dp[pos][mask]
 
    count = 0
 
    # Excluding this element in the subarray 
    count = (count +
             countWays(pos + 1, mask, a, n))
 
    # If there are no common digits 
    # then only this element can be included 
    if (getmask(a[pos]) & mask) == 0:
 
        # Calculate the new mask 
        # if this element is included 
        new_mask = (mask | (getmask(a[pos])))
 
        count = (count +
                 countWays(pos + 1, 
                           new_mask,
                           a, n))
 
    # Store and return the answer 
    dp[pos][mask] = count
    return count
 
# Function to find the count of 
# subarray with all digits unique 
def numberOfSubarrays(a, n):
     
    return countWays(0, 0, a, n)
 
# Driver Code
N = 4
A = [ 1, 12, 23, 34 ]
 
rows = 5000
cols = 1100
 
# Initializing dp
dp = [ [ -1 for i in range(cols) ]
            for j in range(rows) ] 
             
print( numberOfSubarrays(A, N))
 
# This code is contributed by sarthak_eddy.

C#

// C# program to find the count
// of subarrays of an Array
// having all unique digits
using System;
 
public class GFG{
 
// Dynamic programming table
static int [,]dp = new int[5000, (1 << 10) + 5];
 
// Function to obtain
// the mask for any integer
static int getmask(int val)
{
    int mask = 0;
 
    if (val == 0)
        return 1;
 
    while (val > 0) {
        int d = val % 10;
        mask |= (1 << d);
        val /= 10;
    }
    return mask;
}
 
// Function to count the number of ways
static int countWays(int pos, int mask,
            int []a, int n)
{
    // Subarray must not be empty
    if (pos == n)
        return (mask > 0 ? 1 : 0);
 
    // If subproblem has been solved
    if (dp[pos, mask] != -1)
        return dp[pos, mask];
 
    int count = 0;
 
    // Excluding this element in the subarray
    count = count
        + countWays(pos + 1, mask, a, n);
 
    // If there are no common digits
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0) {
 
        // Calculate the new mask
        // if this element is included
        int new_mask
            = (mask | (getmask(a[pos])));
 
        count = count
            + countWays(pos + 1,
            new_mask, a, n);
    }
 
    // Store and return the answer
    return dp[pos, mask] = count;
}
 
// Function to find the count of
// subarray with all digits unique
static int numberOfSubarrays(int []a, int n)
{
    // initializing dp
    for(int i = 0; i < 5000; i++)
    {
        for (int j = 0; j < (1 << 10) + 5; j++) {
            dp[i,j] = -1;
        }
    }
 
    return countWays(0, 0, a, n);
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 4;
    int []A = { 1, 12, 23, 34 };
 
    Console.Write(numberOfSubarrays(A, N));
}
}
// This code contributed by sapnasingh4991

Javascript

<script>
 
// Javascript program to find the count
// of subarrays of an Array
// having all unique digits
 
// Dynamic programming table
var dp = Array.from(Array(5000), ()=>Array((1 << 10) + 5).fill(-1));
 
// Function to obtain
// the mask for any integer
function getmask(val)
{
    var mask = 0;
 
    if (val == 0)
        return 1;
 
    while (val) {
        var d = val % 10;
        mask |= (1 << d);
        val = parseInt(val/10);
    }
    return mask;
}
 
// Function to count the number of ways
function countWays(pos, mask, a, n)
{
    // Subarray must not be empty
    if (pos == n)
        return (mask > 0 ? 1 : 0);
 
    // If subproblem has been solved
    if (dp[pos][mask] != -1)
        return dp[pos][mask];
 
    var count = 0;
 
    // Excluding this element in the subarray
    count = count
            + countWays(pos + 1, mask, a, n);
 
    // If there are no common digits
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0) {
 
        // Calculate the new mask
        // if this element is included
        var new_mask
            = (mask | (getmask(a[pos])));
 
        count = count
                + countWays(pos + 1,
                            new_mask,
                            a, n);
    }
 
    // Store and return the answer
    return dp[pos][mask] = count;
}
 
// Function to find the count of
// subarray with all digits unique
function numberOfSubarrays(a, n)
{
    // initializing dp
    dp = Array.from(Array(5000), ()=>Array((1 << 10) + 5).fill(-1));
 
    return countWays(0, 0, a, n);
}
 
// Driver code
var N = 4;
var A = [1, 12, 23, 34];
document.write( numberOfSubarrays(A, N));
 
</script>

Output:

Time Complexity: O(N * 2¹⁰)

Count the number of unordered triplets with elements in increasing order and product less than or equal to integer X

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