# Count of subarrays of an Array having all unique digits

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

Examples:

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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

Below is the implementation of the above approach:

## C++

 `// C++ program to find the count ` `// of subarrays of an Array ` `// having all unique digits ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check whether ` `// the subarray 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 v) ` `{ ` `  `  `    ``// Storing all digits occurred ` `    ``HashSet digits = ``new` `HashSet(); ` `  `  `    ``// Traversing all the numbers of v ` `    ``for` `(``int` `i = ``0``; i < v.size(); i++) { ` `        ``// Storing all digits of v[i] ` `        ``HashSet d = ``new` `HashSet(); ` `  `  `        ``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 temp = ``new` `Vector(); ` `  `  `        ``// 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 `

Output:

```7
```

Time Complexity: O(N * 2N)

Efficient Approach: This approach depends upon the fact that there exist only 10 unique digits in the Decimal number system. Therefore the longest subarray 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 subarray. 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 subarray 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 subarrays of an Array ` `// having all unique digits ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Dynamic programming table ` `int` `dp[(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) ` `{ ` `    ``// intializing 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) ` `{ ` `    ``// intializing 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) ` `{ ` `    ``// intializing 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 `

Output:

```7
```

Time Complexity: O(N * 210) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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