Count of subsequences of Array with last digit of product as K
Last Updated :
23 Jul, 2025
Given an array A[] of size N and a digit K. Find the number of sub-sequences of the array where the last digit of the product of the elements of the sub-sequence is K.
Example:
Input: A = {2, 3, 4, 2}, K = 2
Output: 4
Explanation: sub-sequences with product's last digit = 2 are: {2}, {2}, {2, 3, 2}, {3, 4}
Input: A = {1, 1, 1, 2}, K = 1
Output: 7
Explanation: The sub-sequences with product's last digit = 2 are: {1}, {1}, {1}, {1, 1}, {1, 1}, {1, 1}, {1, 1, 1}
Approach: The above problem can be solved using recursion based on below idea:
We will use the recursive approach to find each possible sub-sequences, and on the go we will calculate the product of the elements and keep track of the last digit p, then at end we check if p is equal to K we return 1 else return 0.
Illustration:
Note: during multiplication of two numbers say a = 133 and b = 26
133
x 26
------
1798 <--- last digit gets multiplied only once
266X
------
3458 (3*6 = 18, both have 8 as last digits)
So the last digit of the a*b will be equal to the last digit of product of last digits of a and b
Suppose we have numbers 11, 233, 783, 4759, 6
product of the numbers = 57302995266
product of just the last digits, i.e., 1, 3, 3, 9, 6 = 486
Both have last digit as 6
Follow the below steps to solve the problem:
- Base Case: if p is equal to K return 1, else return 0
- There are two options either take the element or don't.
- if we take the element, then the product p gets updated to p*a[n-1], as we are just bothered about the last digit so we can just update p to the last digit of p*a[n-1], i.e., p becomes p*a[n-1]%10.
- other option is don't take the element, so don't update p and do n-1, to look for other possibilities.
- As we need the total number of such sub-sequences we return the sum of the above two calls.
- An edge case is if K=1, then we will get one extra sub-sequence that is an empty sub-sequence, as we initially take p=1 so in an empty sub-sequence we get p==k and 1 is returned. So when K==1 we deduct 1 from the answer.
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: Since Recursion takes exponential time complexity, the above recursive approach can be optimized further using Dynamic Programming. We will construct a look-up table and memoize the recursive code in order to do so.
Below is the implementation of the above approach:
C++
// C++ program for Count the number of
// sub-sequences of an array where the last digit
// of the product of the subsequence is K.
#include <bits/stdc++.h>
using namespace std;
int dp[1001][10];
// recursive utility function
int countSubsequencesUtil(int n, int k, int p,
vector<int>& a)
{
// Base case
if (n == 0) {
if (p == k)
return 1;
else
return 0;
}
// return pre calculated value from
// look-up table
if (dp[n][p] != -1)
return dp[n][p];
// return the total number obtained through
// option1 and option2
return dp[n][p]
= countSubsequencesUtil(n - 1, k,
(p * a[n - 1]) % 10, a)
+ countSubsequencesUtil(n - 1, k, p, a);
}
// Function to Count the number of subsequences
int countSubsequences(vector<int>& a, int k)
{
// initialize the table with -1
memset(dp, -1, sizeof(dp));
int n = a.size();
int ans = countSubsequencesUtil(n, k, 1, a);
// if k is 1 return 1 less
if (k == 1)
return ans - 1;
return ans;
}
// Driver Code
int main()
{
vector<int> a = { 2, 3, 4, 2 };
int k = 2;
cout << countSubsequences(a, k);
return 0;
}
// Java program to count the number of subsequences
// of an array where the last digit of the product
// of the subsequence is K
import java.io.*;
import java.util.*;
public class GFG {
static int[][] dp = new int[1001][10];
// recursive utility function
static int countSubsequencesUtil(int n, int k, int p,
int[] a)
{
// base case
if (n == 0) {
if (p == k)
return 1;
return 0;
}
// return pre calculated value from
// look up table
if (dp[n][p] != -1)
return dp[n][p];
// return the total no. obtained through
// option1 and option2
return dp[n][p]
= countSubsequencesUtil(n - 1, k,
(p * a[n - 1]) % 10, a)
+ countSubsequencesUtil(n - 1, k, p, a);
}
// function to count the number of subsequences
static int countSubsequences(int[] a, int k)
{
// initializing all elements of table with -1
for (int i = 0; i <= 1000; i++) {
for (int j = 0; j < 10; j++) {
dp[i][j] = -1;
}
}
int n = a.length;
int ans = countSubsequencesUtil(n, k, 1, a);
// if k is 1 return 1 less
if (k == 1)
return ans - 1;
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[] a = { 2, 3, 4, 2 };
int k = 2;
System.out.print(countSubsequences(a, k));
}
}
// This code is contributed by phasing17
# Python implementation of above approach
dp = [[-1]*10]*1001
# recursive utility function
def countSubsequencesUtil(n, k, p, a) :
# Base case
if (n == 0) :
if (p == k):
return 1
else:
return 0
# return pre calculated value from
# look-up table
if (dp[n][p] != -1):
return dp[n][p]
# return the total number obtained through
# option1 and option2
dp[n][p] = (countSubsequencesUtil(n - 1, k,(p * a[n - 1]) % 10, a) +
countSubsequencesUtil(n - 1, k, p, a));
return (dp[n][p]);
# Function to Count the number of subsequences
def countSubsequences(a, k) :
n = len(a)
ans = countSubsequencesUtil(n, k, 1, a)
# if k is 1 return 1 less
if (k == 1) :
return (ans - 1)
return ans + 1
# Driver Code
a = [ 2, 3, 4, 2 ]
k = 2
print(countSubsequences(a, k))
# This code is contributed by sanjoy_62.
// C# program for Count the number of
// sub-sequences of an array where the last digit
// of the product of the subsequence is K.
using System;
class GFG {
static int[, ] dp = new int[1001, 10];
// recursive utility function
static int countSubsequencesUtil(int n, int k, int p,
int[] a)
{
// Base case
if (n == 0) {
if (p == k)
return 1;
else
return 0;
}
// return pre calculated value from
// look-up table
if (dp[n, p] != -1)
return dp[n, p];
// return the total number obtained through
// option1 and option2
return dp[n, p]
= countSubsequencesUtil(n - 1, k,
(p * a[n - 1]) % 10, a)
+ countSubsequencesUtil(n - 1, k, p, a);
}
// Function to Count the number of subsequences
static int countSubsequences(int[] a, int k)
{
// initialize the table with -1
for (int i = 0; i < 1001; i++) {
for (int j = 0; j < 10; j++) {
dp[i, j] = -1;
}
}
int n = a.Length;
int ans = countSubsequencesUtil(n, k, 1, a);
// if k is 1 return 1 less
if (k == 1)
return ans - 1;
return ans;
}
// Driver Code
public static void Main()
{
int[] a = { 2, 3, 4, 2 };
int k = 2;
Console.Write(countSubsequences(a, k));
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// JavaScript program for Count the number of
// sub-sequences of an array where the last digit
// of the product of the subsequence is K.
// initialize the dp table with -1
let dp = new Array(1001)
for(let i = 0; i < 1001; i++)
{
dp[i] = new Array(10).fill(-1)
}
// recursive utility function
function countSubsequencesUtil(n, k, p,a)
{
// Base case
if (n == 0) {
if (p == k)
return 1
else
return 0
}
// return pre calculated value from
// look-up table
if (dp[n][p] != -1)
return dp[n][p]
// return the total number obtained through
// option1 and option2
return dp[n][p] = countSubsequencesUtil(n - 1, k, (p * a[n - 1]) % 10, a) + countSubsequencesUtil(n - 1, k, p, a)
}
// Function to Count the number of subsequences
function countSubsequences(a,k)
{
let n = a.length
let ans = countSubsequencesUtil(n, k, 1, a)
// if k is 1 return 1 less
if (k == 1)
return ans - 1;
return ans;
}
// Driver Code
let a = [ 2, 3, 4, 2 ]
let k = 2
document.write(countSubsequences(a, k))
// This code is contributed by shinjanpatra.
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a table to store the solution of the subproblems.
- Initialize the table with base cases
- Fill up the table iteratively
- Return the final solution
Implementation :
C++
// C++ program to count the number of subsequences
// of an array where the last digit of the product
// of the subsequence is k
#include <bits/stdc++.h>
using namespace std;
int dp[1001][10];
// Function to count the number of subsequences
int countSubsequences(vector<int>& a, int k)
{
int n = a.size();
// Initialize dp table
for (int i = 0; i < n; i++) {
// mark the element's remainder with 10 as 1
dp[i][a[i] % 10] = 1;
}
// Fill dp table
for (int i = 1; i < n; i++) {
for (int j = 0; j < 10; j++) {
// add the results of previous cells
dp[i][j] += dp[i - 1][j] + dp[i - 1][(j * a[i]) % 10];
}
}
// Return answer
return dp[n - 1][k];
}
// Driver code
int main()
{
vector<int> a = { 2, 3, 4, 2 };
int k = 2;
// function
cout << countSubsequences(a, k);
return 0;
}
// this code is contributed by bhardwajji
import java.util.*;
public class Main {
static int[][] dp = new int[1001][10];
// Function to count the number of subsequences
static int countSubsequences(List<Integer> a, int k) {
int n = a.size();
// Initialize dp table
for (int i = 0; i < n; i++) {
// mark the element's remainder with 10 as 1
dp[i][a.get(i) % 10] = 1;
}
// Fill dp table
for (int i = 1; i < n; i++) {
for (int j = 0; j < 10; j++) {
// add the results of previous cells
dp[i][j] += dp[i - 1][j] + dp[i - 1][(j * a.get(i)) % 10];
}
}
// Return answer
return dp[n - 1][k];
}
public static void main(String[] args) {
List<Integer> a = Arrays.asList(2, 3, 4, 2);
int k = 2;
// function call
System.out.println(countSubsequences(a, k));
}
}
# Function to count the number of subsequences
def count_subsequences(a, k):
n = len(a)
dp = [[0] * 10 for i in range(n)]
# Initialize dp table
for i in range(n):
# mark the element's remainder with 10 as 1
dp[i][a[i] % 10] = 1
# Fill dp table
for i in range(1, n):
for j in range(10):
# add the results of previous cells
dp[i][j] += dp[i - 1][j] + dp[i - 1][(j * a[i]) % 10]
# Return answer
return dp[n - 1][k]
# Driver code
a = [2, 3, 4, 2]
k = 2
# function call
print(count_subsequences(a, k))
using System;
using System.Collections.Generic;
public class MainClass {
static int[,] dp = new int[1001, 10];
// Function to count the number of subsequences
static int CountSubsequences(List<int> a, int k) {
int n = a.Count;
// Initialize dp table
for (int i = 0; i < n; i++) {
// mark the element's remainder with 10 as 1
dp[i, a[i] % 10] = 1;
}
// Fill dp table
for (int i = 1; i < n; i++) {
for (int j = 0; j < 10; j++) {
// add the results of previous cells
dp[i, j] += dp[i - 1, j] + dp[i - 1, (j * a[i]) % 10];
}
}
// Return answer
return dp[n - 1, k];
}
public static void Main(string[] args) {
List<int> a = new List<int>() {2, 3, 4, 2};
int k = 2;
// function call
Console.WriteLine(CountSubsequences(a, k));
}
}
// Function to count the number of subsequences
function count_subsequences(a, k) {
let n = a.length;
let dp = Array.from(Array(n), () => Array(10).fill(0));
// Initialize dp table
for (let i = 0; i < n; i++) {
// mark the element's remainder with 10 as 1
dp[i][a[i] % 10] = 1;
}
// Fill dp table
for (let i = 1; i < n; i++) {
for (let j = 0; j < 10; j++) {
// add the results of previous cells
dp[i][j] += dp[i - 1][j] + dp[i - 1][(j * a[i]) % 10];
}
}
// Return answer
return dp[n - 1][k];
}
// Driver code
let a = [2, 3, 4, 2];
let k = 2;
// function call
console.log(count_subsequences(a, k));
Output :
4
Time complexity: O(n * 10), where n is the length of the input array.
Auxiliary Space: O(n * 10).
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