Maximum subset sum such that no two elements in set have same digit in them
Last Updated :
18 Apr, 2023
Given an array of N elements. Find the subset of elements which has maximum sum such that no two elements in the subset has common digit present in them.
Examples:
Input : array[] = {22, 132, 4, 45, 12, 223}
Output : 268
Maximum Sum Subset will be = {45, 223} .
All possible digits are present except 1.
But to include 1 either 2 or both 2 and 3 have
to be removed which result in smaller sum value.
Input : array[] = {1, 21, 32, 4, 5 }
Output : 42
- Here we can use Dynamic Programming and Bit Masking to solve this question.
- Consider a 10-bit representation of every number where each bit is 1 if digit corresponding to that bit is present in that number.
- Now maintain a dp[i], which stores the maximum possible sum which can be achieved with all those digits present in the set, corresponding to the bit positions which are 1 in Binary Representation of i.
- Recurrence Relation will be of the form dp[i] = max(dp[i], dp[i^mask] + a[j]) , for all those j from 1 to n such that mask (10-bit Representation of a[j]) satisfy i || mask = i. (Since then only we can assure that all digit available in i are satisfied).
Below is the implementation of the above approach:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
int dp[1024];
// Function to create mask for every number
int get_binary(int u)
{
int ans = 0;
while (u) {
int rem = u % 10;
ans |= (1 << rem);
u /= 10;
}
return ans;
}
// Recursion for Filling DP array
int recur(int u, int array[], int n)
{
// Base Condition
if (u == 0)
return 0;
if (dp[u] != -1)
return dp[u];
int temp = 0;
for (int i = 0; i < n; i++) {
int mask = get_binary(array[i]);
// Recurrence Relation
if ((mask | u) == u) {
dp[u] = max(max(0,
dp[u ^ mask]) + array[i], dp[u]);
}
}
return dp[u];
}
// Function to find Maximum Subset Sum
int solve(int array[], int n)
{
// Initialize DP array
for (int i = 0; i < (1 << 10); i++) {
dp[i] = -1;
}
int ans = 0;
// Iterate over all possible masks of 10 bit number
for (int i = 0; i < (1 << 10); i++) {
ans = max(ans, recur(i, array, n));
}
return ans;
}
// Driver Code
int main()
{
int array[] = { 22, 132, 4, 45, 12, 223 };
int n = sizeof(array) / sizeof(array[0]);
cout << solve(array, n);
}
// Java implementation of above approach
import java.io.*;
class GFG
{
static int []dp = new int [1024];
// Function to create mask for every number
static int get_binary(int u)
{
int ans = 0;
while (u > 0)
{
int rem = u % 10;
ans |= (1 << rem);
u /= 10;
}
return ans;
}
// Recursion for Filling DP array
static int recur(int u, int []array, int n)
{
// Base Condition
if (u == 0)
return 0;
if (dp[u] != -1)
return dp[u];
for (int i = 0; i < n; i++)
{
int mask = get_binary(array[i]);
// Recurrence Relation
if ((mask | u) == u)
{
dp[u] = Math.max(Math.max(0,
dp[u ^ mask]) + array[i], dp[u]);
}
}
return dp[u];
}
// Function to find Maximum Subset Sum
static int solve(int []array, int n)
{
// Initialize DP array
for (int i = 0; i < (1 << 10); i++)
{
dp[i] = -1;
}
int ans = 0;
// Iterate over all possible masks of 10 bit number
for (int i = 0; i < (1 << 10); i++)
{
ans = Math.max(ans, recur(i, array, n));
}
return ans;
}
// Driver Code
static public void main (String[] args)
{
int []array = { 22, 132, 4, 45, 12, 223 };
int n = array.length;
System.out.println(solve(array, n));
}
}
// This code is contributed by anuj_67..
# Python3 implementation of above approach
dp = [0]*1024;
# Function to create mask for every number
def get_binary(u) :
ans = 0;
while (u) :
rem = u % 10;
ans |= (1 << rem);
u //= 10;
return ans;
# Recursion for Filling DP array
def recur(u, array, n) :
# Base Condition
if (u == 0) :
return 0;
if (dp[u] != -1) :
return dp[u];
temp = 0;
for i in range(n) :
mask = get_binary(array[i]);
# Recurrence Relation
if ((mask | u) == u) :
dp[u] = max(max(0, dp[u ^ mask]) + array[i], dp[u]);
return dp[u];
# Function to find Maximum Subset Sum
def solve(array, n) :
i = 0
# Initialize DP array
while(i < (1 << 10)) :
dp[i] = -1;
i += 1
ans = 0;
i = 0
# Iterate over all possible masks of 10 bit number
while(i < (1 << 10)) :
ans = max(ans, recur(i, array, n));
i += 1
return ans;
# Driver Code
if __name__ == "__main__" :
array = [ 22, 132, 4, 45, 12, 223 ] ;
n = len(array);
print(solve(array, n));
# This code is contributed by AnkitRai01
<script>
// Javascript implementation of above approach
let dp = new Array(1024);
dp.fill(-1);
// Function to create mask for every number
function get_binary(u)
{
let ans = 0;
while (u > 0)
{
let rem = u % 10;
ans |= (1 << rem);
u = parseInt(u / 10, 10);
}
return ans;
}
// Recursion for Filling DP array
function recur(u, array, n)
{
// Base Condition
if (u == 0)
return 0;
if (dp[u] != -1)
return dp[u];
for (let i = 0; i < n; i++)
{
let mask = get_binary(array[i]);
// Recurrence Relation
if ((mask | u) == u)
{
dp[u] = Math.max(Math.max(0,
dp[u ^ mask]) + array[i], dp[u]);
}
}
return dp[u];
}
// Function to find Maximum Subset Sum
function solve(array, n)
{
// Initialize DP array
for (let i = 0; i < (1 << 10); i++)
{
dp[i] = -1;
}
let ans = 0;
// Iterate over all possible masks of 10 bit number
for (let i = 0; i < (1 << 10); i++)
{
ans = Math.max(ans, recur(i, array, n));
}
return ans;
}
let array = [ 22, 132, 4, 45, 12, 223 ];
let n = array.length;
document.write(solve(array, n));
</script>
// C# implementation of above approach
using System;
class GFG
{
static int []dp = new int [1024];
// Function to create mask for every number
static int get_binary(int u)
{
int ans = 0;
while (u > 0)
{
int rem = u % 10;
ans |= (1 << rem);
u /= 10;
}
return ans;
}
// Recursion for Filling DP array
static int recur(int u, int []array, int n)
{
// Base Condition
if (u == 0)
return 0;
if (dp[u] != -1)
return dp[u];
for (int i = 0; i < n; i++)
{
int mask = get_binary(array[i]);
// Recurrence Relation
if ((mask | u) == u)
{
dp[u] = Math.Max(Math.Max(0,
dp[u ^ mask]) + array[i], dp[u]);
}
}
return dp[u];
}
// Function to find Maximum Subset Sum
static int solve(int []array, int n)
{
// Initialize DP array
for (int i = 0; i < (1 << 10); i++)
{
dp[i] = -1;
}
int ans = 0;
// Iterate over all possible masks of 10 bit number
for (int i = 0; i < (1 << 10); i++)
{
ans = Math.Max(ans, recur(i, array, n));
}
return ans;
}
// Driver Code
static public void Main ()
{
int []array = { 22, 132, 4, 45, 12, 223 };
int n = array.Length;
Console.WriteLine (solve(array, n));
}
}
// This code is contributed by ajit.
Time Complexity : O(N*(2^10))
Auxiliary Space: O(1024)
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 + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a DP to store the solution of the subproblems.
- Initialize the DP with base cases.
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
- Create a variable ans to store the final result.
- Iterate over Dp and update ans.
- At last return and print ans.
Implementation :
C++
#include <bits/stdc++.h>
using namespace std;
int dp[1024];
// Function to create mask for every number
int get_binary(int u)
{
int ans = 0;
while (u) {
int rem = u % 10;
ans |= (1 << rem);
u /= 10;
}
return ans;
}
// Function to find Maximum Subset Sum
int solve(int array[], int n)
{
// Initialize DP array
for (int i = 0; i < (1 << 10); i++) {
dp[i] = 0;
}
// Fill DP table using bottom-up approach
for (int i = 0; i < n; i++) {
int mask = get_binary(array[i]);
for (int j = (1 << 10) - 1; j >= 0; j--) {
if ((mask | j) == j) {
dp[j] = max(dp[j], dp[j ^ mask] + array[i]);
}
}
}
// Find maximum sum from DP array
int ans = 0;
for (int i = 0; i < (1 << 10); i++) {
ans = max(ans, dp[i]);
}
return ans;
}
// Driver Code
int main()
{
int array[] = { 22, 132, 4, 45, 12, 223 };
int n = sizeof(array) / sizeof(array[0]);
cout << solve(array, n);
}
import java.util.Arrays;
class GFG {
static int dp[] = new int[1024];
// Function to create mask for every number
static int get_binary(int u)
{
int ans = 0;
while (u != 0) {
int rem = u % 10;
ans |= (1 << rem);
u /= 10;
}
return ans;
}
// Function to find Maximum Subset Sum
static int solve(int array[], int n)
{
// Initialize DP array
Arrays.fill(dp, 0);
// Fill DP table using bottom-up approach
for (int i = 0; i < n; i++) {
int mask = get_binary(array[i]);
for (int j = (1 << 10) - 1; j >= 0; j--) {
if ((mask | j) == j) {
dp[j] = Math.max(dp[j], dp[j ^ mask]
+ array[i]);
}
}
}
// Find maximum sum from DP array
int ans = 0;
for (int i = 0; i < (1 << 10); i++) {
ans = Math.max(ans, dp[i]);
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int array[] = { 22, 132, 4, 45, 12, 223 };
int n = array.length;
System.out.println(solve(array, n));
}
}
# Function to create mask for every number
def get_binary(u):
ans = 0
while u:
rem = u % 10
ans |= (1 << rem)
u //= 10
return ans
# Function to find Maximum Subset Sum
def solve(array, n):
# Initialize DP array
dp = [0] * (1 << 10)
# Fill DP table using bottom-up approach
for i in range(n):
mask = get_binary(array[i])
for j in range((1 << 10) - 1, -1, -1):
if (mask | j) == j:
dp[j] = max(dp[j], dp[j ^ mask] + array[i])
# Find maximum sum from DP array
ans = 0
for i in range(1 << 10):
ans = max(ans, dp[i])
return ans
# Driver Code
array = [22, 132, 4, 45, 12, 223]
n = len(array)
print(solve(array, n))
using System;
class Program
{
static int[] dp = new int[1024];
// Function to create mask for every number
static int GetBinary(int u)
{
int ans = 0;
while (u != 0)
{
int rem = u % 10;
ans |= (1 << rem);
u /= 10;
}
return ans;
}
// Function to find Maximum Subset Sum
static int Solve(int[] array, int n)
{
// Initialize DP array
for (int i = 0; i < (1 << 10); i++)
{
dp[i] = 0;
}
// Fill DP table using bottom-up approach
for (int i = 0; i < n; i++)
{
int mask = GetBinary(array[i]);
for (int j = (1 << 10) - 1; j >= 0; j--)
{
if ((mask | j) == j)
{
dp[j] = Math.Max(dp[j], dp[j ^ mask] + array[i]);
}
}
}
// Find maximum sum from DP array
int ans = 0;
for (int i = 0; i < (1 << 10); i++)
{
ans = Math.Max(ans, dp[i]);
}
return ans;
}
// Driver Code
static void Main()
{
int[] array = { 22, 132, 4, 45, 12, 223 };
int n = array.Length;
Console.WriteLine(Solve(array, n));
}
}
// Function to create mask for every number
function getBinary(u) {
let ans = 0;
while (u) {
let rem = u % 10;
ans |= (1 << rem);
u = Math.floor(u / 10);
}
return ans;
}
// Function to find Maximum Subset Sum
function solve(array, n) {
// Initialize DP array
let dp = new Array(1 << 10).fill(0);
// Fill DP table using bottom-up approach
for (let i = 0; i < n; i++) {
let mask = getBinary(array[i]);
for (let j = (1 << 10) - 1; j >= 0; j--) {
if ((mask | j) == j) {
dp[j] = Math.max(dp[j], dp[j ^ mask] + array[i]);
}
}
}
// Find maximum sum from DP array
let ans = 0;
for (let i = 0; i < 1 << 10; i++) {
ans = Math.max(ans, dp[i]);
}
return ans;
}
// Driver Code
let array = [22, 132, 4, 45, 12, 223];
let n = array.length;
console.log(solve(array, n));
Output
268
Time Complexity : O(N*(2^10))
Auxiliary Space: O(2^10)
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