Given two 2D binary arrays, a[][] and b[][] both of size M*N, the task is to pair each row in the array a[][] with any row in the array b[][] such that the total score can be maximized and the score for each pair is calculated as the total indexes at which values of both rows are identical.
Note: Each row of the array b[][] can only be paired with a single row of vector a[][].
Examples:
Input: a[][] = {{1, 1, 0}, {1, 0, 1}, {0, 0, 1}}, b[][] = {{1, 0, 0}, {0, 0, 1}, {1, 1, 0}}
Output: 8
Explanation:
Consider the pairing of rows in the following order, to maximize the total score obtained:
- Row 0 of a[][] paired with row 2 of b[][] has the score of 3.
- Row 1 of a[][] paired with row 0 of b[][] with score of 2.
- Row 2 of a[][] paired with row 1 of b[][] with score of 3.
Therefore, the sum of scores obtained is 3 + 2 + 3 = 8.
Input: a[][] = {{0, 0}, {0, 0}, {0, 0}}, b[][] = {{1, 1}, {1, 1}, {1, 1}}
Output: 0
Naive Approach: The simplest approach to solve the given problem is to generate all possible permutations of the rows of the arrays a[][] and for each permutation of the array a[][], find the sum of scores of each corresponding pair, and if it’s greater than the current answer, update the answer to the value of the current sum of scores. After checking for all the pairs, print the maximum score obtained.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum score void maxScoreSum(vector<vector< int > >& a,
vector<vector< int > >& b)
{ // Stores the maximum sum of scores
int maxSum = 0;
vector< int > pos;
for ( int i = 0; i < a.size(); i++) {
pos.push_back(i);
}
// For each permutation of pos vector
// calculate the score
do {
int curSum = 0;
for ( int i = 0; i < a.size(); i++) {
for ( int j = 0;
j < a[pos[i]].size(); j++) {
// If values at current indexes
// are same then increment the
// current score
curSum += (a[pos[i]][j] == b[i][j]);
maxSum = max(maxSum, curSum);
}
}
} while (next_permutation(pos.begin(), pos.end()));
// Print the maximum score
cout << maxSum;
} // Driver Code int main()
{ int N = 3, M = 3;
vector<vector< int > > a
= { { 1, 1, 0 }, { 1, 0, 1 }, { 0, 0, 1 } };
vector<vector< int > > b
= { { 1, 0, 0 }, { 0, 0, 1 }, { 1, 1, 0 } };
maxScoreSum(a, b);
return 0;
} |
// Java code for the above approach import java.util.*;
class GFG {
public static List<List<Integer> > perms
= new ArrayList<>();
public static void
generatePermutations( int mask, int ind, int N,
List<Integer> current)
{
if (ind == N) {
List<Integer> temp = new ArrayList<>(current);
perms.add(temp);
return ;
}
for ( int i = 0 ; i < N; i++) {
if (((mask >> i) & 1 ) == 0 ) {
current.add(i);
generatePermutations(mask | ( 1 << i),
ind + 1 , N, current);
current.remove(current.size() - 1 );
}
}
}
// Function to find the maximum score
public static void maxScoreSum(List<List<Integer> > a,
List<List<Integer> > b)
{
// Stores the maximum sum of scores
int maxSum = 0 ;
generatePermutations( 0 , 0 , a.size(),
new ArrayList<Integer>());
// For each permutation of {0, a.size() - 1} vector
// calculate the score
for ( int ind = 0 ; ind < perms.size(); ind++) {
List<Integer> pos = perms.get(ind);
int curSum = 0 ;
for ( int i = 0 ; i < a.size(); i++) {
for ( int j = 0 ;
j < a.get(pos.get(i)).size(); j++) {
// If values at current indexes
// are same then increment the
// current score
curSum += (a.get(pos.get(i)).get(j)
== b.get(i).get(j)
? 1
: 0 );
maxSum = Math.max(maxSum, curSum);
}
}
}
// Print the maximum score
System.out.println(maxSum);
}
// Driver Code
public static void main(String[] args)
{
List<List<Integer> > a = new ArrayList<>();
a.add(Arrays.asList( 1 , 1 , 0 ));
a.add(Arrays.asList( 1 , 0 , 1 ));
a.add(Arrays.asList( 0 , 0 , 1 ));
List<List<Integer> > b = new ArrayList<>();
b.add(Arrays.asList( 1 , 0 , 0 ));
b.add(Arrays.asList( 0 , 0 , 1 ));
b.add(Arrays.asList( 1 , 1 , 0 ));
maxScoreSum(a, b);
}
} // This code is contributed by ik_9 |
# Python Program to implement # the above approach def next_permutation(array):
i = len (array) - 1
while (i > 0 and array[i - 1 ] > = array[i]):
i - = 1
if (i < = 0 ):
return False
j = len (array) - 1
while (array[j] < = array[i - 1 ]):
j - = 1
temp = array[i - 1 ]
array[i - 1 ] = array[j]
array[j] = temp
j = len (array) - 1
while (i < j):
temp = array[i];
array[i] = array[j];
array[j] = temp;
i + = 1
j - = 1
return array
# Function to find the maximum score def maxScoreSum(a, b):
# Stores the maximum sum of scores
maxSum = 0
pos = []
for i in range ( len (a)):
pos.append(i)
# For each permutation of pos vector
# calculate the score
while ( True ):
curSum = 0
for i in range ( len (a)):
for j in range ( len (a[pos[i]])):
# If values at current indexes
# are same then increment the
# current score
curSum + = (a[pos[i]][j] = = b[i][j])
maxSum = max (maxSum, curSum)
if (next_permutation(pos) = = False ):
break
# Print the maximum score
print (maxSum)
# Driver Code N, M = 3 , 3
a = [[ 1 , 1 , 0 ], [ 1 , 0 , 1 ], [ 0 , 0 , 1 ]]
b = [[ 1 , 0 , 0 ], [ 0 , 0 , 1 ], [ 1 , 1 , 0 ]]
maxScoreSum(a, b) # This code is contributed by shinjanpatra |
// C# program to implement above approach using System;
using System.Collections.Generic;
class GFG
{ public static List<List< int >> perms = new List<List< int >>();
public static void generate_permuatation( int mask, int ind, int N, List< int > current){
if (ind == N){
List< int > temp = new List< int >(current);
perms.Add(temp);
return ;
}
for ( int i = 0 ; i < N ; i++){
if (((mask >> i) & 1) == 0){
current.Add(i);
generate_permuatation(mask | (1 << i), ind + 1, N, current);
current.RemoveAt(current.Count-1);
}
}
}
// Function to find the maximum score
public static void maxScoreSum(List<List< int >> a, List<List< int >> b)
{
// Stores the maximum sum of scores
int maxSum = 0;
generate_permuatation(0, 0, a.Count, new List< int >());
// For each permutation of {0, a.Count - 1} vector
// calculate the score
for ( int ind = 0 ; ind < perms.Count ; ind++){
List< int > pos = perms[ind];
int curSum = 0;
for ( int i = 0; i < a.Count ; i++) {
for ( int j = 0 ; j < a[pos[i]].Count ; j++) {
// If values at current indexes
// are same then increment the
// current score
curSum += (a[pos[i]][j] == b[i][j] ? 1 : 0);
maxSum = Math.Max(maxSum, curSum);
}
}
}
// Print the maximum score
Console.Write(maxSum);
}
// Driver Code
public static void Main( string [] args){
// int N = 3;
// int M = 3;
List<List< int >> a = new List<List< int >>{
new List< int >{ 1, 1, 0 },
new List< int >{ 1, 0, 1 },
new List< int >{ 0, 0, 1 }
};
List<List< int >> b = new List<List< int >>{
new List< int >{ 1, 0, 0 },
new List< int >{ 0, 0, 1 },
new List< int >{ 1, 1, 0 }
};
maxScoreSum(a, b);
}
} // This code is contributed by subhamgoyal2014. |
<script> // JavaScript Program to implement
// the above approach
function next_permutation(array) {
var i = array.length - 1;
while (i > 0 && array[i - 1] >= array[i]) {
i--;
}
if (i <= 0) {
return false ;
}
var j = array.length - 1;
while (array[j] <= array[i - 1]) {
j--;
}
var temp = array[i - 1];
array[i - 1] = array[j];
array[j] = temp;
j = array.length - 1;
while (i < j) {
temp = array[i];
array[i] = array[j];
array[j] = temp;
i++;
j--;
}
return array;
}
// Function to find the maximum score
function maxScoreSum(a, b) {
// Stores the maximum sum of scores
let maxSum = 0;
let pos = [];
for (let i = 0; i < a.length; i++) {
pos.push(i);
}
// For each permutation of pos vector
// calculate the score
do {
let curSum = 0;
for (let i = 0; i < a.length; i++) {
for (let j = 0;
j < a[pos[i]].length; j++) {
// If values at current indexes
// are same then increment the
// current score
curSum += (a[pos[i]][j] == b[i][j]);
maxSum = Math.max(maxSum, curSum);
}
}
} while (next_permutation(pos));
// Print the maximum score
document.write(maxSum);
}
// Driver Code
let N = 3, M = 3;
let a
= [[1, 1, 0], [1, 0, 1], [0, 0, 1]];
let b
= [[1, 0, 0], [0, 0, 1], [1, 1, 0]];
maxScoreSum(a, b);
// This code is contributed by Potta Lokesh
</script>
|
8
Time Complexity: O(N*M*M!), where M! are the number of permutations and N*M for calculating the score of each pair.
Auxiliary Space: O(M)
Efficient Approach: The above approach can also be optimized using the concept of Bitmasking, The idea is for each row in vector a[][], try all rows in vector b[][] that haven’t been chosen before. Use a bitmask to represent already chosen rows of vector b[][]. To avoid re-computing the same subproblem, memoize the result for each bitmask. Follow the steps below to solve the problem:
- Initialize the variables row as 0, mask as (2M – 1).
- Initialize the vector dp[] of size mask + 1 with values -1.
- If row is greater than equal to a.size() then return 0 and if dp[mask] is not equal to -1 then return dp[mask].
- Initialize the variable ans as 0 to store the answer.
-
Iterate over the range [0, a.size()) using the variable i and perform the following tasks:
- If the bitwise AND of mask and 2i is true then initialize the variable newMask as mask^(1<<i) and curSum as 0.
- Iterate over the range [0, a[i].size()) using the variable j and if a[row][j] equals b[i][j] then increase the value of curSum by 1.
- Set the value of ans as the maximum of ans or curSum + maxScoreSum(a, b, row+1, newmask, dp) recursively.
- After performing the above steps, set the value of dp[mask] as ans and return the value of ans as the answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum defined // score int maxScoreSum(vector<vector< int > >& a,
vector<vector< int > >& b,
int row, int mask,
vector< int >& dp)
{ // If all students are assigned
if (row >= a.size()) {
return 0;
}
if (dp[mask] != -1) {
return dp[mask];
}
int ans = 0;
for ( int i = 0; i < a.size(); i++) {
// Check if row is not paired yet
if (mask & (1 << i)) {
int newMask = mask ^ (1 << i);
int curSum = 0;
// Check for all indexes
for ( int j = 0; j < a[i].size(); j++) {
// If values at current indexes
// are same increase curSum
if (a[row][j] == b[i][j]) {
curSum++;
}
}
// Further recursive call
ans = max(
ans, curSum
+ maxScoreSum(
a, b, row + 1,
newMask, dp));
}
}
// Store the ans for current
// mask and return
return dp[mask] = ans;
} // Utility function to find the maximum // defined score int maxScoreSumUtil(vector<vector< int > >& a,
vector<vector< int > >& b,
int N, int M)
{ int row = 0;
// Create a mask with all set bits
// 1 -> row is not paired yet
// 0 -> row is already paired
int mask = pow (2, M) - 1;
// Initialise dp array with -1
vector< int > dp(mask + 1, -1);
return maxScoreSum(a, b, row, mask, dp);
} // Driver Code int main()
{ int N = 3, M = 3;
vector<vector< int > > a
= { { 1, 1, 0 }, { 1, 0, 1 }, { 0, 0, 1 } };
vector<vector< int > > b
= { { 1, 0, 0 }, { 0, 0, 1 }, { 1, 1, 0 } };
cout << maxScoreSumUtil(a, b, N, M);
return 0;
} |
import java.util.ArrayList;
import java.util.List;
class GFG {
// Function to find the maximum defined
// score
public static int maxScoreSum(List<List<Integer> > a,
List<List<Integer> > b,
int row, int mask,
int [] dp)
{
// If all students are assigned
if (row >= a.size()) {
return 0 ;
}
if (dp[mask] != - 1 ) {
return dp[mask];
}
int ans = 0 ;
for ( int i = 0 ; i < a.size(); i++) {
// Check if row is not paired yet
if ((mask & ( 1 << i)) > 0 ) {
int newMask = mask ^ ( 1 << i);
int curSum = 0 ;
// Check for all indexes
for ( int j = 0 ; j < a.get(i).size(); j++) {
// If values at current indexes
// are same increase curSum
if (a.get(row).get(j)
== b.get(i).get(j)) {
curSum++;
}
}
// Further recursive call
ans = Math.max(
ans, curSum
+ maxScoreSum(a, b, row + 1 ,
newMask, dp));
}
}
// Store the ans for current
// mask and return
return dp[mask] = ans;
}
// Utility function to find the maximum
// defined score
public static int
maxScoreSumUtil(List<List<Integer> > a,
List<List<Integer> > b, int N, int M)
{
int row = 0 ;
// Create a mask with all set bits
// 1 -> row is not paired yet
// 0 -> row is already paired
int mask = ( int )Math.pow( 2 , M) - 1 ;
// Initialise dp array with -1
int [] dp = new int [ 1 << M];
for ( int i = 0 ; i <= mask; i++) {
dp[i] = - 1 ;
}
return maxScoreSum(a, b, row, mask, dp);
}
// Driver Code
public static void main(String[] args)
{
int N = 3 ;
int M = 3 ;
List<List<Integer> > a
= new ArrayList<List<Integer> >() {
{
add( new ArrayList<Integer>() {
{
add( 1 );
add( 1 );
add( 0 );
}
});
add( new ArrayList<Integer>() {
{
add( 1 );
add( 0 );
add( 1 );
}
});
add( new ArrayList<Integer>() {
{
add( 0 );
add( 0 );
add( 1 );
}
});
}
};
List<List<Integer> > b
= new ArrayList<List<Integer> >() {
{
add( new ArrayList<Integer>() {
{
add( 1 );
add( 0 );
add( 0 );
}
});
add( new ArrayList<Integer>() {
{
add( 0 );
add( 0 );
add( 1 );
}
});
add( new ArrayList<Integer>() {
{
add( 1 );
add( 1 );
add( 0 );
}
});
}
};
System.out.println(maxScoreSumUtil(a, b, N, M));
}
} |
# Python 3 program for the above approach # Function to find the maximum defined # score def maxScoreSum(a, b, row, mask, dp):
# If all students are assigned
if (row > = len (a)):
return 0
if (dp[mask] ! = - 1 ):
return dp[mask]
ans = 0
for i in range ( len (a)):
# Check if row is not paired yet
if (mask & ( 1 << i)):
newMask = mask ^ ( 1 << i)
curSum = 0
# Check for all indexes
for j in range ( len (a[i])):
# If values at current indexes
# are same increase curSum
if (a[row][j] = = b[i][j]):
curSum + = 1
# Further recursive call
ans = max (
ans, curSum
+ maxScoreSum(
a, b, row + 1 ,
newMask, dp))
# Store the ans for current
# mask and return
dp[mask] = ans
return dp[mask]
# Utility function to find the maximum # defined score def maxScoreSumUtil(a,
b,
N, M):
row = 0
# Create a mask with all set bits
# 1 -> row is not paired yet
# 0 -> row is already paired
mask = pow ( 2 , M) - 1
# Initialise dp array with -1
dp = [ - 1 ] * (mask + 1 )
return maxScoreSum(a, b, row, mask, dp)
# Driver Code if __name__ = = "__main__" :
N = 3
M = 3
a = [[ 1 , 1 , 0 ], [ 1 , 0 , 1 ], [ 0 , 0 , 1 ]]
b = [[ 1 , 0 , 0 ], [ 0 , 0 , 1 ], [ 1 , 1 , 0 ]]
print (maxScoreSumUtil(a, b, N, M))
# This code is contributed by ukasp.
|
// C# program to implement above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to find the maximum defined
// score
public static int maxScoreSum(List<List< int >> a, List<List< int >> b, int row, int mask, List< int > dp)
{
// If all students are assigned
if (row >= a.Count){
return 0;
}
if (dp[mask] != -1){
return dp[mask];
}
int ans = 0;
for ( int i = 0 ; i < a.Count ; i++) {
// Check if row is not paired yet
if ((mask & (1 << i)) > 0){
int newMask = mask ^ (1 << i);
int curSum = 0;
// Check for all indexes
for ( int j = 0; j < a[i].Count ; j++) {
// If values at current indexes
// are same increase curSum
if (a[row][j] == b[i][j]) {
curSum++;
}
}
// Further recursive call
ans = Math.Max(ans, curSum + maxScoreSum(a, b, row + 1, newMask, dp));
}
}
// Store the ans for current
// mask and return
return dp[mask] = ans;
}
// Utility function to find the maximum
// defined score
public static int maxScoreSumUtil(List<List< int >> a, List<List< int >> b, int N, int M)
{
int row = 0;
// Create a mask with all set bits
// 1 -> row is not paired yet
// 0 -> row is already paired
int mask = ( int )Math.Pow(2, M) - 1;
// Initialise dp array with -1
List< int > dp = new List< int >();
for ( int i = 0 ; i <= mask ; i++){
dp.Add(-1);
}
return maxScoreSum(a, b, row, mask, dp);
}
// Driver Code
public static void Main( string [] args){
int N = 3;
int M = 3;
List<List< int >> a = new List<List< int >>{
new List< int >{ 1, 1, 0 },
new List< int >{ 1, 0, 1 },
new List< int >{ 0, 0, 1 }
};
List<List< int >> b = new List<List< int >>{
new List< int >{ 1, 0, 0 },
new List< int >{ 0, 0, 1 },
new List< int >{ 1, 1, 0 }
};
Console.Write(maxScoreSumUtil(a, b, N, M));
}
} // This code is contributed by subhamgoyal2014. |
<script> // JavaScript code for the approach // Function to find the maximum defined // score function maxScoreSum(a, b, row, mask, dp){
// If all students are assigned
if (row >= a.length)
return 0
if (dp[mask] != -1)
return dp[mask]
let ans = 0
for (let i=0;i<a.length;i++){
// Check if row is not paired yet
if (mask & (1 << i)){
newMask = mask ^ (1 << i)
curSum = 0
// Check for all indexes
for (let j=0;j<a[i].length;j++){
// If values at current indexes
// are same increase curSum
if (a[row][j] == b[i][j])
curSum += 1
}
// Further recursive call
ans = Math.max(ans, curSum + maxScoreSum(a, b, row + 1,newMask, dp))
}
}
// Store the ans for current
// mask and return
dp[mask] = ans
return dp[mask]
} // Utility function to find the maximum // defined score function maxScoreSumUtil(a,b,N, M){
let row = 0
// Create a mask with all set bits
// 1 -> row is not paired yet
// 0 -> row is already paired
let mask = Math.pow(2, M) - 1
// Initialise dp array with -1
let dp = new Array(mask + 1).fill(-1)
return maxScoreSum(a, b, row, mask, dp)
} // Driver Code let N = 3 let M = 3 let a = [[1, 1, 0], [1, 0, 1], [0, 0, 1]] let b = [[1, 0, 0], [0, 0, 1], [1, 1, 0]] document.write(maxScoreSumUtil(a, b, N, M), "</br>" )
// This code is contributed by shinjanpatra </script> |
8
Time Complexity: O(2M*M*N)
Auxiliary Space: O(2M)