Given two arrays a[] and b[][2] with both having size N. N operations have to be performed, and initially counter will be 0. Additionally, there is a M bonus score for a hitting streak of b[i][0] and the bonus score received will be b[i][1], the task is to find the maximum score by performing either of the below operations optimally for all i from 1 to N.
- Operation 1: increases the counter’s value by 1 and receives a[i] score
- Operation 2: resets the counter’s value to 0, without receiving any score.
Examples:
Input: a[] = {2, 7, 1, 8, 2, 8}, b[][2] = {{2, 10}, {3, 1}, {5, 5}}
Output: 48
Explanation:
- For i = 1, Change the counter’s value from 0 to 1 and receive a 2 score.
- For i = 2, Change the counter’s value from 1 to 2 and receive a 7 score and get 10 scores as a streak bonus as well.
- For i = 3, Change the counter’s value from 2 to 0.
- For i = 4, Change the counter’s value from 0 to 1 and receive 8 scores.
- For i = 5, Change the counter’s value from 1 to 2 and receive 2 scores and get a 10 score as a streak bonus.
- For i = 6, Change the counter’s value from 2 to 3 and receive 8 scores. and get 1 score as a streak bonus.
Total score : 2 + 7 + 10 + 8 + 2 +10 + 8 +1 = 48.
Input: a[] = {1000000000, 1000000000, 1000000000}, b[][2] = {{1, 1000000000}, {3, 1000000000}}
Output: 5000000000
Naive approach: The problem can be solved based on The basis of the following idea:
Basic way to solve this problem is to generate all 2N combinations by recursive brute force.
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following idea:
Dynamic programming along with hashing can be used to solve to this problem.
- dp[i][j] represents maximum score till i’th operation with current counter j.
- recurrence relation: dp[i][j] = max(dp[i – 1][j + 1] + a[i] + hash[j + 1], dp[i – 1][0])
It can be observed that there are N * N states but the recursive function is called exponential times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done using recursive structure intact and just store the value in a HashMap and whenever the function is called, return the value store without computing .
Follow the steps below to solve the problem:
- Creating a hashmap array for mapping values of b[i][0] to b[i][1] named hash[].
- Create a recursive function that takes two parameters i representing i’th operation and j representing counter at i’th operation.
- Call the recursive function for both increasing the current counter by 1 and resetting the counter.
- Check the base case if all N operations are over then return 0.
- Create a 2d array of dp[5001][5001] initially filled with -1.
- If the answer for a particular state is computed then save it in dp[i][j].
- If the answer for a particular state is already computed then just return dp[i][j].
Below is the implementation of the above approach.
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std;
// To avoid integer overflow #define int long long // dp table initialized with - 1 int dp[5001][5001];
// Recursive function to count maximum score // by performing following operations int recur( int i, int j, int a[], int hash[], int N)
{ // Base case
if (i == N) {
return 0;
}
// If answer for current state is already
// calculated then just return dp[i][j]
if (dp[i][j] != -1)
return dp[i][j];
// Calling recursive function for
// performing operation 1
int ans = recur(i + 1, j + 1, a, hash, N) + a[i]
+ hash[j + 1];
// Calling recursive function for
// performing operation 2
ans = max(ans, recur(i + 1, 0, a, hash, N) + 0LL);
// Save and return dp value
return dp[i][j] = ans;
} // Function to count maximum score // by performing following operations void findMaximumScore( int a[], int b[][2], int N, int M)
{ // Filling dp table with -1
memset (dp, -1, sizeof (dp));
// Creating Hash table
int hash[N + 1] = { 0 };
// Mapping hash table with values
for ( int i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
cout << recur(0, 0, a, hash, N) << endl;
} // Driver Code int32_t main() { // Input 1
int a[] = { 2, 7, 1, 8, 2, 8 },
b[][2] = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
int N = sizeof (a) / sizeof (a[0]);
int M = sizeof (b) / sizeof (b[0]);
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
int a1[] = { 1000000000, 1000000000, 1000000000 },
b1[][2] = { { 1, 1000000000 }, { 3, 1000000000 } };
int N1 = sizeof (a1) / sizeof (a1[0]);
int M1 = sizeof (b1) / sizeof (b1[0]);
// Function Call
findMaximumScore(a1, b1, N1, M1);
return 0;
} |
// Java code for the above approach import java.util.Arrays;
class Main
{ // To avoid integer overflow
static long MAX = 1000000007 ;
// dp table initialized with - 1
static long [][] dp = new long [ 5001 ][ 5001 ];
// Recursive function to count maximum score
// by performing following operations
static long recur( int i, int j, int [] a, int [] hash, int N)
{
// Base case
if (i == N) {
return 0 ;
}
// If answer for current state is already
// calculated then just return dp[i][j]
if (dp[i][j] != - 1 ) {
return dp[i][j];
}
// Calling recursive function for
// performing operation 1
long ans = recur(i + 1 , j + 1 , a, hash, N) + a[i]
+ hash[j + 1 ];
// Calling recursive function for
// performing operation 2
ans = Math.max(ans, recur(i + 1 , 0 , a, hash, N) + 0 );
// Save and return dp value
return dp[i][j] = ans;
}
// Function to count maximum score
// by performing following operations
static void findMaximumScore( int [] a, int [][] b, int N, int M)
{
// Filling dp table with -1
for ( long [] row : dp) {
Arrays.fill(row, - 1 );
}
// Creating Hash table
int [] hash = new int [N + 1 ];
// Mapping hash table with values
for ( int i = 0 ; i < M; i++) {
hash[b[i][ 0 ]] += b[i][ 1 ];
}
System.out.println(recur( 0 , 0 , a, hash, N));
}
// Driver Code
public static void main(String[] args) {
// Input 1
int [] a = { 2 , 7 , 1 , 8 , 2 , 8 };
int [][] b = { { 2 , 10 }, { 3 , 1 }, { 5 , 5 } };
int N = a.length;
int M = b.length;
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
int [] a1 = { 1000000000 , 1000000000 , 1000000000 };
int [][] b1 = { { 1 , 1000000000 }, { 3 , 1000000000 } };
int N1 = a1.length;
int M1 = b1.length;
// Function Call
findMaximumScore(a1, b1, N1, M1);
}
} // This code is contributed by Potta Lokesh |
#Python code to implement the approach # dp table initialized with - 1 dp = [[ - 1 for i in range ( 5001 )] for j in range ( 5001 )]
# Recursive function to count maximum score # by performing following operations def recur(i,j,a, hash ,N):
# Base case
if (i = = N):
return 0
# If answer for current state is already
# calculated then just return dp[i][j]
if (dp[i][j]! = - 1 ):
return dp[i][j]
# Calling recursive function for
# performing operation 1
ans = recur(i + 1 ,j + 1 ,a, hash ,N) + a[i] + hash [j + 1 ]
# Calling recursive function for
# performing operation 2
ans = max (ans,recur(i + 1 , 0 ,a, hash ,N))
# Save and return dp value
dp[i][j] = ans
return dp[i][j]
# Function to count maximum score # by performing following operations def findMaximumScore(a,b,N,M):
# Filling dp table with -1
for i in range ( len (dp)):
for j in range ( len (dp[ 0 ])):
dp[i][j] = - 1
# Creating Hash table
hash = [ 0 ] * (N + 1 )
# Mapping hash table with values
for i in range (M):
hash [b[i][ 0 ]] + = b[i][ 1 ]
print (recur( 0 , 0 ,a, hash ,N))
# Driver Code # Input 1 a = [ 2 , 7 , 1 , 8 , 2 , 8 ]
b = [[ 2 , 10 ],[ 3 , 1 ],[ 5 , 5 ]]
N = len (a)
M = len (b)
# Function call findMaximumScore(a,b,N,M) # Input 2 a1 = [ 1000000000 , 1000000000 , 1000000000 ]
b1 = [[ 1 , 1000000000 ],[ 3 , 1000000000 ]]
N1 = len (a1)
M1 = len (b1)
# Function call findMaximumScore(a1,b1,N1,M1) # This code is contributed by Pushpesh Raj. |
using System;
using System.Linq;
class GFG
{ // To avoid integer overflow
static long MAX = 1000000007;
// dp table initialized with - 1
static long [, ] dp = new long [5001, 5001];
// Recursive function to count maximum score
// by performing following operations
static long recur( int i, int j, int [] a, int [] hash,
int N)
{
// Base case
if (i == N) {
return 0;
}
// If answer for current state is already
// calculated then just return dp[i][j]
if (dp[i, j] != -1) {
return dp[i, j];
}
// Calling recursive function for
// performing operation 1
long ans = recur(i + 1, j + 1, a, hash, N) + a[i]
+ hash[j + 1];
// Calling recursive function for
// performing operation 2
ans = Math.Max(ans,
recur(i + 1, 0, a, hash, N) + 0);
// Save and return dp value
return dp[i, j] = ans;
}
// Function to count maximum score
// by performing following operations
static void findMaximumScore( int [] a, int [, ] b, int N,
int M)
{
// Filling dp table with -1
for ( int i = 0; i < 5001; i++) {
for ( int j = 0; j < 5001; j++) {
dp[i, j] = -1;
}
}
// Creating Hash table
int [] hash = new int [N + 1];
// Mapping hash table with values
for ( int i = 0; i < M; i++) {
hash[b[i, 0]] += b[i, 1];
}
Console.WriteLine(recur(0, 0, a, hash, N));
}
// Driver Code
public static void Main( string [] args)
{
// Input 1
int [] a = { 2, 7, 1, 8, 2, 8 };
int [, ] b = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
int N = a.Length;
int M = b.GetLength(0);
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
int [] a1 = { 1000000000, 1000000000, 1000000000 };
int [, ] b1
= { { 1, 1000000000 }, { 3, 1000000000 } };
int N1 = a1.Length;
int M1 = b1.GetLength(0);
// Function Call
findMaximumScore(a1, b1, N1, M1);
}
} // This code is contributed by ik_9 |
// Javascript code to implement the approach // dp table initialized with - 1 let dp= new Array(5001);
for (let i=0; i<5001; i++)
dp[i]= new Array(5001);
// Recursive function to count maximum score // by performing following operations function recur(i, j, a, hash, N)
{ // Base case
if (i == N) {
return 0;
}
// If answer for current state is already
// calculated then just return dp[i][j]
if (dp[i][j] != -1)
return dp[i][j];
// Calling recursive function for
// performing operation 1
let ans = recur(i + 1, j + 1, a, hash, N) + a[i]
+ hash[j + 1];
// Calling recursive function for
// performing operation 2
ans = Math.max(ans, recur(i + 1, 0, a, hash, N) + 0);
// Save and return dp value
return dp[i][j] = ans;
} // Function to count maximum score // by performing following operations function findMaximumScore(a, b, N, M)
{ // Filling dp table with -1
for (let i=0; i<5001; i++)
for (let j=0; j<5001; j++)
dp[i][j]=-1;
// Creating Hash table
let hash = new Array(N+1).fill(0);
// Mapping hash table with values
for (let i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
document.write(recur(0, 0, a, hash, N))
} // Driver Code // Input 1
let a = [ 2, 7, 1, 8, 2, 8 ],
b = [[ 2, 10 ], [ 3, 1 ], [ 5, 5 ]];
let N = a.length;
let M = b.length;
// Function Call
findMaximumScore(a, b, N, M);
document.write( "<br>" );
// Input 2
let a1 = [ 1000000000, 1000000000, 1000000000 ],
b1 = [ [ 1, 1000000000 ], [ 3, 1000000000 ] ];
let N1 = a1.length;
let M1 = b1.length;
// Function Call
findMaximumScore(a1, b1, N1, M1);
|
48 5000000000
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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.
Implementation :
#include <bits/stdc++.h> using namespace std;
// To avoid integer overflow #define int long long // dp table int dp[5001][5001];
// Function to count maximum score // by performing following operations void findMaximumScore( int a[], int b[][2], int N, int M)
{ // Creating Hash table
int hash[N + 1] = { 0 };
// Mapping hash table with values
for ( int i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
// Base case initialization
for ( int i = 0; i <= N; i++) {
dp[N][i] = 0;
}
// Tabulation
for ( int i = N - 1; i >= 0; i--) {
for ( int j = 0; j <= i; j++) {
// Operation 1
int ans = dp[i + 1][j + 1] + a[i] + hash[j + 1];
// Operation 2
ans = max(ans, dp[i + 1][0]);
dp[i][j] = ans;
}
}
cout << dp[0][0] << endl;
} // Driver Code int32_t main() { // Input 1
int a[] = { 2, 7, 1, 8, 2, 8 },
b[][2] = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
int N = sizeof (a) / sizeof (a[0]);
int M = sizeof (b) / sizeof (b[0]);
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
int a1[] = { 1000000000, 1000000000, 1000000000 },
b1[][2] = { { 1, 1000000000 }, { 3, 1000000000 } };
int N1 = sizeof (a1) / sizeof (a1[0]);
int M1 = sizeof (b1) / sizeof (b1[0]);
// Function Call
findMaximumScore(a1, b1, N1, M1);
return 0;
} |
import java.util.Arrays;
public class MaximumScore {
// Function to count maximum score
static void findMaximumScore( int [] a, int [][] b, int N, int M) {
// To avoid integer overflow
long [][] dp = new long [N + 1 ][N + 1 ];
// Creating Hash table
long [] hash = new long [N + 1 ];
// Mapping hash table with values
for ( int i = 0 ; i < M; i++) {
hash[b[i][ 0 ]] += b[i][ 1 ];
}
// Base case initialization
for ( int i = 0 ; i <= N; i++) {
dp[N][i] = 0 ;
}
// Tabulation
for ( int i = N - 1 ; i >= 0 ; i--) {
for ( int j = 0 ; j <= i; j++) {
// Operation 1
long ans = dp[i + 1 ][j + 1 ] + a[i] + hash[j + 1 ];
// Operation 2
ans = Math.max(ans, dp[i + 1 ][ 0 ]);
dp[i][j] = ans;
}
}
System.out.println(dp[ 0 ][ 0 ]);
}
public static void main(String[] args) {
// Input 1
int [] a = { 2 , 7 , 1 , 8 , 2 , 8 };
int [][] b = { { 2 , 10 }, { 3 , 1 }, { 5 , 5 } };
int N = a.length;
int M = b.length;
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
int [] a1 = { 1000000000 , 1000000000 , 1000000000 };
int [][] b1 = { { 1 , 1000000000 }, { 3 , 1000000000 } };
int N1 = a1.length;
int M1 = b1.length;
// Function Call
findMaximumScore(a1, b1, N1, M1);
}
} |
import numpy as np
# Function to count maximum score # by performing following operations def findMaximumScore(a, b, N, M):
# Creating Hash table
hash = np.zeros(N + 1 )
# Mapping hash table with values
for i in range (M):
hash [b[i][ 0 ]] + = b[i][ 1 ]
# Base case initialization
dp = np.zeros((N + 1 , N + 1 ))
# Tabulation
for i in range (N - 1 , - 1 , - 1 ):
for j in range (i + 1 ):
ans = dp[i + 1 ][j + 1 ] + a[i] + hash [j + 1 ]
ans = max (ans, dp[i + 1 ][ 0 ])
dp[i][j] = ans
print (dp[ 0 ][ 0 ])
# Input 1 a = [ 2 , 7 , 1 , 8 , 2 , 8 ]
b = [[ 2 , 10 ], [ 3 , 1 ], [ 5 , 5 ]]
N = len (a)
M = len (b)
findMaximumScore(a, b, N, M) # Input 2 a1 = [ 1000000000 , 1000000000 , 1000000000 ]
b1 = [[ 1 , 1000000000 ], [ 3 , 1000000000 ]]
N1 = len (a1)
M1 = len (b1)
findMaximumScore(a1, b1, N1, M1) |
using System;
public class GFG
{ public static void FindMaximumScore( int [] a, int [][] b, int N, int M)
{
// Creating Hash table
long [] hash = new long [N + 1];
// Mapping hash table with values
for ( int i = 0; i < M; i++)
{
hash[b[i][0]] += b[i][1];
}
// dp table
long [,] dp = new long [5001, 5001];
// Base case initialization
for ( int i = 0; i <= N; i++)
{
dp[N, i] = 0;
}
// Tabulation
for ( int i = N - 1; i >= 0; i--)
{
for ( int j = 0; j <= i; j++)
{
// Operation 1
long ans = dp[i + 1, j + 1] + a[i] + hash[j + 1];
// Operation 2
ans = Math.Max(ans, dp[i + 1, 0]);
dp[i, j] = ans;
}
}
Console.WriteLine(dp[0, 0]);
}
public static void Main()
{
// Input 1
int [] a = { 2, 7, 1, 8, 2, 8 };
int [][] b = { new int [] { 2, 10 }, new int [] { 3, 1 }, new int [] { 5, 5 } };
int N = a.Length;
int M = b.Length;
// Function Call
FindMaximumScore(a, b, N, M);
// Input 2
int [] a1 = { 1000000000, 1000000000, 1000000000 };
int [][] b1 = { new int [] { 1, 1000000000 }, new int [] { 3, 1000000000 } };
int N1 = a1.Length;
int M1 = b1.Length;
// Function Call
FindMaximumScore(a1, b1, N1, M1);
}
} |
// Function to count maximum score // by performing following operations function findMaximumScore(a, b, N, M) {
// Creating Hash table
var hash = new Array(N + 1).fill(0);
// Mapping hash table with values
for ( var i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
// Base case initialization
var dp = new Array(N + 1).fill(0).map(() => new Array(N + 1).fill(0));
// Tabulation
for ( var i = N - 1; i >= 0; i--) {
for ( var j = 0; j <= i; j++) {
var ans = dp[i + 1][j + 1] + a[i] + hash[j + 1];
ans = Math.max(ans, dp[i + 1][0]);
dp[i][j] = ans;
}
}
console.log(dp[0][0]);
} // Input 1 var a = [2, 7, 1, 8, 2, 8];
var b = [[2, 10], [3, 1], [5, 5]];
var N = a.length;
var M = b.length;
findMaximumScore(a, b, N, M); // Input 2 var a1 = [1000000000, 1000000000, 1000000000];
var b1 = [[1, 1000000000], [3, 1000000000]];
var N1 = a1.length;
var M1 = b1.length;
findMaximumScore(a1, b1, N1, M1); |
48 5000000000
Time Complexity: O(N^2)
Auxiliary Space: O(N^2)
Efficient approach : Space optimization
In previous approach the dp[i][j] is depend upon the current and previous row of 2D matrix. So to optimize space we use two vectors curr and next that keep track of current and next row of DP.
Implementation :
// C++ code for the above approach #include <bits/stdc++.h> using namespace std;
// To avoid integer overflow #define int long long // Function to count maximum score // by performing following operations void findMaximumScore( int a[], int b[][2], int N, int M)
{ // Creating Hash table
int hash[N + 1] = { 0 };
// to store the values of next row of DP
vector< int > next(N + 1, 0);
// Mapping hash table with values
for ( int i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
// iterating over DP to get the current
// value from previous computations
for ( int i = N - 1; i >= 0; i--) {
vector< int > curr(N + 1);
for ( int j = 0; j <= i; j++) {
// Operation 1
int ans = next[j + 1] + a[i] + hash[j + 1];
// Operation 2
ans = max(ans, next[0]);
curr[j] = ans;
}
// assigning values to iterate further
next = curr;
}
// print final answer
cout << next[0] << endl;
} // Driver Code int32_t main() { // Input 1
int a[] = { 2, 7, 1, 8, 2, 8 },
b[][2] = { { 2, 10 }, { 3, 1 }, { 5, 5 } };
int N = sizeof (a) / sizeof (a[0]);
int M = sizeof (b) / sizeof (b[0]);
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
int a1[] = { 1000000000, 1000000000, 1000000000 },
b1[][2] = { { 1, 1000000000 }, { 3, 1000000000 } };
int N1 = sizeof (a1) / sizeof (a1[0]);
int M1 = sizeof (b1) / sizeof (b1[0]);
// Function Call
findMaximumScore(a1, b1, N1, M1);
return 0;
} |
import java.util.Arrays;
public class Main {
// Function to count maximum score
// by performing following operations
static void findMaximumScore( long a[], long b[][], int N, int M) {
// Creating Hash table
long hash[] = new long [N + 1 ];
// to store the values of the next row of DP
long [] next = new long [N + 1 ];
// Mapping hash table with values
for ( int i = 0 ; i < M; i++) {
hash[( int )b[i][ 0 ]] += b[i][ 1 ];
}
// iterating over DP to get the current
// value from previous computations
for ( int i = N - 1 ; i >= 0 ; i--) {
long [] curr = new long [N + 1 ];
for ( int j = 0 ; j <= i; j++) {
// Operation 1
long ans = next[j + 1 ] + a[i] + hash[j + 1 ];
// Operation 2
ans = Math.max(ans, next[ 0 ]);
curr[j] = ans;
}
// assigning values to iterate further
next = Arrays.copyOf(curr, curr.length);
}
// print final answer
System.out.println(next[ 0 ]);
}
// Driver Code
public static void main(String[] args) {
// Input 1
long a[] = { 2 , 7 , 1 , 8 , 2 , 8 };
long b[][] = { { 2 , 10 }, { 3 , 1 }, { 5 , 5 } };
int N = a.length;
int M = b.length;
// Function Call
findMaximumScore(a, b, N, M);
// Input 2
long a1[] = { 1000000000 , 1000000000 , 1000000000 };
long b1[][] = { { 1 , 1000000000 }, { 3 , 1000000000 } };
int N1 = a1.length;
int M1 = b1.length;
// Function Call
findMaximumScore(a1, b1, N1, M1);
}
} |
# Python code for the above approach # Function to count maximum score # by performing following operations def findMaximumScore(a, b, N, M):
# Creating Hash table
hash = [ 0 ] * (N + 1 )
# to store the values of the next row of DP
next_row = [ 0 ] * (N + 1 )
# Mapping hash table with values
for i in range (M):
hash [b[i][ 0 ]] + = b[i][ 1 ]
# iterating over DP to get the current
# value from previous computations
for i in range (N - 1 , - 1 , - 1 ):
curr = [ 0 ] * (N + 1 )
for j in range (i + 1 ):
# Operation 1
ans = next_row[j + 1 ] + a[i] + hash [j + 1 ]
# Operation 2
ans = max (ans, next_row[ 0 ])
curr[j] = ans
# assigning values to iterate further
next_row = curr
# print final answer
print (next_row[ 0 ])
# Driver Code # Input 1 a = [ 2 , 7 , 1 , 8 , 2 , 8 ]
b = [[ 2 , 10 ], [ 3 , 1 ], [ 5 , 5 ]]
N = len (a)
M = len (b)
# Function Call findMaximumScore(a, b, N, M) # Input 2 a1 = [ 1000000000 , 1000000000 , 1000000000 ]
b1 = [[ 1 , 1000000000 ], [ 3 , 1000000000 ]]
N1 = len (a1)
M1 = len (b1)
# Function Call findMaximumScore(a1, b1, N1, M1) # This code is contributed by Vaibhav Nandan |
using System;
public class GFG {
// Function to count maximum score
// by performing the following operations
public static void
FindMaximumScore( long [] a, long [][] b, int N, int M)
{
// Creating a hash table
long [] hash = new long [N + 1];
// To store the values of the next row of DP
long [] next = new long [N + 1];
// Mapping hash table with values
for ( int i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
// Iterating over DP to get the current
// value from previous computations
for ( int i = N - 1; i >= 0; i--) {
long [] curr = new long [N + 1];
for ( int j = 0; j <= i; j++) {
// Operation 1
long ans = next[j + 1] + a[i] + hash[j + 1];
// Operation 2
ans = Math.Max(ans, next[0]);
curr[j] = ans;
}
// Assigning values to iterate further
next = curr;
}
// Print the final answer
Console.WriteLine(next[0]);
}
// Driver Code
public static void Main()
{
// Input 1
long [] a = { 2, 7, 1, 8, 2, 8 };
long [][] b
= { new long [] { 2, 10 }, new long [] { 3, 1 },
new long [] { 5, 5 } };
int N = a.Length;
int M = b.Length;
// Function Call
FindMaximumScore(a, b, N, M);
// Input 2
long [] a1 = { 1000000000, 1000000000, 1000000000 };
long [][] b1 = { new long [] { 1, 1000000000 },
new long [] { 3, 1000000000 } };
int N1 = a1.Length;
int M1 = b1.Length;
// Function Call
FindMaximumScore(a1, b1, N1, M1);
}
} |
// JS code for the above approach // Function to count maximum score // by performing following operations function findMaximumScore(a, b, N, M)
{ // Creating Hash table
let hash = new Array(N + 1).fill(0);
// to store the values of next row of DP
let next = new Array(N + 1).fill(0);
// Mapping hash table with values
for (let i = 0; i < M; i++) {
hash[b[i][0]] += b[i][1];
}
// iterating over DP to get the current
// value from previous computations
for (let i = N - 1; i >= 0; i--) {
let curr = new Array(N + 1);
for (let j = 0; j <= i; j++) {
// Operation 1
let ans = next[j + 1] + a[i] + hash[j + 1];
// Operation 2
ans = Math.max(ans, next[0]);
curr[j] = ans;
}
// assigning values to iterate further
next = curr;
}
// print final answer
console.log(next[0]);
} // Driver Code // Input 1 let a = [ 2, 7, 1, 8, 2, 8 ]; let b = [ [ 2, 10 ], [ 3, 1 ], [ 5, 5 ] ]; let N = a.length; let M = b.length; // Function Call findMaximumScore(a, b, N, M); // Input 2 let a1 = [ 1000000000, 1000000000, 1000000000 ]; let b1 = [ [ 1, 1000000000 ], [ 3, 1000000000 ] ]; let N1 = a1.length; let M1 = b1.length; // Function Call findMaximumScore(a1, b1, N1, M1); |
48 5000000000
Time Complexity: O(N^2)
Auxiliary Space: O(N)
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