Given integers N and K representing the number of batches and number of students in each batch respectively, and a 2D array ratings[][] of size N * K where each row has ratings for every K students and Q queries of type {a, b}. The task for each query is to count the number of groups of N students possible by selecting a student from each batch, such that sum of ratings in each group lies in the range [a, b] inclusively.
Examples:
Input: N = 2, K = 3, ratings[][]= { {1, 2, 3}, {4, 5, 6} }, Q = 2, Queries[][]={ {6, 6}, {1, 6} }
Output: 2 3
Explanation:
All possible groups of size N(=2) are:
1 + 4 = 5
1 + 5 = 6
1 + 6 = 7
2 + 4 = 6
2 + 5 = 7
2 + 6 = 8
3 + 4 = 7
3 + 5 = 8
3 + 6 = 9
Query 1: The groups whose sum in range of (6, 6) inclusive are (1 + 5), (2 + 4) is 2.
Query 2: The groups whose sum in range of (1, 6) inclusive are (1 + 4), (1 + 5), (2 + 4) is 3.Input: N = 3, K = 3, ratings[][]={ {1, 2, 3}, {4, 5, 6}, {7, 8, 9} }, Q = 2, Queries[][]={ {10, 13}, {1, 7} }
Output: 4 0
Explanation:
Out of All possible groups of size N(=3):
Query 1: The groups whose sum in range (10, 13) inclusive is (1 + 4 + 7), (1 + 5 + 7), (2 + 4 + 7), (1 + 4 + 8) is 4.
Query 2: The groups whose sum in range of (1, 7) inclusive is 0.
Naive Approach: The simplest approach is to use recursion to generate all possible groups of size N. At each step of recursion calculate the sum that lies within the range of a query and find the number of groups that lie in the given range.
Time Complexity: O(NK)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by using Dynamic Programming. The idea is that if the number of times the sum S appears in the ith row is known, then the Prefix Sum Technique can be used to answer all the queries in constant time. So in this way, the Overlapping Subproblems are calculated only once reducing the exponential time into polynomial time. Below are the steps:
- Initialize auxiliary array dp[][] where dp[i][sum] is the number of times a sum is present in the ith row.
- For each batch, i iterate through all possible sum S, and for each j students, if sum S is greater than the current rating ratings[i][j] then update the current dp state dp[i][S] as:
dp[i][S] = dp[i][S] + dp[i – 1][sum – rating[i][j]]
- After the above steps, dp[N – 1][sum], which is the number of times the sum appears in the (N – 1)th row.
- To answer each queries efficiently find the prefix sum of the last row i.e., dp[N – 1][S] for all values of sum S.
- Now, the number of ways to form groups in the given range [a, b] is dp[N – 1][b] – dp[N – 1][a – 1].
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Given n batches and k students #define n 2 #define k 3 // Function to count number of // ways to get given sum groups void numWays( int ratings[n][k], int queries[][2])
{ // Initialise dp array
int dp[n][10000 + 2];
// Mark all 1st row values as 1
// since the mat[0][i] is all
// possible sums in first row
for ( int i = 0; i < k; i++)
dp[0][ratings[0][i]] += 1;
// Fix the ith row
for ( int i = 1; i < n; i++) {
// Fix the sum
for ( int sum = 0; sum <= 10000; sum++)
{
// Iterate through all
// values of ith row
for ( int j = 0; j < k; j++)
{
// If sum can be obtained
if (sum >= ratings[i][j])
dp[i][sum]
+= dp[i - 1][sum - ratings[i][j]];
}
}
}
// Find the prefix sum of last row
for ( int sum = 1; sum <= 10000; sum++)
{
dp[n - 1][sum] += dp[n - 1][sum - 1];
}
// Traverse each query
for ( int q = 0; q < 2; q++)
{
int a = queries[q][0];
int b = queries[q][1];
// No of ways to form groups
cout << dp[n - 1][b] - dp[n - 1][a - 1] << " " ;
}
} // Driver Code int main()
{ // Given ratings
int ratings[n][k] = { { 1, 2, 3 }, { 4, 5, 6 } };
// Given Queries
int queries[][2] = { { 6, 6 }, { 1, 6 } };
// Function Call
numWays(ratings, queries);
return 0;
} |
// Java program for the above approach import java.util.*;
public class Main {
// Function to count number of
// ways to get given sum groups
public static void
numWays( int [][] ratings,
int queries[][],
int n, int k)
{
// Initialise dp array
int dp[][] = new int [n][ 10000 + 2 ];
// Mark all 1st row values as 1
// since the mat[0][i] is all
// possible sums in first row
for ( int i = 0 ; i < k; i++)
dp[ 0 ][ratings[ 0 ][i]] += 1 ;
// Fix the ith row
for ( int i = 1 ; i < n; i++)
{
// Fix the sum
for ( int sum = 0 ; sum <= 10000 ; sum++)
{
// Iterate through all
// values of ith row
for ( int j = 0 ; j < k; j++)
{
// If sum can be obtained
if (sum >= ratings[i][j])
dp[i][sum]
+= dp[i - 1 ]
[sum - ratings[i][j]];
}
}
}
// Find the prefix sum of last row
for ( int sum = 1 ; sum <= 10000 ; sum++) {
dp[n - 1 ][sum] += dp[n - 1 ][sum - 1 ];
}
// Traverse each query
for ( int q = 0 ; q < queries.length; q++) {
int a = queries[q][ 0 ];
int b = queries[q][ 1 ];
// No of ways to form groups
System.out.print(dp[n - 1 ][b] - dp[n - 1 ][a - 1 ]
+ " " );
}
}
// Driver Code
public static void main(String args[])
{
// Given N batches and K students
int N = 2 , K = 3 ;
// Given ratings
int ratings[][] = { { 1 , 2 , 3 }, { 4 , 5 , 6 } };
// Given Queries
int queries[][] = { { 6 , 6 }, { 1 , 6 } };
// Function Call
numWays(ratings, queries, N, K);
}
} |
# Python3 program for the # above approach # Function to count number of # ways to get given sum groups def numWays(ratings, queries,
n, k):
# Initialise dp array
dp = [[ 0 for i in range ( 10002 )]
for j in range (n)];
# Mark all 1st row values as 1
# since the mat[0][i] is all
# possible sums in first row
for i in range (k):
dp[ 0 ][ratings[ 0 ][i]] + = 1 ;
# Fix the ith row
for i in range ( 1 , n):
# Fix the sum
for sum in range ( 10001 ):
# Iterate through all
# values of ith row
for j in range (k):
# If sum can be obtained
if ( sum > = ratings[i][j]):
dp[i][ sum ] + = dp[i - 1 ][ sum -
ratings[i][j]];
# Find the prefix sum of
# last row
for sum in range ( 1 , 10001 ):
dp[n - 1 ][ sum ] + = dp[n - 1 ][ sum - 1 ];
# Traverse each query
for q in range ( len (queries)):
a = queries[q][ 0 ];
b = queries[q][ 1 ];
# No of ways to form groups
print (dp[n - 1 ][b] -
dp[n - 1 ][a - 1 ],
end = " " );
# Driver Code if __name__ = = '__main__' :
# Given N batches and
# K students
N = 2 ;
K = 3 ;
# Given ratings
ratings = [[ 1 , 2 , 3 ],
[ 4 , 5 , 6 ]];
queries = [[ 6 , 6 ],
[ 1 , 6 ]];
# Function Call
numWays(ratings, queries, N, K);
# This code is contributed by 29AjayKumar |
// C# program for the above approach using System;
class GFG{
// Function to count number of // ways to get given sum groups public static void numWays( int [,] ratings,
int [,]queries,
int n, int k)
{ // Initialise dp array
int [,]dp = new int [n, 10000 + 2];
// Mark all 1st row values as 1
// since the mat[0,i] is all
// possible sums in first row
for ( int i = 0; i < k; i++)
dp[0, ratings[0, i]] += 1;
// Fix the ith row
for ( int i = 1; i < n; i++)
{
// Fix the sum
for ( int sum = 0; sum <= 10000; sum++)
{
// Iterate through all
// values of ith row
for ( int j = 0; j < k; j++)
{
// If sum can be obtained
if (sum >= ratings[i, j])
dp[i, sum] += dp[i - 1,
sum - ratings[i, j]];
}
}
}
// Find the prefix sum of last row
for ( int sum = 1; sum <= 10000; sum++)
{
dp[n - 1, sum] += dp[n - 1, sum - 1];
}
// Traverse each query
for ( int q = 0; q < queries.GetLength(0); q++)
{
int a = queries[q, 0];
int b = queries[q, 1];
// No of ways to form groups
Console.Write(dp[n - 1, b] -
dp[n - 1, a - 1] + " " );
}
} // Driver Code public static void Main(String []args)
{ // Given N batches and K students
int N = 2, K = 3;
// Given ratings
int [,]ratings = { { 1, 2, 3 }, { 4, 5, 6 } };
// Given Queries
int [,]queries = { { 6, 6 }, { 1, 6 } };
// Function Call
numWays(ratings, queries, N, K);
} } // This code is contributed by Amit Katiyar |
<script> // Javascript program for the above approach // Given n batches and k students var n = 2;
var k = 3;
// Function to count number of // ways to get given sum groups function numWays(ratings, queries)
{ // Initialise dp array
var dp = Array.from(
Array(n), ()=>Array(10002).fill(0));
// Mark all 1st row values as 1
// since the mat[0][i] is all
// possible sums in first row
for ( var i = 0; i < k; i++)
dp[0][ratings[0][i]] += 1;
// Fix the ith row
for ( var i = 1; i < n; i++)
{
// Fix the sum
for ( var sum = 0; sum <= 10000; sum++)
{
// Iterate through all
// values of ith row
for ( var j = 0; j < k; j++)
{
// If sum can be obtained
if (sum >= ratings[i][j])
dp[i][sum]+= dp[i - 1][sum - ratings[i][j]];
}
}
}
// Find the prefix sum of last row
for ( var sum = 1; sum <= 10000; sum++)
{
dp[n - 1][sum] += dp[n - 1][sum - 1];
}
// Traverse each query
for ( var q = 0; q < 2; q++)
{
var a = queries[q][0];
var b = queries[q][1];
// No of ways to form groups
document.write(dp[n - 1][b] -
dp[n - 1][a - 1] + " " );
}
} // Driver Code // Given ratings var ratings = [ [ 1, 2, 3 ], [ 4, 5, 6 ] ];
// Given Queries var queries = [ [ 6, 6 ], [ 1, 6 ] ];
// Function Call numWays(ratings, queries); // This code is contributed by famously </script> |
2 3
Time Complexity: O(N*maxSum*K), where maxSum is the maximum sum.
Auxiliary Space: O(N*maxSum), where maxSum is the maximum sum.
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 prev that keep track of current and previous row of DP.
Implementation Steps:
- Initialize two vectors curr and prev to keep track of only current and previous row of Dp with 0.
- Now iterative over subproblems and get the current computation.
- While iteration initialize curr vector.
- Now compute the current value by the help of prev vector and store that value in curr.
- After every iteration store values of curr vector in prev vector for further iteration.
- At last return the answer stored in curr[0].
Implementation:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Given n batches and k students #define n 2 #define k 3 // Function to count number of // ways to get given sum groups void numWays( int ratings[n][k], int queries[][2])
{ // Initialise dp array
// int dp[n][10000 + 2];
vector< int > prev(10000 + 2);
vector< int > curr(10000 + 2);
// Mark all 1st row values as 1
// since the prev[i] is all
// possible sums in first row
for ( int i = 0; i < k; i++)
prev[ratings[0][i]] += 1;
// Fix the ith row
for ( int i = 1; i < n; i++) {
// Fix the sum
for ( int sum = 0; sum <= 10000; sum++) {
// Iterate through all
// values of ith row
for ( int j = 0; j < k; j++) {
// If sum can be obtained
if (sum >= ratings[i][j])
curr[sum] += prev[sum - ratings[i][j]];
}
}
// assigning values of curr vector to prev vector
// for further iteration
prev = curr;
}
// Find the prefix sum of last row
for ( int sum = 1; sum <= 10000; sum++) {
curr[sum] += curr[sum - 1];
}
// Traverse each query
for ( int q = 0; q < 2; q++) {
int a = queries[q][0];
int b = queries[q][1];
// No of ways to form groups
cout << curr[b] - curr[a - 1] << " " ;
}
} // Driver Code int main()
{ // Given ratings
int ratings[n][k] = { { 1, 2, 3 }, { 4, 5, 6 } };
// Given Queries
int queries[][2] = { { 6, 6 }, { 1, 6 } };
// Function Call
numWays(ratings, queries);
return 0;
} |
import java.util.Arrays;
public class Main {
// Given n batches and k students
static final int n = 2 ;
static final int k = 3 ;
// Function to count the number of ways to get given sum
// groups
static void numWays( int [][] ratings, int [][] queries)
{
// Initialize dp array
int [] prev = new int [ 10000 + 2 ];
int [] curr = new int [ 10000 + 2 ];
// Mark all 1st row values as 1 since the prev[i] is
// all possible sums in the first row
for ( int i = 0 ; i < k; i++) {
prev[ratings[ 0 ][i]] += 1 ;
}
// Fix the ith row
for ( int i = 1 ; i < n; i++) {
// Fix the sum
for ( int s = 0 ; s <= 10000 ; s++) {
// Iterate through all values of ith row
for ( int j = 0 ; j < k; j++) {
// If sum can be obtained
if (s >= ratings[i][j]) {
curr[s] += prev[s - ratings[i][j]];
}
}
}
// Reset prev array to zeros for the next
// iteration
Arrays.fill(prev, 0 );
// Assigning values of curr array to prev array
// for further iteration
System.arraycopy(curr, 0 , prev, 0 , curr.length);
Arrays.fill(curr,
0 ); // Reset curr array to zeros for
// the next iteration
}
// Find the prefix sum of the last row
for ( int s = 1 ; s <= 10000 ; s++) {
prev[s] += prev[s - 1 ];
}
// Traverse each query
for ( int q = 0 ; q < 2 ; q++) {
int a = queries[q][ 0 ];
int b = queries[q][ 1 ];
// Number of ways to form groups
System.out.print((prev[b] - prev[a - 1 ]) + " " );
}
}
// Driver Code
public static void main(String[] args)
{
// Given ratings
int [][] ratings = { { 1 , 2 , 3 }, { 4 , 5 , 6 } };
// Given Queries
int [][] queries = { { 6 , 6 }, { 1 , 6 } };
// Function Call
numWays(ratings, queries);
}
} |
# Python program for the above approach # Given n batches and k students n = 2
k = 3
# Function to count number of # ways to get given sum groups def numWays(ratings, queries):
# Initialise dp array
prev = [ 0 ] * ( 10000 + 2 )
curr = [ 0 ] * ( 10000 + 2 )
# Mark all 1st row values as 1
# since the prev[i] is all
# possible sums in first row
for i in range (k):
prev[ratings[ 0 ][i]] + = 1
# Fix the ith row
for i in range ( 1 , n):
# Fix the sum
for s in range ( 0 , 10001 ):
# Iterate through all
# values of ith row
for j in range (k):
# If sum can be obtained
if s > = ratings[i][j]:
curr[s] + = prev[s - ratings[i][j]]
# assigning values of curr list to prev list
# for further iteration
prev = curr.copy()
curr = [ 0 ] * ( 10000 + 2 )
# Find the prefix sum of last row
for s in range ( 1 , 10001 ):
prev[s] + = prev[s - 1 ]
# Traverse each query
for q in range ( 2 ):
a = queries[q][ 0 ]
b = queries[q][ 1 ]
# No of ways to form groups
print (prev[b] - prev[a - 1 ], end = ' ' )
# Driver Code if __name__ = = '__main__' :
# Given ratings
ratings = [[ 1 , 2 , 3 ], [ 4 , 5 , 6 ]]
# Given Queries
queries = [[ 6 , 6 ], [ 1 , 6 ]]
# Function Call
numWays(ratings, queries)
|
using System;
class Program
{ const int n = 2;
const int k = 3;
static void NumWays( int [,] ratings, int [,] queries)
{
int [] prev = new int [10000 + 2];
int [] curr = new int [10000 + 2];
for ( int i = 0; i < k; i++)
{
prev[ratings[0, i]] += 1;
}
for ( int i = 1; i < n; i++)
{
for ( int sum = 0; sum <= 10000; sum++)
{
curr[sum] = 0; // Fix: Reset curr array to zero before using it
for ( int j = 0; j < k; j++)
{
if (sum >= ratings[i, j])
{
curr[sum] += prev[sum - ratings[i, j]];
}
}
}
Array.Copy(curr, prev, curr.Length);
}
for ( int sum = 1; sum <= 10000; sum++)
{
curr[sum] += curr[sum - 1];
}
for ( int q = 0; q < 2; q++)
{
int a = queries[q, 0];
int b = queries[q, 1];
Console.Write(curr[b] - curr[a - 1] + " " );
}
}
static void Main()
{
int [,] ratings = { { 1, 2, 3 }, { 4, 5, 6 } };
int [,] queries = { { 6, 6 }, { 1, 6 } };
NumWays(ratings, queries);
// Output: 2 3
}
} |
// Given n batches and k students const n = 2; const k = 3; // Function to count number of ways to get given sum groups function numWays(ratings, queries) {
// Initialize dp array
let prev = new Array(10000 + 2).fill(0);
let curr = new Array(10000 + 2).fill(0);
// Mark all 1st row values as 1 since the prev[i] is all possible sums in first row
for (let i = 0; i < k; i++) prev[ratings[0][i]] += 1;
// Fix the ith row
for (let i = 1; i < n; i++) {
// Fix the sum
for (let sum = 0; sum <= 10000; sum++) {
// Iterate through all values of ith row
for (let j = 0; j < k; j++) {
// If sum can be obtained
if (sum >= ratings[i][j]) curr[sum] += prev[sum - ratings[i][j]];
}
}
// Assigning values of curr vector to prev vector for further iteration
prev = [...curr];
}
// Find the prefix sum of last row
for (let sum = 1; sum <= 10000; sum++) {
curr[sum] += curr[sum - 1];
}
// Traverse each query
for (let q = 0; q < queries.length; q++) {
let a = queries[q][0];
let b = queries[q][1];
// No of ways to form groups
console.log(curr[b] - curr[a - 1]);
}
} // Given ratings const ratings = [ [1, 2, 3],
[4, 5, 6],
]; // Given Queries const queries = [ [6,6],
[1,6]
]; // Function Call numWays(ratings, queries); |
Output
2 3
Time Complexity: O(N*maxSum*K), where maxSum is the maximum sum.
Auxiliary Space: O(maxSum)