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] the 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++
// 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 groupsvoid 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 Codeint 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
// Java program for the above approachimport 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
# Python3 program for the # above approach# Function to count number of# ways to get given sum groupsdef 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 Codeif __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#
// C# program for the above approachusing System;class GFG{// Function to count number of// ways to get given sum groupspublic 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 Codepublic 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 |
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.
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