Given an array A[] containing N elements and an integer K. The task is to calculate the product of all elements of subsequences of size K except the minimum and the maximum elements for each subsequence.
Note: Since the answer can be very large so print the final answer as mod of 109 + 7.
Examples:
Input : arr[] = {1, 2, 3 4}, K = 3
Output : 36
Subsequences of length 3 are:
{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}
Excluding minimum and maximum elements from
each of the above subsequences, product will be:
(2 * 2 * 3 * 3) = 36.
Input : arr[] = {10, 5, 16, 6}, k=3
Output : 3600
Naive Approach: A simple approach is to generate all possible subsequences one by one and multiply all elements except maximum and minimum and further multiplying all of them. Since there will be a total of (n)C(K) subsequences all having K – 2 elements to be multiplied which is tedious work to do.
Efficient Approach: The idea is to first sort the array since it doesn’t matter if we consider subsequences or subsets.
Now count the occurrence of each element one by one.
In total, a number can occur in (n-1)C(K-1) subsequences out of which (i)C(K-1) times it will occur as maximum element and (n-i-1)C(K-1) times it will occur as a minimum element of that subsequence.
Hence, in total 
element will occur:
(n-1)C(K-1) - (i)C(K-1) - (n-i-1)C(K-1) times. (let's say it x)
So, at first we’ll be calculating x for each element a[i] and then multiply a[i] x times. i.e (![a[i]^x mod(10^9 + 7)](https://neveropen.co.za/wp-content/uploads/2023/11/dominic_rubhabha_quicklatex.com-8baa253819b7ac2e61ef1a77733df6bd_l3.png "Rendered by QuickLaTeX.com")
).
Since, It’s too difficult to calculate this for large arrays, so we’ll use Fermat’s Little Theorem.
Below is the implementation of the above approach:
C++
// C++ program to find product of all // Subsequences of size K except the // minimum and maximum Elements#include <bits/stdc++.h>using namespace std;#define MOD 1000000007#define ll long long#define max 101// 2D array to store value of // combinations nCrll C[max - 1][max - 1];ll power(ll x, unsigned ll y){ unsigned ll res = 1; x = x % MOD; while (y > 0) { if (y & 1) { res = (res * x) % MOD; } y = y >> 1; x = (x * x) % MOD; } return res % MOD;}// Function to pre-calculate value of all // combinations nCrvoid combi(int n, int k){ int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= min(i, k); j++) { if (j == 0 || j == i) C[i][j] = 1; else C[i][j] = (C[i - 1][j - 1] % MOD + C[i - 1][j] % MOD) % MOD; } }}// Function to calculate product of all subsequences // except the minimum and maximum elementsunsigned ll product(ll a[], int n, int k){ unsigned ll ans = 1; // Sorting array so that it becomes easy // to calculate the number of times an // element will come in first or last place sort(a, a + n); // An element will occur 'powa' times in total // of which 'powla' times it will be last element // and 'powfa' times it will be first element ll powa = C[n - 1][k - 1]; for (int i = 0; i < n; i++) { ll powla = C[i][k - 1]; ll powfa = C[n - i - 1][k - 1]; // In total it will come // powe = powa-powla-powfa times ll powe = ((powa % MOD) - (powla + powfa) % MOD + MOD) % MOD; // Multiplying a[i] powe times using // Fermat Little Theorem under MODulo // MOD for fast exponentiation unsigned ll mul = power(a[i], powe) % MOD; ans = ((ans % MOD) * (mul % MOD)) % MOD; } return ans % MOD;}// Driver Codeint main(){ // pre-calculation of all combinations combi(100, 100); ll arr[] = { 1, 2, 3, 4 }; int n = sizeof(arr) / sizeof arr[0]; int k = 3; unsigned ll ans = product(arr, n, k); cout << ans << endl; return 0;} |
Java
// Java program to find product of all // Subsequences of size K except the // minimum and maximum Elementsimport java.util.Arrays;class GFG { static int MOD= 1000000007;static int max =101;// 2D array to store value of // combinations nCrstatic long C[][] = new long[max ][max];static long power(long x, long y){ long res = 1; x = x % MOD; while (y > 0) { if (y % 2== 1) { res = (res * x) % MOD; } y = y >> 1; x = (x * x) % MOD; } return res % MOD;}// Function to pre-calculate value of all // combinations nCrstatic void combi(int n, int k){ int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i, k); j++) { if (j == 0 || j == i) C[i][j] = 1; else C[i][j] = (C[i - 1][j - 1] % MOD + C[i - 1][j] % MOD) % MOD; } }}// Function to calculate product of all subsequences // except the minimum and maximum elementsstatic long product(long a[], int n, int k){ long ans = 1; // Sorting array so that it becomes easy // to calculate the number of times an // element will come in first or last place Arrays.sort(a); // An element will occur 'powa' times in total // of which 'powla' times it will be last element // and 'powfa' times it will be first element long powa = C[n - 1][k - 1]; for (int i = 0; i < n; i++) { long powla = C[i][k - 1]; long powfa = C[n - i - 1][k - 1]; // In total it will come // powe = powa-powla-powfa times long powe = ((powa % MOD) - (powla + powfa) % MOD + MOD) % MOD; // Multiplying a[i] powe times using // Fermat Little Theorem under MODulo // MOD for fast exponentiation long mul = power(a[i], powe) % MOD; ans = ((ans % MOD) * (mul % MOD)) % MOD; } return ans % MOD;}// Driver Codepublic static void main(String[] args){ // pre-calculation of all combinations combi(100, 100); long arr[] = { 1, 2, 3, 4 }; int n = arr.length; int k = 3; long ans = product(arr, n, k); System.out.println(ans);}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python 3 program to find product of all # Subsequences of size K except the # minimum and maximum ElementsMOD = 1000000007max = 101# 2D array to store value of # combinations nCrC = [[0 for i in range(max)] for j in range(max)]def power(x,y): res = 1 x = x % MOD while (y > 0): if (y & 1): res = (res * x) % MOD y = y >> 1 x = (x * x) % MOD return res % MOD# Function to pre-calculate value of all # combinations nCrdef combi(n, k): for i in range(n + 1): for j in range(min(i, k) + 1): if (j == 0 or j == i): C[i][j] = 1 else: C[i][j] = (C[i - 1][j - 1] % MOD + C[i - 1][j] % MOD) % MOD# Function to calculate product of all subsequences # except the minimum and maximum elementsdef product(a, n, k): ans = 1 # Sorting array so that it becomes easy # to calculate the number of times an # element will come in first or last place a.sort(reverse = False) # An element will occur 'powa' times in total # of which 'powla' times it will be last element # and 'powfa' times it will be first element powa = C[n - 1][k - 1] for i in range(n): powla = C[i][k - 1] powfa = C[n - i - 1][k - 1] # In total it will come # powe = powa-powla-powfa times powe = ((powa % MOD) - (powla + powfa) % MOD + MOD) % MOD # Multiplying a[i] powe times using # Fermat Little Theorem under MODulo # MOD for fast exponentiation mul = power(a[i], powe) % MOD ans = ((ans % MOD) * (mul % MOD)) % MOD return ans % MOD# Driver Codeif __name__ == '__main__': # pre-calculation of all combinations combi(100, 100) arr = [1, 2, 3, 4] n = len(arr) k = 3 ans = product(arr, n, k) print(ans)# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find product of all // Subsequences of size K except the // minimum and maximum Elementsusing System;class GFG{static int MOD = 1000000007;static int max = 101;// 2D array to store value of // combinations nCrstatic long [,]C = new long[max, max];static long power(long x, long y){ long res = 1; x = x % MOD; while (y > 0) { if (y % 2 == 1) { res = (res * x) % MOD; } y = y >> 1; x = (x * x) % MOD; } return res % MOD;}// Function to pre-calculate value // of all combinations nCrstatic void combi(int n, int k){ int i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= Math.Min(i, k); j++) { if (j == 0 || j == i) C[i, j] = 1; else C[i, j] = (C[i - 1, j - 1] % MOD + C[i - 1, j] % MOD) % MOD; } }}// Function to calculate product of // all subsequences except // the minimum and maximum elementsstatic long product(long []a, int n, int k){ long ans = 1; // Sorting array so that it becomes easy // to calculate the number of times an // element will come in first or last place Array.Sort(a); // An element will occur 'powa' times // in total of which 'powla' times it // will be last element and 'powfa' times // it will be first element long powa = C[n - 1, k - 1]; for (int i = 0; i < n; i++) { long powla = C[i, k - 1]; long powfa = C[n - i - 1, k - 1]; // In total it will come // powe = powa-powla-powfa times long powe = ((powa % MOD) - (powla + powfa) % MOD + MOD) % MOD; // Multiplying a[i] powe times using // Fermat Little Theorem under MODulo // MOD for fast exponentiation long mul = power(a[i], powe) % MOD; ans = ((ans % MOD) * (mul % MOD)) % MOD; } return ans % MOD;}// Driver Codestatic public void Main (){ // pre-calculation of all combinations combi(100, 100); long []arr = { 1, 2, 3, 4 }; int n = arr.Length; int k = 3; long ans = product(arr, n, k); Console.WriteLine(ans);}}// This code contributed by ajit |
Javascript
<script> // Javascript program to find product of all // Subsequences of size K except the // minimum and maximum Elements let MOD= 1000000007; let max =101; // 2D array to store value of // combinations nCr let C = new Array(max); for(let i = 0; i < max; i++) { C[i] = new Array(max); for(let j = 0; j < max; j++) { C[i][j] = 0; } } function power(x, y) { let res = 1; x = x % MOD; while (y > 0) { if (y % 2== 1) { res = (res * x) % MOD; } y = y >> 1; x = (x * x) % MOD; } return res % MOD; } // Function to pre-calculate value of all // combinations nCr function combi(n, k) { let i, j; for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i, k); j++) { if (j == 0 || j == i) C[i][j] = 1; else C[i][j] = (C[i - 1][j - 1] % MOD + C[i - 1][j] % MOD) % MOD; } } } // Function to calculate product of all subsequences // except the minimum and maximum elements function product(a, n, k) { let ans = 1; // Sorting array so that it becomes easy // to calculate the number of times an // element will come in first or last place a.sort(function(a, b){return a - b}); // An element will occur 'powa' times in total // of which 'powla' times it will be last element // and 'powfa' times it will be first element let powa = C[n - 1][k - 1]; for (let i = 0; i < n; i++) { let powla = C[i][k - 1]; let powfa = C[n - i - 1][k - 1]; // In total it will come // powe = powa-powla-powfa times let powe = ((powa % MOD) - (powla + powfa) % MOD + MOD) % MOD; // Multiplying a[i] powe times using // Fermat Little Theorem under MODulo // MOD for fast exponentiation let mul = power(a[i], powe) % MOD; ans = ((ans % MOD) * (mul % MOD)) % MOD; } return ans % MOD; } // pre-calculation of all combinations combi(100, 100); let arr = [ 1, 2, 3, 4 ]; let n = arr.length; let k = 3; let ans = product(arr, n, k); document.write(ans);</script> |
Output:
36
Time Complexity: O(nlogn + max*max), where n is the size of the given array and max is the defined constant.
Auxiliary Space: O(max*max), where max is the defined constant.
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