Number of K length subsequences with minimum sum
Given an array arr[] of size N and an integer K, the task is to find the number of K length subsequences of this array such that the sum of these subsequences is minimum possible.
Examples:
Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 1
Subsequences of lengths 2 are (1, 2), (1, 3), (1, 4),
(2, 3), (2, 4) and (3, 4).
The minimum sum is 3 and the only subsequence
with this sum is (1, 2).Input: arr[] = {2, 1, 2, 2, 2, 1}, K = 3
Output: 4
Approach: The minimum possible sum of a subsequence of length K from the given array is the sum of the K smallest elements of the array. Let X be the maximum element among the K smallest elements of the array, and let the number of times it occurs among the K smallest elements of the array be Y and its total occurrence in the complete array be cntX. Now, there are cntXCY ways to select this element in the K smallest elements which is the count of required subsequences.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the value of // Binomial Coefficient C(n, k) int binomialCoeff(int n, int k) { int C[n + 1][k + 1]; int i, j; // Calculate value of Binomial Coefficient // in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= min(i, k); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using previously // stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } return C[n][k]; } // Function to return the count // of valid subsequences int cntSubSeq(int arr[], int n, int k) { // Sort the array sort(arr, arr + n); // Maximum among the minimum K elements int num = arr[k - 1]; // Y will store the frequency of num // in the minimum K elements int Y = 0; for (int i = k - 1; i >= 0; i--) { if (arr[i] == num) Y++; } // cntX will store the frequency of // num in the complete array int cntX = Y; for (int i = k; i < n; i++) { if (arr[i] == num) cntX++; } return binomialCoeff(cntX, Y); } // Driver code int main() { int arr[] = { 1, 2, 3, 4 }; int n = sizeof(arr) / sizeof(int); int k = 2; cout << cntSubSeq(arr, n, k); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the value of // Binomial Coefficient C(n, k) static int binomialCoeff(int n, int k) { int C[][] = new int [n + 1][k + 1]; int i, j; // Calculate value of Binomial Coefficient // in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i, k); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using previously // stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } return C[n][k]; } // Function to return the count // of valid subsequences static int cntSubSeq(int arr[], int n, int k) { // Sort the array Arrays.sort(arr); // Maximum among the minimum K elements int num = arr[k - 1]; // Y will store the frequency of num // in the minimum K elements int Y = 0; for (int i = k - 1; i >= 0; i--) { if (arr[i] == num) Y++; } // cntX will store the frequency of // num in the complete array int cntX = Y; for (int i = k; i < n; i++) { if (arr[i] == num) cntX++; } return binomialCoeff(cntX, Y); } // Driver code public static void main (String[] args) { int arr[] = { 1, 2, 3, 4 }; int n = arr.length; int k = 2; System.out.println(cntSubSeq(arr, n, k)); } } // This code is contributed by AnkitRai01 |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the value of // Binomial Coefficient C(n, k) static int binomialCoeff(int n, int k) { int [,]C = new int [n + 1, k + 1]; int i, j; // Calculate value of Binomial Coefficient // in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.Min(i, k); j++) { // Base Cases if (j == 0 || j == i) C[i, j] = 1; // Calculate value using previously // stored values else C[i, j] = C[i - 1, j - 1] + C[i - 1, j]; } } return C[n, k]; } // Function to return the count // of valid subsequences static int cntSubSeq(int []arr, int n, int k) { // Sort the array Array.Sort(arr); // Maximum among the minimum K elements int num = arr[k - 1]; // Y will store the frequency of num // in the minimum K elements int Y = 0; for (int i = k - 1; i >= 0; i--) { if (arr[i] == num) Y++; } // cntX will store the frequency of // num in the complete array int cntX = Y; for (int i = k; i < n; i++) { if (arr[i] == num) cntX++; } return binomialCoeff(cntX, Y); } // Driver code public static void Main (String[] args) { int []arr = { 1, 2, 3, 4 }; int n = arr.Length; int k = 2; Console.WriteLine(cntSubSeq(arr, n, k)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach # Function to return the value of # Binomial Coefficient C(n, k) def binomialCoeff(n, k) : C = [[0 for i in range(n + 1)] for j in range(k + 1)] # Calculate value of Binomial Coefficient # in bottom up manner for i in range (0, n + 1 ): for j in range (0, min(i, k) + 1): # Base Cases if (j == 0 or j == i): C[i][j] = 1 # Calculate value using previously # stored values else : C[i][j] = C[i - 1][j - 1] + C[i - 1][j] return C[n][k] # Function to return the count # of valid subsequences def cntSubSeq(arr, n, k) : # Sort the array arr.sort() # Maximum among the minimum K elements num = arr[k - 1]; # Y will store the frequency of num # in the minimum K elements Y = 0; for i in range (k - 1, -1, 1) : if (arr[i] == num): Y += 1 # cntX will store the frequency of # num in the complete array cntX = Y; for i in range (k, n): if (arr[i] == num) : cntX += 1 return binomialCoeff(cntX, Y) # Driver code arr = [ 1, 2, 3, 4 ] n = len(arr) k = 2print(cntSubSeq(arr, n, k)) # This code is contributed by ihritik |
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Output:
1
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