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# Number of K length subsequences with minimum sum

• Difficulty Level : Hard
• Last Updated : 17 Aug, 2021

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 the minimum possible.

Examples:

Input: arr[] = {1, 2, 3, 4}, K = 2
Output:
Subsequences of length 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, the 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 ``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;``}`

## 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`

## 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`

## 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 ``=` `2``print``(cntSubSeq(arr, n, k))` `# This code is contributed by ihritik`

## Javascript

 ``
Output:
`1`

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