# Minimum count of numbers required from given array to represent S

Given an integer S and an array arr[], the task is to find the minimum number of elements whose sum is S, such that any element of the array can be chosen any number of times to get sum S.

Examples:

Input: arr[] = {25, 10, 5}, S = 30
Output: 2
Explanation:
In the given array there are many possible solutions such as –
5 + 5 + 5 + 5 + 5 + 5 = 30, or
10 + 10 + 10 = 30, or
25 + 5 = 30
Hence, the minimum possible solution is 2

Input: arr[] = {2, 1, 4, 3, 5, 6}, Sum= 6
Output: 1
Explantion:
In the given array there are many possible solutions such as –
2 + 2 + 2 = 6, or
2 + 4 = 6, or
6 = 6,
Hence, the minimum possible solution is 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The idea is to find every possible sequence recursively such that their sum is equal to the given S and also keep track of the minimum sequence such that their sum is given S. In this way, the minimum possible solution can be calculated easily.

Below is the implmentation of the above approach:

## C++

 `// C++ implmentation to find the ` `// minimum number of sequence ` `// required from array such that ` `// their sum is equal to given S ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the ` `// minimum elements required to ` `// get the sum of given value S ` `int` `printAllSubsetsRec(``int` `arr[], ` `                       ``int` `n, ` `                       ``vector<``int``> v, ` `                       ``int` `sum) ` `{ ` `    ``// Condition if the ` `    ``// sequence is found ` `    ``if` `(sum == 0) { ` `        ``return` `(``int``)v.size(); ` `    ``} ` ` `  `    ``if` `(sum < 0) ` `        ``return` `INT_MAX; ` ` `  `    ``// Condition when no ` `    ``// such sequence found ` `    ``if` `(n == 0) ` `        ``return` `INT_MAX; ` ` `  `    ``// Calling for without choosing ` `    ``// the current index value ` `    ``int` `x = printAllSubsetsRec( ` `        ``arr, ` `        ``n - 1, v, sum); ` ` `  `    ``// Calling for after choosing ` `    ``// the current index value ` `    ``v.push_back(arr[n - 1]); ` `    ``int` `y = printAllSubsetsRec( ` `        ``arr, n, v, ` `        ``sum - arr[n - 1]); ` `    ``return` `min(x, y); ` `} ` ` `  `// Function for every array ` `int` `printAllSubsets(``int` `arr[], ` `                    ``int` `n, ``int` `sum) ` `{ ` `    ``vector<``int``> v; ` `    ``return` `printAllSubsetsRec(arr, n, ` `                              ``v, sum); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 1, 4, 3, 5, 6 }; ` `    ``int` `sum = 6; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << printAllSubsets(arr, n, sum) ` `         ``<< endl; ` `    ``return` `0; ` `} `

Output:

```1
```

Performance Analysis:

• Time Complexity: As in the above approach, there are two choose for every number in each step which takes O(2N) time, Hence the Time Complexity will be O(2N).
• Space Complexity: As in the above approach, there is no extra space used, Hence the space complexity will be O(1).

Effiecient Approach: As in the above approach there is overlapping subproblems, So the idea is to use Dynamic Programming paradigm to solve this problem. Create a DP table of N * S to store the pre-computed answer for the previous sequence that is the minimum length sequence required to get the sum as S – arr[i] and in this way the finally after calculating for every value of the array, the answer to the problem will be dp[N][S], where m is the length of the array and S is the given sum.

Below is the implmentation of the above approach:

## C++

 `// C++ implmentation to find the ` `// minimum number of sequence ` `// required from array such that ` `// their sum is equal to given S ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the count of ` `// minimum length of the sequence ` `int` `Count(``int` `S[], ``int` `m, ``int` `n) ` `{ ` `    ``vector > table( ` `        ``m + 1, ` `        ``vector<``int``>( ` `            ``n + 1, 0)); ` ` `  `    ``// Loop to intialize the array ` `    ``// as infinite in the row 0 ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``table[i] = INT_MAX - 1; ` `    ``} ` ` `  `    ``// Loop to find the solution ` `    ``// by pre-computation for the ` `    ``// sequence ` `    ``for` `(``int` `i = 1; i <= m; i++) { ` ` `  `        ``for` `(``int` `j = 1; j <= n; j++) { ` `            ``if` `(S[i - 1] > j) { ` `                ``table[i][j] ` `                    ``= table[i - 1][j]; ` `            ``} ` `            ``else` `{ ` ` `  `                ``// Minimum possible ` `                ``// for the previous ` `                ``// minimum value ` `                ``// of the sequence ` `                ``table[i][j] ` `                    ``= min( ` `                        ``table[i - 1][j], ` `                        ``table[i][j - S[i - 1]] + 1); ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `table[m][n]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 9, 6, 5, 1 }; ` `    ``int` `m = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << Count(arr, m, 11); ` `    ``return` `0; ` `} `

## Java

 `// Java implmentation to find the  ` `// minimum number of sequence  ` `// required from array such that  ` `// their sum is equal to given S  ` `import` `java.util.*;  ` ` `  `class` `GFG{  ` ` `  `// Function to find the count of  ` `// minimum length of the sequence  ` `static` `int` `Count(``int` `S[], ``int` `m, ``int` `n)  ` `{  ` `    ``int` `[][]table = ``new` `int``[m + ``1``][n + ``1``];  ` ` `  `    ``// Loop to intialize the array  ` `    ``// as infinite in the row 0  ` `    ``for``(``int` `i = ``1``; i <= n; i++)  ` `    ``{  ` `    ``table[``0``][i] = Integer.MAX_VALUE - ``1``;  ` `    ``}  ` ` `  `    ``// Loop to find the solution  ` `    ``// by pre-computation for the  ` `    ``// sequence  ` `    ``for``(``int` `i = ``1``; i <= m; i++)  ` `    ``{  ` `    ``for``(``int` `j = ``1``; j <= n; j++)  ` `    ``{  ` `        ``if` `(S[i - ``1``] > j)  ` `        ``{  ` `            ``table[i][j] = table[i - ``1``][j];  ` `        ``}  ` `        ``else` `        ``{  ` `                 `  `            ``// Minimum possible for the  ` `            ``// previous minimum value  ` `            ``// of the sequence  ` `            ``table[i][j] = Math.min(table[i - ``1``][j],  ` `                            ``table[i][j - S[i - ``1``]] + ``1``);  ` `        ``}  ` `    ``}  ` `    ``}  ` `    ``return` `table[m][n];  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `arr[] = { ``9``, ``6``, ``5``, ``1` `};  ` `    ``int` `m = arr.length;  ` `     `  `    ``System.out.print(Count(arr, m, ``11``));  ` `}  ` `}  ` ` `  `// This code is contributed by gauravrajput1  `

## Python3

 `# Python3 implmentation to find the  ` `# minimum number of sequence  ` `# required from array such that  ` `# their sum is equal to given S  ` ` `  `# Function to find the count of  ` `# minimum length of the sequence  ` `def` `Count(S, m, n):  ` `    ``table ``=` `[[``0` `for` `i ``in` `range``(n ``+` `1``)]  ` `                ``for` `i ``in` `range``(m ``+` `1``)]  ` ` `  `    ``# Loop to intialize the array  ` `    ``# as infinite in the row 0  ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``table[``0``][i] ``=` `10``*``*``9` `-` `1` ` `  `    ``# Loop to find the solution  ` `    ``# by pre-computation for the  ` `    ``# sequence  ` `    ``for` `i ``in` `range``(``1``, m ``+` `1``):  ` ` `  `        ``for` `j ``in` `range``(``1``, n ``+` `1``):  ` `            ``if` `(S[i ``-` `1``] > j):  ` `                ``table[i][j] ``=` `table[i ``-` `1``][j]  ` `            ``else``:  ` ` `  `                ``# Minimum possible  ` `                ``# for the previous  ` `                ``# minimum value  ` `                ``# of the sequence  ` `                ``table[i][j] ``=` `min``(table[i ``-` `1``][j],  ` `                                ``table[i][j ``-` `S[i ``-` `1``]] ``+` `1``)  ` ` `  `    ``return` `table[m][n]  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `'__main__'``:  ` `    ``arr``=` `[``9``, ``6``, ``5``, ``1``]  ` `    ``m ``=` `len``(arr)  ` `    ``print``(Count(arr, m, ``11``))  ` ` `  `# This code is contributed by Mohit Kumar  `

## C#

 `// C# implmentation to find the ` `// minimum number of sequence ` `// required from array such that ` `// their sum is equal to given S ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find the count of ` `// minimum length of the sequence ` `static` `int` `Count(``int``[] S, ``int` `m, ``int` `n) ` `{ ` `    ``int``[,] table = ``new` `int``[m + 1, n + 1]; ` ` `  `    ``// Loop to intialize the array ` `    ``// as infinite in the row 0 ` `    ``for``(``int` `i = 1; i <= n; i++)  ` `    ``{ ` `       ``table[0, i] = ``int``.MaxValue - 1; ` `    ``} ` ` `  `    ``// Loop to find the solution ` `    ``// by pre-computation for the ` `    ``// sequence ` `    ``for``(``int` `i = 1; i <= m; i++) ` `    ``{ ` `       ``for``(``int` `j = 1; j <= n; j++)  ` `       ``{ ` `          ``if` `(S[i - 1] > j) ` `          ``{ ` `              ``table[i, j] = table[i - 1, j]; ` `          ``} ` `          ``else` `          ``{ ` `               `  `              ``// Minimum possible for the ` `              ``// previous minimum value ` `              ``// of the sequence ` `              ``table[i, j] = Math.Min(table[i - 1, j], ` `                            ``table[i, j - S[i - 1]] + 1); ` `          ``} ` `       ``} ` `    ``} ` `    ``return` `table[m, n]; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int``[] arr = { 9, 6, 5, 1 }; ` `    ``int` `m = 4; ` ` `  `    ``Console.WriteLine(Count(arr, m, 11)); ` `} ` `} ` ` `  `// This code is contributed by jrishabh99 `

Output:

```2
```

Performance Analysis:

• Time Complexity: As in the above approach, there are two loop for the calulation of the minimum length sequence required which takes O(N2) time, Hence the Time Complexity will be O(N2).
• Space Complexity: As in the above approach, there is a extra dp table used, Hence the space complexity will be O(N2).

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