# Distribution of a Number in Array within a Range

Given integers S, N, K, L and R where S has to be distributed in an array of size N such that each element must be from the range [L, R] and the sum of K elements of the array should be greater than the sum of the remaining N – K elements whose sum is equal to Sk and these elements are in non-increasing order.

Examples:

Input: N = 5, K = 3, L = 1, R = 3, S = 13, Sk = 9
Output: 3 3 3 2 2

Input: N = 5, K = 3, L = 1, R = 3, S = 15, Sk = 9
Output: 3 3 3 3 3

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

Approach: If Sk can be distributed into k elements equally then store Sk/k into all the first k elements of the array, otherwise first element of the array will be (Sk/k) + (Sk % k) and remaining k – 1 elements will be (Sk – Sk % k) % k – 1. Similarly, distribute the S-Sk into n-k elements.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function for the distribution of the number ` `void` `distribution(``int` `n, ``int` `k, ``int` `l, ` `                  ``int` `r, ``int` `S, ``int` `Sk) ` `{ ` `    ``int` `a[n]; ` `    ``int` `len = k, temp, rem, s; ` `    ``int` `diff = S - Sk; ` `    ``for` `(``int` `i = 0; i < len; i++) { ` `        ``// Distribute the number ` `        ``// among k elements ` `        ``temp = Sk / k; ` `        ``rem = Sk % k; ` `        ``if` `(temp + rem >= l && temp + rem <= r) { ` `            ``a[i] = temp; ` `        ``} ` `        ``else` `if` `(temp + rem > r) { ` `            ``a[i] = r; ` `        ``} ` `        ``else` `if` `(temp + rem < r) { ` `            ``cout << ``"-1"``; ` `            ``return``; ` `        ``} ` `        ``Sk = Sk - a[i]; ` `        ``k = k - 1; ` `    ``} ` ` `  `    ``// If there is some remaining ` `    ``// sum to distribute ` `    ``if` `(Sk > 0) { ` `        ``cout << ``"-1"``; ` `        ``return``; ` `    ``} ` ` `  `    ``// If there are elements remaining ` `    ``// to distribute i.e. (n - k) ` `    ``if` `(len) { ` `        ``k = n - len; ` `        ``for` `(``int` `i = len; i < n; i++) { ` `            ``// Divide the remaining sum into ` `            ``// n-k elements ` `            ``temp = diff / k; ` `            ``rem = diff % k; ` `            ``if` `(temp + rem >= l && temp + rem <= r) { ` `                ``a[i] = temp; ` `            ``} ` `            ``else` `if` `(temp + rem > r) { ` `                ``a[i] = r; ` `            ``} ` `            ``else` `if` `(temp + rem < r) { ` `                ``cout << ``"-1"``; ` `                ``return``; ` `            ``} ` `            ``diff = diff - a[i]; ` `            ``k = k - 1; ` `        ``} ` `        ``if` `(diff) { ` `            ``cout << ``"-1"``; ` `            ``return``; ` `        ``} ` `    ``} ` ` `  `    ``// Print the distribution ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cout << a[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5, k = 3, l = 1, ` `        ``r = 5, S = 13, Sk = 9; ` ` `  `    ``distribution(n, k, l, r, S, Sk); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function for the distribution of the number  ` `    ``static` `void` `distribution(``int` `n, ``int` `k, ``int` `l,  ` `                            ``int` `r, ``int` `S, ``int` `Sk)  ` `    ``{  ` `        ``int` `a[] = ``new` `int``[n];  ` `        ``int` `len = k, temp, rem, s;  ` `        ``int` `diff = S - Sk;  ` `         `  `        ``for` `(``int` `i = ``0``; i < len; i++) ` `        ``{ ` `             `  `            ``// Distribute the number  ` `            ``// among k elements  ` `            ``temp = Sk / k;  ` `            ``rem = Sk % k;  ` `            ``if` `(temp + rem >= l && temp + rem <= r) ` `            ``{  ` `                ``a[i] = temp;  ` `            ``}  ` `            ``else` `if` `(temp + rem > r) ` `            ``{  ` `                ``a[i] = r;  ` `            ``}  ` `            ``else` `if` `(temp + rem < r) ` `            ``{  ` `                ``System.out.print(``"-1"``);  ` `                ``return``;  ` `            ``}  ` `            ``Sk = Sk - a[i];  ` `            ``k = k - ``1``;  ` `        ``}  ` `     `  `        ``// If there is some remaining  ` `        ``// sum to distribute  ` `        ``if` `(Sk > ``0``) ` `        ``{  ` `            ``System.out.print(``"-1"``);  ` `            ``return``;  ` `        ``}  ` `     `  `        ``// If there are elements remaining  ` `        ``// to distribute i.e. (n - k)  ` `        ``if` `(len != ``0``) ` `        ``{  ` `            ``k = n - len;  ` `            ``for` `(``int` `i = len; i < n; i++)  ` `            ``{ ` `                 `  `                ``// Divide the remaining sum into  ` `                ``// n-k elements  ` `                ``temp = diff / k;  ` `                ``rem = diff % k;  ` `                ``if` `(temp + rem >= l && temp + rem <= r) ` `                ``{  ` `                    ``a[i] = temp;  ` `                ``}  ` `                ``else` `if` `(temp + rem > r) ` `                ``{  ` `                    ``a[i] = r;  ` `                ``}  ` `                ``else` `if` `(temp + rem < r)  ` `                ``{  ` `                    ``System.out.print(``"-1"``);  ` `                    ``return``;  ` `                ``}  ` `                ``diff = diff - a[i];  ` `                ``k = k - ``1``;  ` `            ``}  ` `            ``if` `(diff != ``0``)  ` `            ``{  ` `                ``System.out.print(``"-1"``);  ` `                ``return``;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Print the distribution  ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{  ` `            ``System.out.print(a[i] + ``" "``);  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `n = ``5``, k = ``3``, l = ``1``,  ` `            ``r = ``5``, S = ``13``, Sk = ``9``;  ` `     `  `        ``distribution(n, k, l, r, S, Sk);  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python implementation of the approach  ` ` `  `# Function for the distribution of the number ` `def` `distribution(n, k, l, r, S, Sk): ` `    ``a ``=` `[``0``] ``*` `n; ` `    ``len` `=` `k; ` `    ``temp, rem, s ``=` `0``, ``0``, ``0``; ` `    ``diff ``=` `S ``-` `Sk; ` ` `  `    ``for` `i ``in` `range``(``len``): ` ` `  `        ``# Distribute the number ` `        ``# among k elements ` `        ``temp ``=` `Sk ``/` `k; ` `        ``rem ``=` `Sk ``%` `k; ` `        ``if` `(temp ``+` `rem >``=` `l ``and` `temp ``+` `rem <``=` `r): ` `            ``a[i] ``=` `temp; ` `        ``elif``(temp ``+` `rem > r): ` `            ``a[i] ``=` `r; ` `        ``elif``(temp ``+` `rem < r): ` `            ``print``(``"-1"``); ` `            ``return``; ` `         `  `        ``Sk ``=` `Sk ``-` `a[i]; ` `        ``k ``=` `k ``-` `1``; ` `     `  `    ``# If there is some remaining ` `    ``# sum to distribute ` `    ``if` `(Sk > ``0``): ` `        ``print``(``"-1"``); ` `        ``return``; ` `     `  `    ``# If there are elements remaining ` `    ``# to distribute i.e. (n - k) ` `    ``if` `(``len` `!``=` `0``): ` `        ``k ``=` `n ``-` `len``; ` `        ``for` `i ``in` `range``(``len``, n): ` ` `  `            ``# Divide the remaining sum into ` `            ``# n-k elements ` `            ``temp ``=` `diff ``/` `k; ` `            ``rem ``=` `diff ``%` `k; ` `            ``if` `(temp ``+` `rem >``=` `l ``and` `temp ``+` `rem <``=` `r): ` `                ``a[i] ``=` `temp; ` `            ``elif``(temp ``+` `rem > r): ` `                ``a[i] ``=` `r; ` `            ``elif``(temp ``+` `rem < r): ` `                ``print``(``"-1"``); ` `                ``return``; ` `             `  `            ``diff ``=` `diff ``-` `a[i]; ` `            ``k ``=` `k ``-` `1``; ` `         `  `        ``if` `(diff !``=` `0``): ` `            ``print``(``"-1"``); ` `            ``return``; ` `         `  `    ``# Prthe distribution ` `    ``for` `i ``in` `range``(n): ` `        ``print``(``int``(a[i]), end``=``" "``); ` `     `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n, k, l, r, S, Sk ``=` `5``, ``3``, ``1``, ``5``, ``13``, ``9``; ` ` `  `    ``distribution(n, k, l, r, S, Sk); ` `     `  `# This code is contributed by PrinciRaj1992 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function for the distribution of the number  ` `    ``static` `void` `distribution(``int` `n, ``int` `k, ``int` `l,  ` `                            ``int` `r, ``int` `S, ``int` `Sk)  ` `    ``{  ` `        ``int` `[]a = ``new` `int``[n];  ` `        ``int` `len = k, temp, rem;  ` `        ``int` `diff = S - Sk;  ` `         `  `        ``for` `(``int` `i = 0; i < len; i++) ` `        ``{ ` `             `  `            ``// Distribute the number  ` `            ``// among k elements  ` `            ``temp = Sk / k;  ` `            ``rem = Sk % k;  ` `            ``if` `(temp + rem >= l && temp + rem <= r) ` `            ``{  ` `                ``a[i] = temp;  ` `            ``}  ` `            ``else` `if` `(temp + rem > r) ` `            ``{  ` `                ``a[i] = r;  ` `            ``}  ` `            ``else` `if` `(temp + rem < r) ` `            ``{  ` `                ``Console.Write(``"-1"``);  ` `                ``return``;  ` `            ``}  ` `            ``Sk = Sk - a[i];  ` `            ``k = k - 1;  ` `        ``}  ` `     `  `        ``// If there is some remaining  ` `        ``// sum to distribute  ` `        ``if` `(Sk > 0) ` `        ``{  ` `            ``Console.Write(``"-1"``);  ` `            ``return``;  ` `        ``}  ` `     `  `        ``// If there are elements remaining  ` `        ``// to distribute i.e. (n - k)  ` `        ``if` `(len != 0) ` `        ``{  ` `            ``k = n - len;  ` `            ``for` `(``int` `i = len; i < n; i++)  ` `            ``{ ` `                 `  `                ``// Divide the remaining sum into  ` `                ``// n-k elements  ` `                ``temp = diff / k;  ` `                ``rem = diff % k;  ` `                ``if` `(temp + rem >= l && temp + rem <= r) ` `                ``{  ` `                    ``a[i] = temp;  ` `                ``}  ` `                ``else` `if` `(temp + rem > r) ` `                ``{  ` `                    ``a[i] = r;  ` `                ``}  ` `                ``else` `if` `(temp + rem < r)  ` `                ``{  ` `                    ``Console.Write(``"-1"``);  ` `                    ``return``;  ` `                ``}  ` `                ``diff = diff - a[i];  ` `                ``k = k - 1;  ` `            ``}  ` `            ``if` `(diff != 0)  ` `            ``{  ` `                ``Console.Write(``"-1"``);  ` `                ``return``;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Print the distribution  ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{  ` `            ``Console.Write(a[i] + ``" "``);  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{  ` `        ``int` `n = 5, k = 3, l = 1,  ` `            ``r = 5, S = 13, Sk = 9;  ` `     `  `        ``distribution(n, k, l, r, S, Sk);  ` `    ``}  ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```3 3 3 2 2
```

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