# Generate array with minimum sum which can be deleted in P steps

• Difficulty Level : Medium
• Last Updated : 25 Aug, 2021

Given two numbers N and P. The task is to generate an array of all positive elements, and in one operation you can choose a minimum number in the array and subtract it from all array elements. If the array element becomes 0 then you will remove it.
You have to print the minimum possible sum of the array and one possible array such that after applying exactly P steps the array will vanish.

Examples:

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Input : N = 4, P = 2
Output :
The Minimum Possible Sum is: 5
The Array Elements are: 1 2 1 1
Explanation:
The array can be [1, 2, 1, 1] after 1st step it becomes [0, 1, 0, 0] and it becomes [1] and after step 2 it will be vanished.Thus the sum is 5 and it is minimum possible value.
Input : N = 3 , P = 1
Output
The Minimum Possible Sum is: 3
The Array Elements are: 1 1 1

Approach: The problem can be solved by following a greedy approach. First, we will place first P natural numbers, and for the rest (N – P) positions we will fill it with 1 because we have to minimize the sum.
So the sum will be P*(P+1)/2 + (N – P).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to find the required array``void` `findArray(``int` `N, ``int` `P)``{``    ``// calculating minimum possible sum``    ``int` `ans = (P * (P + 1)) / 2 + (N - P);` `    ``// Array``    ``int` `arr[N + 1];` `    ``// place first P natural elements``    ``for` `(``int` `i = 1; i <= P; i++)``        ``arr[i] = i;` `    ``// Fill rest of the elements with 1``    ``for` `(``int` `i = P + 1; i <= N; i++)``        ``arr[i] = 1;` `    ``cout << ``"The Minimum Possible Sum is: "` `<< ans << ``"\n"``;``    ``cout << ``"The Array Elements are: \n"``;` `    ``for` `(``int` `i = 1; i <= N; i++)``        ``cout << arr[i] << ``' '``;``}` `// Driver Code``int` `main()``{``    ``int` `N = 5, P = 3;` `    ``findArray(N, P);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function to find the required array``    ``static` `void` `findArray(``int` `N, ``int` `P)``    ``{``        ``// calculating minimum possible sum``        ``int` `ans = (P * (P + ``1``)) / ``2` `+ (N - P);` `        ``// Array``        ``int` `arr[] = ``new` `int``[N + ``1``];` `        ``// place first P natural elements``        ``for` `(``int` `i = ``1``; i <= P; i++)``        ``{``            ``arr[i] = i;``        ``}` `        ``// Fill rest of the elements with 1``        ``for` `(``int` `i = P + ``1``; i <= N; i++)``        ``{``            ``arr[i] = ``1``;``        ``}` `        ``System.out.print(``"The Minimum Possible Sum is: "` `+``                                                ``ans + ``"\n"``);``        ``System.out.print(``"The Array Elements are: \n"``);` `        ``for` `(``int` `i = ``1``; i <= N; i++)``        ``{``            ``System.out.print(arr[i] + ``" "``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `N = ``5``, P = ``3``;` `        ``findArray(N, P);``    ``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of above approach` `# Function to find the required array``def` `findArray(N, P):``    ` `    ``# calculating minimum possible sum``    ``ans ``=` `(P ``*` `(P ``+` `1``)) ``/``/` `2` `+` `(N ``-` `P);` `    ``# Array``    ``arr ``=` `[``0``] ``*` `(N ``+` `1``);` `    ``# place first P natural elements``    ``for` `i ``in` `range``(``1``, P ``+` `1``):``        ``arr[i] ``=` `i;` `    ``# Fill rest of the elements with 1``    ``for` `i ``in` `range``(P ``+` `1``, N ``+` `1``):``        ``arr[i] ``=` `1``;` `    ``print``(``"The Minimum Possible Sum is: "``, ans);``    ``print``(``"The Array Elements are: "``);` `    ``for` `i ``in` `range``(``1``, N ``+` `1``):``        ``print``(arr[i], end ``=` `" "``);` `# Driver Code``N ``=` `5``;``P ``=` `3``;``findArray(N, P);` `# This code is contributed by mits`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to find the required array``    ``static` `void` `findArray(``int` `N, ``int` `P)``    ``{``        ``// calculating minimum possible sum``        ``int` `ans = (P * (P + 1)) / 2 + (N - P);` `        ``// Array``        ``int` `[]arr = ``new` `int``[N + 1];` `        ``// place first P natural elements``        ``for` `(``int` `i = 1; i <= P; i++)``        ``{``            ``arr[i] = i;``        ``}` `        ``// Fill rest of the elements with 1``        ``for` `(``int` `i = P + 1; i <= N; i++)``        ``{``            ``arr[i] = 1;``        ``}` `        ``Console.Write(``"The Minimum Possible Sum is: "` `+``                                                ``ans + ``"\n"``);``        ``Console.Write(``"The Array Elements are: \n"``);` `        ``for` `(``int` `i = 1; i <= N; i++)``        ``{``            ``Console.Write(arr[i] + ``" "``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `N = 5, P = 3;` `        ``findArray(N, P);``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## PHP

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## Javascript

 ``
Output:
```The Minimum Possible Sum is: 8
The Array Elements are:
1 2 3 1 1```

Time Complexity: O(N)

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