# Generate an array B[] from the given array A[] which satisfies the given conditions

Given an array A[] of N integers such that A + A + A + … A[N – 1] = 0. The task is to generate an array B[] such that B[i] is either ⌊A[i] / 2⌋ or ⌈A[i] / 2⌉ for all valid i and B + B + B + … + B[N – 1] = 0.

Examples:

Input: A[] = {1, 2, -5, 3, -1}
Output: 0 1 -2 1 0

Input: A[] = {3, -5, -7, 9, 2, -2}
Output: 1 -2 -4 5 1 -1

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

Approach: For even integers, it is safe to assume that B[i] will be A[i] / 2 but for odd integers, to maintain the sum equal to zero, take the ceil of exactly half of odd integers and floor of exactly other half odd integers. Since Odd – Odd = Even and Even – Even = Even and 0 is also Even, it can be said that A[] will always contain even number of odd integers so that the sum can be 0. So for a valid input, there will always be an answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to print ` `// the array elements ` `void` `printArr(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Function to generate and print ` `// the required array ` `void` `generateArr(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To switch the ceil and floor ` `    ``// function alternatively ` `    ``bool` `flip = ``true``; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If the number is odd then print the ceil ` `        ``// or floor value after division by 2 ` `        ``if` `(arr[i] & 1) { ` ` `  `            ``// Use the ceil and floor alternatively ` `            ``if` `(flip ^= ``true``) ` `                ``cout << ``ceil``((``float``)(arr[i]) / 2.0) << ``" "``; ` `            ``else` `                ``cout << ``floor``((``float``)(arr[i]) / 2.0) << ``" "``; ` `        ``} ` ` `  `        ``// If arr[i] is even then it will ` `        ``// be completely divisible by 2 ` `        ``else` `{ ` `            ``cout << arr[i] / 2 << ``" "``; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, -5, -7, 9, 2, -2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``generateArr(arr, n); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the approach ` `from` `math ``import` `ceil, floor ` ` `  `# Utility function to print ` `# the array elements ` `def` `printArr(arr, n): ` `    ``for` `i ``in` `range``(n): ` `        ``print``(arr[i], end ``=` `" "``) ` ` `  `# Function to generate and print ` `# the required array ` `def` `generateArr(arr, n): ` ` `  `    ``# To switch the ceil and floor ` `    ``# function alternatively ` `    ``flip ``=` `True` ` `  `    ``# For every element of the array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# If the number is odd then prthe ceil ` `        ``# or floor value after division by 2 ` `        ``if` `(arr[i] & ``1``): ` ` `  `            ``# Use the ceil and floor alternatively ` `            ``flip ^``=` `True` `            ``if` `(flip): ` `                ``print``(``int``(ceil((arr[i]) ``/` `2``)),  ` `                                   ``end ``=` `" "``) ` `            ``else``: ` `                ``print``(``int``(floor((arr[i]) ``/` `2``)),  ` `                                    ``end ``=` `" "``) ` ` `  `        ``# If arr[i] is even then it will ` `        ``# be completely divisible by 2 ` `        ``else``: ` `            ``print``(``int``(arr[i] ``/` `2``), end ``=` `" "``) ` ` `  `# Driver code ` `arr ``=` `[``3``, ``-``5``, ``-``7``, ``9``, ``2``, ``-``2``] ` `n ``=` `len``(arr) ` ` `  `generateArr(arr, n) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```1 -2 -4 5 1 -1
```

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Improved By : mohit kumar 29