# Smallest number that never becomes negative when processed against array elements

Last Updated : 19 Sep, 2023

Given an array of size n your goal is to find a number such that when the number is processed against each array element starting from the 0th index till the (n-1)-th index under the conditions given below, it never becomes negative.

1. If the number is greater than an array element, then it is increased by the difference of the number and the array element.
2. If the number is smaller than an array element, then it is decreased by the difference of the number and the array element.

Examples:

```Input : arr[] = {3 4 3 2 4}
Output : 4
Explanation :
If we process 4 from left to right
in given array, we get following :
When processed with 3, it becomes 5.
When processed with 5, it becomes 6
When processed with 3, it becomes 9
When processed with 2, it becomes 16
When processed with 4, it becomes 28
We always get a positive number. For
all values lower than 4, it would
become negative for some value of the
array.

Input: arr[] = {4 4}
Output : 3
Explanation :
When processed with 4, it becomes 2
When processed with next 4, it becomes 1```

Simple Approach: A simple approach is to find the maximum element in the array and test against each number starting from 1 till the maximum element, that it crosses the whole array with 0 value or not.

Implementation:

## C++

 `// C++ program to find the smallest number ` `// that never becomes positive when processed ` `// with given array elements. ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `ll suitable_num(ll a[], ``int` `n) ` `{ ` `    ``// Finding max element in the array ` `    ``ll max = *max_element(a, a + n);  ` ` `  `    ``for` `(``int` `x = 1; x < max; x++) { ` ` `  `        ``// Creating copy of i since it's   ` `        ``// getting modified at later steps. ` `        ``int` `num = x;  ` ` `  `        ``// Checking that num doesn't becomes ` `        ``// negative. ` `        ``int` `j; ` `        ``for` `(j = 0; j < n; j++)  ` `        ``{   ` `            ``if` `(num > a[j]) ` `                ``num += (num - a[j]); ` `            ``else` `if` `(a[j] > num) ` `                ``num -= (a[j] - num); ` `            ``if` `(num < 0) ` `                ``break``; ` `        ``} ` ` `  `        ``if` `(j == n)  ` `            ``return` `x;         ` `    ``} ` ` `  `    ``return` `max; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``ll a[] = { 3, 4, 3, 2, 4 }; ` `    ``int` `n = ``sizeof``(a)/(``sizeof``(a[0]));  ` `    ``cout << suitable_num(a, n); ` `    ``return` `0; ` `} `

## Java

 `// A Java program to find the smallest number ` `// that never becomes positive when processed ` `// with given array elements. ` `import` `java.util.Arrays; ` `public` `class` `Largest_Number_NotNegate  ` `{ ` `    ``static` `long` `suitable_num(``long` `a[], ``int` `n) ` `    ``{ ` `        ``// Finding max element in the array ` `        ``long` `max = Arrays.stream(a).max().getAsLong(); ` ` `  `        ``for` `(``int` `x = ``1``; x < max; x++) { ` ` `  `            ``// Creating copy of i since it's ` `            ``// getting modified at later steps. ` `            ``int` `num = x; ` ` `  `            ``// Checking that num doesn't becomes ` `            ``// negative. ` `            ``int` `j; ` `            ``for` `(j = ``0``; j < n; j++) { ` `                ``if` `(num > a[j]) ` `                    ``num += (num - a[j]); ` `                ``else` `if` `(a[j] > num) ` `                    ``num -= (a[j] - num); ` `                ``if` `(num < ``0``) ` `                    ``break``; ` `            ``} ` ` `  `            ``if` `(j == n) ` `                ``return` `x; ` `        ``} ` ` `  `        ``return` `max; ` `    ``} ` `     `  `    ``// Driver program to test above method ` `    ``public` `static` `void` `main(String[] args) { ` ` `  `        ``long` `a[] = { ``3``, ``4``, ``3``, ``2``, ``4` `}; ` `        ``int` `n = a.length; ` `        ``System.out.println(suitable_num(a, n)); ` `    ``} ` `} ` `// This code is contributed by Sumit Ghosh `

## Python3

 `# Python program to find the smallest number ` `# that never becomes positive when processed ` `# with given array elements. ` `def` `suitable_num(a): ` `    ``mx ``=` `max``(a) ` `     `  `    ``for` `x ``in` `range``(``1``, mx): ` `         `  `        ``# Creating copy of i since it's   ` `        ``# getting modified at later steps. ` `        ``num ``=` `x ` `         `  `        ``# Checking that num doesn't becomes ` `        ``# negative. ` `        ``j ``=` `0``; ` `        ``while` `j < ``len``(a): ` `            ``if` `num > a[j]: ` `                ``num ``+``=` `num ``-` `a[j] ` `            ``else` `if` `a[j] > num: ` `                ``num ``-``=` `(a[j] ``-` `num) ` `            ``if` `num < ``0``: ` `                ``break` `            ``j ``+``=` `1` `        ``if` `j ``=``=` `len``(a): ` `            ``return` `x ` `    ``return` `mx ` ` `  `# Driver code ` `a ``=``[ ``3``, ``4``, ``3``, ``2``, ``4` `] ` `print``(suitable_num(a)) ` ` `  `# This code is contributed by Sachin Bisht `

## C#

 `// A C# program to find the smallest number ` `// that never becomes positive when processed ` `// with given array elements. ` `using` `System; ` `using` `System.Linq; ` `using` `System.Collections; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` `    ``static` `long` `suitable_num(``long` `[]a, ``int` `n) ` `    ``{ ` `        ``// Finding max element in the array ` `        ``long` `max = a.Max(); ` ` `  `        ``for` `(``int` `x = 1; x < max; x++) ` `        ``{ ` ` `  `            ``// Creating copy of i since it's ` `            ``// getting modified at later steps. ` `            ``long` `num = x; ` ` `  `            ``// Checking that num doesn't becomes ` `            ``// negative. ` `            ``int` `j; ` `            ``for` `(j = 0; j < n; j++) ` `            ``{ ` `                ``if` `(num > a[j]) ` `                    ``num += (num - a[j]); ` `                ``else` `if` `(a[j] > num) ` `                    ``num -= (a[j] - num); ` `                ``if` `(num < 0) ` `                    ``break``; ` `            ``} ` ` `  `            ``if` `(j == n) ` `                ``return` `x; ` `        ``} ` `        ``return` `max; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``long` `[]a = { 3, 4, 3, 2, 4 }; ` `        ``int` `n = a.Length; ` `        ``Console.Write(suitable_num(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Javascript

 ``

Output

`4`

Time Complexity: O (n^2)
Auxiliary Space: O (1)

Efficient Approach: Efficient approach to solving this problem would be to use the fact that when you reach the last array element, the value with which we started can be at least 0, which means suppose the last array element is a[n-1] then the value at a[n-2] must be greater than or equal to a[n-1]/2.

Implementation:

## C++

 `// Efficient C++ program to find the smallest  ` `// number that never becomes positive when  ` `// processed with given array elements. ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `ll suitable_num(ll a[], ``int` `n) ` `{ ` `    ``ll num = 0; ` ` `  `    ``// Calculating the suitable number at each step. ` `    ``for` `(``int` `i = n - 1; i >= 0; i--)  ` `        ``num = round((a[i] + num) / 2.0); ` ` `  `    ``return` `num; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``ll a[] = { 3, 4, 3, 2, 4 }; ` `    ``int` `n = ``sizeof``(a)/(``sizeof``(a[0]));  ` `    ``cout << suitable_num(a, n); ` `    ``return` `0; ` `} `

## Java

 `// Efficient Java program to find the smallest  ` `// number that never becomes positive when  ` `// processed with given array elements. ` `public` `class` `Largest_Number_NotNegate { ` `    ``static` `long` `suitable_num(``long` `a[], ``int` `n) { ` `        ``long` `num = ``0``; ` ` `  `        ``// Calculating the suitable number at each step. ` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) ` `            ``num = Math.round((a[i] + num) / ``2.0``); ` ` `  `        ``return` `num; ` `    ``} ` ` `  `    ``// Driver Program to test above function ` `    ``public` `static` `void` `main(String[] args) { ` ` `  `        ``long` `a[] = { ``3``, ``4``, ``3``, ``2``, ``4` `}; ` `        ``int` `n = a.length; ` `        ``System.out.println(suitable_num(a, n)); ` ` `  `    ``} ` `} ` `// This code is contributed by Sumit Ghosh `

## Python3

 `# Efficient Python program to find the smallest  ` `# number that never becomes positive when  ` `# processed with given array elements. ` `def` `suitable_num(a): ` `    ``num ``=` `0` `     `  `    ``# Calculating the suitable number at each step. ` `    ``i ``=` `len``(a) ``-` `1` `    ``while` `i >``=` `0``: ` `        ``num ``=` `round``((a[i] ``+` `num) ``/` `2.0``) ` `     `  `        ``i ``-``=` `1` `    ``return` `int``(num) ` ` `  `# Driver code ` `a ``=` `[ ``3``, ``4``, ``3``, ``2``, ``4` `] ` `print` `(suitable_num(a)) ` ` `  `# This code is contributed by Sachin Bisht `

## C#

 `// Efficient C# program to find the smallest  ` `// number that never becomes positive when  ` `// processed with given array elements. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `static` `long` `suitable_num(``long``[] a, ``int` `n)  ` `{ ` `    ``long` `num = 0; ` ` `  `    ``// Calculating the suitable number  ` `    ``// at each step. ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) ` `        ``num = (``long``)Math.Round((a[i] + num) / 2.0,  ` `                                ``MidpointRounding.AwayFromZero); ` ` `  `    ``return` `num; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{ ` `    ``long``[] a = { 3, 4, 3, 2, 4 }; ` `    ``int` `n = a.Length; ` `    ``Console.Write(suitable_num(a, n)); ` `} ` `} ` ` `  `// This code is contributed by ita_c `

## PHP

 `= 0; ``\$i``--)  ` `        ``\$num` `= ``round``((``\$a``[``\$i``] + ``\$num``) / 2.0); ` ` `  `    ``return` `\$num``; ` `} ` ` `  `// Driver code ` `\$a` `= ``array``( 3, 4, 3, 2, 4 ); ` `\$n` `= sizeof(``\$a``);  ` `echo` `suitable_num(``\$a``, ``\$n``); ` ` `  `// This code is contributed by ita_c ` `?> `

## Javascript

 ``

Output

`4`

Time Complexity: O(n)
Auxiliary Space: O(1)