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

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.

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

## C++

`// C++ program to find the smallest number ` `// that never becomes positive when processed ` `// with given array elements. ` `#include <bits/stdc++.h> ` `#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; ` `} ` |

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## 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 ` |

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## Python

`# 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] ` ` ` `elif` `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 ` |

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Output:

4

Time Complexity: O (n^2)

Auxiliary Space: O (1)

**Efficient Approach:** Efficient approach to solve 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 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.

## C++

`// Efficient C++ program to find the smallest ` `// number that never becomes positive when ` `// processed with given array elements. ` `#include <bits/stdc++.h> ` `#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; ` `} ` |

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## 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 ` |

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## Python

`# 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 ` |

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## C#

// Efficient Java 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

?>

**Output:**

4

**Time Complexity:** O (n)

**Auxiliary Space:** O (1)

This article is contributed by **Aditya Gupta**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Reference: https://www.hackerrank.com/challenges/chief-hopper

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