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. 

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 <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;
}

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

<script>
// A Javascript program to find the smallest number
// that never becomes positive when processed
// with given array elements.

    function suitable_num(a,n)
    {
        // Finding max element in the array
        let max = Math.max(...a)

        for (let x = 1; x < max; x++) {

            // Creating copy of i since it's
            // getting modified at later steps.
            let num = x;

            // Checking that num doesn't becomes
            // negative.
            let 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
    let a=[3, 4, 3, 2, 4];
    let n = a.length;
    document.write(suitable_num(a, n));

    //This code is contributed by avanitrachhadiya2155

</script>

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 <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;
}

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

<?php
// Efficient PHP program to find the smallest
// number that never becomes positive when
// processed with given array elements.

function suitable_num(&$a, $n)
{
    $num = 0;

    // Calculating the suitable number
    // at each step.
    for ($i = $n - 1; $i >= 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

<script>

// Efficient Javascript program to find the smallest
// number that never becomes positive when
// processed with given array elements.
function suitable_num(a,n)
{
    let num = 0;

    // Calculating the suitable
    // number at each step.
    for(let i = n - 1; i >= 0; i--)
        num = Math.round((a[i] + num) / 2.0);

    return num;
}

// Driver code
let a = [ 3, 4, 3, 2, 4 ];
let n = a.length;

document.write(suitable_num(a, n));

// This code is contributed by rag2127

</script>

Output

4

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