Finding the minimum number which when added to running sum of array is 1

Perform the accumulation in reverse, meaning:

Start with 1 and subtract values from the end of the array walking back to the the start of the array. Whenever you get a value that is less than 1, correct it to 1 before continuing the subtractions.

The value you get at completion of this algorithm is the minimum value of x you are looking for.

Implementation in a runnable JavaScript snippet:

function getMinX(arr) {
    let x = 1;
    for (let i = arr.length - 1; i >= 0; i--) {
        x = x - arr[i];
        if (x < 1) x = 1;
    }
    return x;
}

let arr = [-2, 3, 1, -5];
console.log(getMinX(arr)); // 4

Some languages support functional programming style, in which you can use a fold. In JavaScript that would be coded as:

const getMinX = arr => arr.reduceRight((x, val) => Math.max(1, x-val), 1);

console.log(getMinX([-2, 3, 1, -5])); // 4

You should

1. Calculate rolling sums for the input array of integers (time O(n))
   E.g.: arr = [-2, 3, 1, -5] => arr_rolling = [0, -2, 1, 2, -3]
2. Get the minimum element among the sums -- min_sum (time O(n))
   E.g.: arr_rolling = [0, -2, 1, 2, -3] => min_sum = -3
3. X = 1 - min_sum (time O(1))
   E.g.: min_sum = -3 => X = 4

The final time complexity is O(n).

Once you understand the algorithm, you may notice that there's no need to store rolling sums and you can calculate rolling sum minimum "on the fly". Once you notice it, the memory usage becomes O(1).

You can go through the array left-to-right adjusting for x as you go (Python):

In [2]: arr = [-2, 3, 1, -5]

In [3]: accum = x = 0

In [4]: for item in arr:
   ...:     accum += item
   ...:     if accum < 1:
   ...:         x += 1 - accum
   ...:         accum = 1
   ...:         

In [5]: accum
Out[5]: 1

In [6]: x
Out[6]: 4

def minX(arr):
    temp,count1,counter,found= 1,2,len(arr)-1,False
    while  counter !=-1 :
        temp+=(arr[counter] * -1)
        counter -=1
        if temp < 1 :
            temp = 0
            temp += count1
            count1 += 1
            counter = len(arr)-1
    return temp

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#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int minX(vector<int> &arr)
{
    vector<int> temp;
    int sumwX = 1;
    for(auto it = arr.begin(); it != arr.end(); ++it)
    {
        sumwX += *it;
        temp.push_back(sumwX);
    }
    
    return (1 - (*min_element(temp.begin(), temp.end()) - 1 ));
}

int main()
{
    
    vector<int> M = {-5,4,-2,3,29,-1,-6,-1,0,-5};
    vector<int> N = {-2,3,1,-5};
    vector<int> V = {-5,4,-2,3,1,-1,-6,-1,0,-5};
    cout << minX(M) << endl;
    cout << minX(N) << endl;
    cout << minX(V) << endl;
    return 0;
}