Make a code have less Complexity

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>


int sum_array(int* array, int cnt)
{
    int res = 0;
    int i;
    for ( i = 0; i < cnt ; ++i)
        res += array[i];
    return res;
}

int main()
{
    FILE* input = fopen("share.in","r");

    int N = 0;
    fscanf(input,"%d",&N);

    int *array = (int*)malloc(N * sizeof(int));

    for (int i = 0; i < N; i++)
        fscanf(input,"%d",&array[i]);

    fclose(input);

    int Min = 0;
    int bestA = 0, bestB = 0, bestMin = INT_MAX;
    int A, B;
    int i;
    for ( A = 0; A < N - 2; ++A)
    {
        for ( B = A + 1; B < N - 1; ++B)
        {
            int ProfitA = sum_array(array, A + 1);
            int ProfitB = sum_array(array + A + 1, B - A );
            int ProfitC = sum_array(array + B + 1, N - 1 - B );

            //here the values are "current" - valid
            Min = (ProfitA > ProfitB) ? ProfitA : ProfitB;
            Min = (ProfitC > Min) ? ProfitC : Min;

            if( Min < bestMin )
                bestA = A, bestB = B, bestMin = Min;

        }
    }
    printf("%d\n", bestMin);
    free(array);
    return 0;
}

int idx0 = 0;
int idx1 = 1;
int idx2 = N-1;

int sum0 = arr[0];
int sum1 = arr[1];
int sum2 = arr[N - 1];

int max = (sum0 > sum1) ? sum0 : sum1;
    max = (sum2 > Max) ? sum2 : Max;

for (i=0; i < N - 1; i++){
    sum += arr[i];
}


while ((idx1 + 1) < idx2) {
    const int left = (sum0 + arr[idx1]);
    const int right = (sum2 + arr[idx2-1]);

    if (left <= right && left <= max) {
        sum0 = left;
        sum1 -= arr[idx1++];
        max = (sum1 > sum2) ? sum1 : sum2; //left cannot be greater
    } else if (right < left && right <= max) {
        sum2 = right;
        sum1 -= arr[--idx2];
        max = (sum1 > sum2) ? sum1 : sum2; //right cannot be greater
    } else break;

    //something like return make_pair(idx[1],idx[2]);
    //and what I return? I return the value I want? The minimun one?

}