House Robber Problem C easy to understand Sol

My first basic sol where no confusion needed and just easy to understand :)
condition is: adjacent houses cannot be robbed
Solution: So s0 can be first house, s1 can be second house,

• s2 to (first+3rd) houses
• In a loop starting from 4th house, (where this 4th house can be added to s0 and s1 as not adjacent)
• Keep adding the current element to s0 and current element to s1
  compare the sums and assign to s,
• Keep assigning the max value from s
• Keep moving s0 to s1, s1 to s2, s2 to s..

/* adjacent houses cannot be robbed 
   so s0 can be first house, s1 can be second house,
   s2 to (first+3rd) houses
   
   in a loop starting from 4th house,
   keep adding the current element to s0 and current element to s1
   compare the sums and assign to s,
      Keep assigning the max value from s
      Keep moving s0 to s1, s1 to s2, s2 to s..
*/
int rob(int* nums, int ns){
    int ii, mx=0, s0, s1, s2,s;
    
    if (!ns) return 0;
    if (ns == 1) return nums[0];
    s0 = nums[0];
    s1 = nums[1];
    if (ns == 2) return (s0 > s1 ? s0 : s1);
    s2 = nums[0] + nums[2];
    mx = (s2 > s1) ? s2 : s1;
    for (ii=3; ii<ns; ii++){
        if ((s1 +nums[ii]) > (s0 + nums[ii]))
            s = s1 + nums[ii];
        else 
            s = s0 + nums[ii];
        if (mx < s) mx = s;
        s0=s1; s1=s2; s2 =s;
    }
    return mx;
}

Next optimized solution, where tried to some temp variables, additions.. :

/* this solution is to reduce the temp. variables
   , one extra addition and one extra assignment
   s2 can be reduced,
   s1 + nums[ii] and s0 + nums[ii] this comparison will take place 
   when s0 is moved to s1 in next step.
   So we can directly initialize s0 to 0, s1 to first house
   in loop compare s0+current house with s1 and keep assigning max to s
      Keep updating max with max value of s
      Keep moving s1 to s0, s to s1
 */
int rob(int* nums, int ns){
    int ii, mx=0, s0, s1, s;
    
    if (!ns) return 0;
    if (ns == 1) return nums[0];
    s0 = 0;
    s1 = nums[0];
    for (ii=1; ii<ns; ii++){
        if (s1 > (s0 + nums[ii]))
            s = s1;
        else 
            s = s0 + nums[ii];
        if (mx < s) mx = s;
        s0=s1; s1=s;
    }
    return mx;
}