1523. Count Odd Numbers in an Interval Range

/* JAVA BEST SOLUTION UNDERSTNADEABLE O(1) */
class Solution {
  public int countOdds(int low, int high) {
      int range = high-low; //the no of odd numbers between range will be range/2
      if(low%2!=0 & high%2!=0) return range/2 + 1;  //low and high are odd
      if(low%2==0 & high%2==0) return range/2;      //low and high are even
      else return range/2 + 1;                      //one is odd and other even
  }
}

LOGIC - USE THIS, turn it itno cases and find out the answer



/* JAVA ACCEPTED SOLUTION 0(1) */
class Solution {
  public int countOdds(int low, int high) {
    // If low is even, make it odd.
    if ((low & 1) == 0) {
      low++;
    }

    // low could become greater than high due to incrementation
    // if it is, the answer is 0; otherwise, use the formula.
    return low > high ? 0 : (high - low) / 2 + 1;
  }
}
LOGIC: Let's start with an odd integer, say x; what would be the next odd integer? It would be x + 2; as explained above, there is exactly one even integer between 
every two odd integers, and hence all odd integers are equally spaced with a gap of one integer. 
By trying out some examples, we can deduce that the count of odd integers between odd x and a greater integer y where x is odd would be (y−x)/2 + 1.




/* JAVA ONE LINER SOLUTION */
class Solution {
    public int countOdds(int low, int high) {
        return ((high + 1) / 2) - (low / 2);  
    }
}



/* JAVA O(n) SOLUTION -- TLE */
class Solution {
    public int countOdds(int low, int high) {
        int count =0;
        while(low<=high){
            if(low%2!=0) count++;
            low++;
        }
        return count;     
    }
}