Solving codingBat Post4 with one loop in Java -

the question solving this problem codingbat in java.

problem statement:

given non-empty array of ints, return new array containing elements original array come after last 4 in original array. original array contain @ least 1 4. note valid in java create array of length 0.

post4({2, 4, 1, 2}) → {1, 2}

post4({4, 1, 4, 2}) → {2}

post4({4, 4, 1, 2, 3}) → {1, 2, 3}

here solution:

public int[] post4(int[] nums) {  int lastfour=-1; int[] post4={};    for(int i=nums.length-1;i>=0;i--)   {      if((nums[i]==4))      {        lastfour=i;    //find index of last 4 in array        break;      }      }     int newlen=(nums.length-lastfour)-1;    post4=new int[newlen];   //reassign post4 array required length    for(int j=0;j<newlen;j++)   {      post4[j]=nums[lastfour+1];   //assign values orig. array after last 4      lastfour++;   }    return post4;    } 

but have used 2 loops. should solved using @ max 1 loop. not use collections or wrappers classes.

• before iteration create result array, lets length 0.
• each time find 4 create new result array size based on index of 4 , length of nums store rest of elements.
• if number not 4 place in result array (don't place if result arrays length 0 because means didn't find 4 yet, or last element of nums array).

here example solution

public int[] post4(int[] nums) {      int[] result = new int[0];      int j = 0;     (int = 0; i<nums.length; i++){        if (nums[i] == 4) {            result = new int[nums.length - i-1];            j=0;        }        else             if (result.length>0) result[j++] = nums[i];      }      return result; }