This Question falls under harder array questions, here it goes

Return an array that contains exactly the same numbers as the given array, but rearranged so that every 4 is immediately followed by a 5. Do not move the 4's, but every other number may move. The array contains the same number of 4's and 5's, and every 4 has a number after it that is not a 4. In this version, 5's may appear anywhere in the original array.

fix45({5, 4, 9, 4, 9, 5}) → {9, 4, 5, 4, 5, 9}

fix45({1, 4, 1, 5}) → {1, 4, 5, 1}

fix45({1, 4, 1, 5, 5, 4, 1}) → {1, 4, 5, 1, 1, 4, 5}

Here is the to-do list or rather algorithm of the solution i came up with.

- Loop through every element of the array if it is a 4 get the index of it.
- Check whether it has a 5 after it. if it does , we do not need to worry about it.
- Now search for a 5 in the array which is not attached to a 4. because we would not want to alter already fixed 4 s.
- After we found the index of a 5 swap it with the preceding element of 4.
- We have to consider boundaries of loops so that our code won’t throw Array index out of bounds exceptions.

1: public int[] fix45(int[] nums) {2: int len=nums.length;3:4: for(int i=0 , j=0 ; i < len-1 ; i++){5: if(nums[i] == 4){6: if(nums[i+1]==5) continue;7: while(j<len){8: if(j==0 && nums[j]==5 ) break;9: else if(nums[j]==5 && nums[j-1] != 4) break;10: j++;11: }// end while12:13: nums[j]=nums[i+1]; // swap values14: nums[i+1]=5;15: nums[i]=4;16: j=0;17: i++; // skip 518: }// end if19:20: }// end for21:22: return nums;23: }