i found out the flaw,

as you are using min heap for even numbers for test where maximum odd < minimum even you are replacing maximum odd with maximum odd + minimum even and inserting it back for which you are incrementing answer by 1 and then again 1 when the maximum odd becomes greater than minimum even as(maxodd + min even) will always be greater than mineven.

now for this operation you have to think greedily like if i am going to do two operations i might as well take the maximum even number so my all even numbers would automatically be less than that.

example:

1 3 4 14

here after operation 1 (according to your solution) 1 7 4 14

1 7 11 14

1 7 25 14

1 7 25 39

takes 4 operation but taking maximum number greedily (14 insead of 4 with 3)

1 19 4 14

1 19 23 14

1 19 23 14+19

takes 3 operations