O(n) solution:
odd_index=-1;
even_index=-1;
//now find first odd no.'s index
for(i=0;i<n;i++)
if(a[i]%2!=0)
{
odd_index=i;
break;
}
//if odd_index==-1 return; (only even no.s are there)
//select any even no.'s index after odd_index and swap
for(i=odd_index+1;i<n;i++)
{
if(a[i]%2 == 0)
{
even_index = i;
swap(a[odd_index],a[even_index]);
odd_index++; //no. next to a[odd_index] will always be odd after
swapping
}
}
complexity:O(n)+O(n)=O(n);
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