Answer: 10:10
We are given a double hashing scheme with a hash table of size m=11 (slots 0 to 10). The hash function for the i-th probe is:
h(k,i)=[h1(k)+i⋅h2(k)]mod11
where h1(k)=kmod11 and h2(k)=1+(kmod7). Let's insert the keys step-by-step:
- Key 63: h1(63)=63mod11=8. Slot 8 is empty. Store 63 at slot 8.
- Key 50: h1(50)=50mod11=6. Slot 6 is empty. Store 50 at slot 6.
- Key 25: h1(25)=25mod11=3. Slot 3 is empty. Store 25 at slot 3.
- Key 79: h1(79)=79mod11=2. Slot 2 is empty. Store 79 at slot 2.
- Key 67: h1(67)=67mod11=1. Slot 1 is empty. Store 67 at slot 1.
- Key 24:
- h1(24)=24mod11=2
- h2(24)=1+(24mod7)=1+3=4
- Probe i=0: h(24,0)=2 (occupied by 79)
- Probe i=1: h(24,1)=(2+1×4)mod11=6 (occupied by 50)
- Probe i=2: h(24,2)=(2+2×4)mod11=10 (empty)
Store 24 at slot 10.
Thus, key 24 gets stored at slot 10.