3. (30 points) Translate the following c code to MIPS assembly code. Try to use a minimum number of instructions. Register allocations i Ss0 j Ss1 base of A[] Ss2 base of B[] Ss3 3.1. (10 points) if (i != j) i = i + 3 else j i + 1 j = i + j; 3.2. (10 points) A [j+1] = B[i+2] + B [5]; 3.3. (10 points) i = 0; while (A[i] != B[i]) { i++; }