Question

You are given two 32-bit numbers, N andM, and two bit positions, i and j. Write a method to insert M into Nsuch that M starts at bit j and ends at bit i. You can assume that the bits j through i have enough space to fit all ofM. That is, ifM= 10011, you can assume that there are at least 5 bits between j and i. You would not, for example, have j-3 and i=2, because M could not fully fit between bit 3 and bit 2.

EXAMPLE:

Input: N = 16000000000, M = 10011, i = 2, j = 6

Output: N = 10001001100

Solution

This is a basic bit manipulation question. The key is to use binary maks. Two things to note:

1. The ‘~’ means negate. So (~0) is a sequence of 1 (the value equals to -1).

2. When shifting bits, DO NOT USE ‘>>’ because it’s signed shift. Instead, use ‘>>>’ (unsigned right shift operator).

Code

written by me

public static int myAnswer(int n, int m, int i, int j) {
    int rr = (~0 >>> (31 - j));
    int ll = (~0 << i);
    // printBinary(rr);
    // printBinary(ll);

    int middleMask = ll & rr;
    int twoEndMask = ll ^ rr;

    n = n & twoEndMask;
    m = (m << i) & middleMask;

    return n | m;
}