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05.01. Insert Into Bits

Description

You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to insert M into N such 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 of M. That is, if M = 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.

Example1:


 Input: N = 10000000000, M = 10011, i = 2, j = 6

 Output: N = 10001001100

Example2:


 Input: N = 0, M = 11111, i = 0, j = 4

 Output: N = 11111

Solutions

Solution 1: Bit Manipulation

First, we clear the bits from the $i$-th to the $j$-th in $N$, then we left shift $M$ by $i$ bits, and finally perform a bitwise OR operation on $M$ and $N$.

The time complexity is $O(\log n)$, where $n$ is the size of $N$. The space complexity is $O(1)$.

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class Solution:
    def insertBits(self, N: int, M: int, i: int, j: int) -> int:
        for k in range(i, j + 1):
            N &= ~(1 << k)
        return N | M << i

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class Solution {
    public int insertBits(int N, int M, int i, int j) {
        for (int k = i; k <= j; ++k) {
            N &= ~(1 << k);
        }
        return N | M << i;
    }
}

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class Solution {
public:
    int insertBits(int N, int M, int i, int j) {
        for (int k = i; k <= j; ++k) {
            N &= ~(1 << k);
        }
        return N | M << i;
    }
};

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func insertBits(N int, M int, i int, j int) int {
    for k := i; k <= j; k++ {
        N &= ^(1 << k)
    }
    return N | M<<i
}

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function insertBits(N: number, M: number, i: number, j: number): number {
    for (let k = i; k <= j; ++k) {
        N &= ~(1 << k);
    }
    return N | (M << i);
}

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class Solution {
    func insertBits(_ N: Int, _ M: Int, _ i: Int, _ j: Int) -> Int {
        var result = N

        for k in i...j {
            result &= ~(1 << k)
        }

        return result | (M << i)
    }
}

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