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  1. Flipping bits
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Flipping bits

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  • samuelejoshi
     + 0 comments

    Java - How can I look at a problem and find the most simple solution? Is there a guideline on what to think about before coding. For example, logical operators first?

        public static long flippingBits(long n) {
        // Convert to binary
        long q = n;
        String binary = "";

        while (q > 0) {
            long remainder = q % 2;
            binary = remainder + binary;
            q = q / 2;
        }
        System.out.println("Binary of " + n + " = " + binary);

        // Calculate how many zeros to add at the start to get 32-bit
        int zerosToAdd = 32 - binary.length();
        String paddedBinary = "0".repeat(zerosToAdd) + binary;
        System.out.println("32-bit padded: " + paddedBinary);

        // Flip the bits
        StringBuilder flipped = new StringBuilder();
        for (int i = 0; i < paddedBinary.length(); i++) {
            if (paddedBinary.charAt(i) == '0') {
                flipped.append('1');
            } else {
                flipped.append('0');
            }
        }

        // Convert StringBuilder to String
        String flippedStr = flipped.toString();

        // Parse as base-2
        long decimal = Long.parseLong(flippedStr, 2);


        return decimal;

    }

  • ivanbilych
     + 0 comments

    C++ solution:

    long flippingBits(long n) {
    return ~static_cast<uint32_t>(n);
}

  • exelure
     + 0 comments

    Golang

    func flippingBits(n int64) int64 {
    return n ^ int64(0xFFFFFFFF)
}

  • reyyeung11
     + 0 comments
    def flippingBits(n):
    l = format(n,'032b')
    m = []
    for i in range(len(l)):
        m.append(1-int(l[i]))

    sum = 0
    for i in range(len(m)):
        sum += int(m[i])*(2**(31-i))
    return sum

  • yaya00k8
     + 0 comments

    Definitely less elegant than other solutions, but for my fellow beginners:

    def flippingBits(n):

        # turn 'n' into binary string
        l=format(n, '032b')
        m=[]

        # flip all bits
        for i in range(len(l)):
                m.append(1-int(l[i]))

        num=0
        if m[-1]==1:
                num+=1
        for i in range(len(m)-1):

                # start from the back of the string, skip m [ -1 ]
                curr=int(m[-i-2])
                num+=(2*curr)**(i+1)
        return(num)