C++ solution:

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

Golang

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

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

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)

In Python 3:

def flippingBits(n):
    x = int('0b11111111111111111111111111111111', base=0)
    return x^n