find the time complexity

Question

Consider the following function

int foo(int n) {
    int count1=0, count2=0;
    if (n < 0) n = -n;
    if (n == 0) return 1;
    If (n == 1) return 0;
    while (n) {
        if (n & 1) count1++;
        n = n >> 1;
        if (n & 1) count2++;
        n = n>>1;
    }
    return foo(abs(count1 - count2));
}

What is the time complexity for the above function?

Comments

$\begin{align*} T(n) = T\left ( \log_2 \left ( n^\frac{1}{2} \right ) \right ) + \log_4 n \end{align*}$

I found complex structure. what is the given answer ?

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it is O(log₄n)

actually i tried by putting n= 1024..then on evry while() iteration,it gets reduced by 1/4.

so,answer is O(log ₄ n)

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and forget about recursive call or what ?? :) :) :)

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but yes...recursive call is taking very small value ...

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actually at the end of while loop you will get diff between count 1 and count2 like 1 or 0 ..so it will hardly take one more fucntion call..

am i correct??

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if the number is n = $01010101010101010101$.in binary.etc..something...abs(coont1-count2) $\approx$ $\log_2 \sqrt{n}$..

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even if you take n= 0101010101 suppose,

then after 5 iterations count1 will be 5 but count 2 will be 4 only because in last iteration,n became 0 before 

if (n & 1) count2++;

so count2++ did nt get exceute.

thn it exits while loop.and abs(count1 -count2 ) =1

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why are you incrementing count2...even after taking that sequence of bits ??