time complexitySat, 11 Feb 2023
def bar(val, r): count = 0 ncount = 0 v = val for j in range(2 * r): if v % 2 == 1: count = count + 1 else: ncount = ncount + 1 v = v // 2 if count == r: return 1 else: return 0
O(n)
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