Android Developer Interview Questions

8,366 android developer interview questions shared by candidates

The remote coderpad challenge was to map some json to an object. The json object contained a part of a url path and had children with deeper parts of the path and so on. Had to write an algorithm to traverse the object's tree and print all full paths.
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Android Developer

Interviewed at Groupon

2.9
Oct 4, 2019

The remote coderpad challenge was to map some json to an object. The json object contained a part of a url path and had children with deeper parts of the path and so on. Had to write an algorithm to traverse the object's tree and print all full paths.

Fibonacci Series: The Fibonacci Sequence is the series of numbers where the next number is found by adding up the two numbers before it. F(0) = 0 F(1) = 1 F(n) = F(n-1) + F(n-2) Which of the following is the correct and the most efficient implementation for fibonacci numbers. int fib(int n) { int f[n+1]; int i; f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { f[i] = f[i-1] + f[i-2]; } return f[n]; } int fib(int n) { int f[n+1]; int i; if( n <= 1) return n; f[0] = 0; f[1] = 1; for (i = 2; i < n; i++) { f[i] = f[i-1] + f[i-2]; } return f[n]; } int fib(int n) { if (n <= 1) return n; return fib(n-1) + fib(n-2); } int fib(int n) { int a = 0, b = 1, c, i; if( n == 0) return a; for (i = 2; i <= n; i++) { c = a + b; a = b; b = c; } return b; }
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Android Developer

Interviewed at Smartprix

3.4
Aug 3, 2016

Fibonacci Series: The Fibonacci Sequence is the series of numbers where the next number is found by adding up the two numbers before it. F(0) = 0 F(1) = 1 F(n) = F(n-1) + F(n-2) Which of the following is the correct and the most efficient implementation for fibonacci numbers. int fib(int n) { int f[n+1]; int i; f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { f[i] = f[i-1] + f[i-2]; } return f[n]; } int fib(int n) { int f[n+1]; int i; if( n <= 1) return n; f[0] = 0; f[1] = 1; for (i = 2; i < n; i++) { f[i] = f[i-1] + f[i-2]; } return f[n]; } int fib(int n) { if (n <= 1) return n; return fib(n-1) + fib(n-2); } int fib(int n) { int a = 0, b = 1, c, i; if( n == 0) return a; for (i = 2; i <= n; i++) { c = a + b; a = b; b = c; } return b; }

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