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10,446 developer android interview questions shared by candidates

The questions were about android development and what makes a high-quality codebase, and SE design patterns, and some important concepts of android development. Live coding: normal coding questions in leetcode & hackerrank, nothing too difficult
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Android Software Engineer

Interviewed at Delivery Hero

3.4★
Oct 18, 2020

The questions were about android development and what makes a high-quality codebase, and SE design patterns, and some important concepts of android development. Live coding: normal coding questions in leetcode & hackerrank, nothing too difficult

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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