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The problems are not hard.

3 Answers

But you may need well preparation.

o(nlogn) solution based on binary search below. An o(n) solution based on hashtable is also possible but consume o(n) memory. public static void main (String[] args) throws java.lang.Exception { Pair p = findPairBySum(15, new int[] {1,2,3,4,5,6,7,8,9,10}); System.out.println(p.x); System.out.println(p.y); } public static Pair findPairBySum(int sum, int[] sortedArr) { for (int i=0; i= 0) return new Pair(first, expected); } return null; } public static int getIndex(int num, int[] sortedArr, int startIndex, int endIndex) { if (startIndex > endIndex) return -1; int pos = startIndex + (endIndex - startIndex) / 2; int valAtMiddle = sortedArr[pos]; if (valAtMiddle == num) return pos; if (num < valAtMiddle) return getIndex(num, sortedArr, startIndex, pos-1); return getIndex(num, sortedArr, pos+1, endIndex); } public static class Pair { public int x; public int y; public Pair(int x, int y) { this.x = x; this.y = y; } }

Note that the array is sorted already. O(n) time with O(1) space is feasible.

What can you bring to the company

1 Answer

Find the top K frequent elements.

1 Answer

u like to work here

1 Answer

Group discussion in Cantonese about how to improve the service quality of MTR

1 Answer

Self introdction in English

1 Answer

Ask about previous job experience

1 Answer

Try to prioritize the schedule of an Engineer

1 Answer

Why do you choose joining this industry not choosing other industry?

1 Answer

The technical test about some algorithm and Java technology.

1 Answer
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