Write a function that takes in an array and repeats an integer that appears the most.

Question

Jayasagar · Accepted Answer

Using map , i think one loop is sufficient.

private static int mostReapeatingNumber(int[] is) {

		HashMap map = new HashMap();
		int tempHighestCount = 0;
		int keyHighest = 0;		
		
		for (int index=0; index  tempHighestCount) {
					tempHighestCount = numCount;
					keyHighest = number;	
				}
			}
		}
		
                return keyHighest;
	}

Zhable Baba · Answer

I think there's no need to have a map. Just maintain variables prev_max_run, prev_max_num, prev_num, curr_num and curr_run. In the loop if the prev_num was equal to curr_num increment curr_run. When you find the num is different check curr_run with prev_run. If curr_run > prev_run, prev_max_num = curr_num.

Gofishus · Answer

Confusing. In your example, 2 appears the most. Do you mean the integer that repeats the most consecutively? Cause that would be 3. 
Anyways, in either case, you can go through the array adding all the key-value pairs (number and times) to a hashmap and then access the hashmap in constant time. O(n).

Praveen Chettypally · Answer

class FindMostOccurences
    {
        public static DictionaryEntry MostOccurences(int[] Array)
        {
            Hashtable ht = new Hashtable();
            for (int i = 0; i  Int32.Parse(de.Value.ToString()))
                {
                    {
                        de.Key = item.Key;
                        de.Value = item.Value;
                    }
                }
            }
            return de;
        }
    }

Anonymous · Answer

if:  Array [2][2][3][3][3][2][1][2][1] it should print [3]

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